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# Differential Equation of Higher Order

Differential Equation of Higher Order

Contents

Describe the nth order Linear Differential Equation 1

Find the solution of Higher Order Linear Differential Equation 2

Describe Complementary Functions 3

Find the Complementary Function of Higher-order Differential Equations 5

Describe Particular Integral 6

Find Particular Integral of eax 9

Find Particular Integral of xn 10

Find Particular Integral of sin(ax) 12

Find Particular Integral of cos(ax) 13

Find Particular Integral of eax f(x) 15

Find Particular Integral of sec(ax) 17

Describe Homogeneous Linear Differential Equations 19

Find the Solution of Homogeneous Linear Differential Equations 21

Describe Simultaneous Differential Equations 23

Find the Solution of Simultaneous Differential Equations 25

Describe the Wronskian for Linearly Dependent or Linearly Independent solutions 26

Describe method of Variation of Parameters 26

Describe method of Variation of Parameters 26

Find the Solution of Differential Equation using method of Variation of Parameters 26

Find the Complete Solution of the Differential Equation whose one Solution is known 26

Find the Solution of Differential Equation by using Removal of First Derivative method 26

Find the Solution of a Differential Equation by Changing the Independent Variable 26

Apply Differential Equations in Electrical Circuits 26

Apply Differential Equations in Simple Harmonic Motion 26

Apply Differential Equations Simple Population Model 26

# Describe the nth order Linear Differential Equation

An nth order linear differential equation is a differential equation that involves the nth derivative of the dependent variable y with respect to the independent variable x. It can be written in the form:

an(x) y(n) + a(n-1)(x) y(n-1) + … + a1(x) y’ + a0(x) y = f(x)

where an(x), a(n-1)(x), …, a1(x), a0(x) and f(x) are functions of x. The highest order derivative, y(n), is multiplied by an(x), which is usually assumed to be non-zero.

For example, the following are examples of nth order linear differential equations:

y” + y’ + 2y = x

y”’ – 3y” + 2y’ + 4y = e(2x)

2y(4) + 3y” – 4y’ + y = sin(x)

The solution to an nth order linear differential equation involves finding a function y that satisfies the equation for all values of x. The solution typically involves finding the values of the constants of integration that appear in the general solution.

Solving nth order linear differential equations is a complex topic that involves a variety of techniques, including finding the characteristic equation and using methods such as undetermined coefficients, variation of parameters, and Laplace transforms.

# Find the solution of Higher Order Linear Differential Equation

A linear differential equation of nth order can be expressed as:

y⁽ⁿ⁾(x) + a₁y⁽ⁿ⁻¹⁾(x) + a₂y⁽ⁿ⁻²⁾(x) + … + aₙ₋₁y'(x) + aₙy(x) = f(x)

where a₁, a₂, …, aₙ are constants and f(x) is a given function. The equation is said to be homogeneous if f(x) = 0.

To solve this type of equation, we first need to find the characteristic equation:

rⁿ + a₁r⁽ⁿ⁻¹⁾ + a₂r⁽ⁿ⁻²⁾ + … + aₙ₋₁r + aₙ = 0

We then find the roots of the characteristic equation, which can be real, complex or repeated. Let the roots be denoted by r₁, r₂, …, rₙ. Then the general solution of the homogeneous equation is given by:

y(x) = C₁e(r₁x) + C₂e(r₂x) + … + Cₙe(rₙx)

where C₁, C₂, …, Cₙ are arbitrary constants to be determined based on initial or boundary conditions.

If f(x) is not zero, we can find a particular solution by using the method of undetermined coefficients or variation of parameters, and the general solution is given by the sum of the homogeneous solution and the particular solution.

Let’s consider an example:

y⁽⁴⁾ – 3y⁽³⁾ – 9y” + 27y = 0

The characteristic equation is:

r⁴ – 3r³ – 9r² + 27 = 0

which can be factored as:

(r – 3)³(r + 1) = 0

So the roots are r₁ = 3 (repeated three times) and r₂ = -1. The homogeneous solution is:

y(x) = C₁e(3x) + C₂xe(3x) + C₃x²e(3x) + C₄e(-x)

To find a particular solution, we can guess that it has the form yp = A cos 3x + B sin 3x + C. Substituting this into the differential equation and solving for the coefficients A, B and C gives:

yp = 2/3 cos 3x – 1/3 sin 3x – 2/9

Therefore, the general solution to the differential equation is:

y(x) = C₁e(3x) + C₂xe(3x) + C₃x²e(3x) + C₄e(-x) + 2/3 cos 3x – 1/3 sin 3x – 2/9

The arbitrary constants C₁, C₂, C₃, C₄ can be determined based on initial or boundary conditions.

# Describe Complementary Functions

In the context of linear differential equations, the complementary function (also known as the homogeneous solution) is a particular solution that satisfies the homogeneous version of the differential equation. A homogeneous differential equation is an equation in which all terms contain the dependent variable and its derivatives, and there are no non-derivative terms.

The complementary function is an important concept in finding the general solution to a linear differential equation. The general solution to a linear differential equation is the sum of the complementary function and the particular integral. The complementary function represents the solution to the homogeneous equation and the particular integral represents a specific solution to the non-homogeneous equation.

For example, consider the second-order homogeneous differential equation:

y”(x) + 2y'(x) + y(x) = 0

The complementary function is found by assuming that the solution is of the form y(x) = e(rx) and substituting this into the equation.

y”(x) + 2y'(x) + y(x) = r2e(rx) + 2re(rx) + e(rx) = 0

Simplifying the equation yields the characteristic equation:

r2 + 2r + 1 = (r+1)2 = 0

The solution to the characteristic equation is r = -1, which means that the complementary function is:

yc(x) = c1e(-x) + c2xe(-x)

where c1 and c2 are constants determined by the initial or boundary conditions.

The general solution to the original differential equation is then:

y(x) = yc(x) + yp(x)

where yp(x) is a particular integral that satisfies the non-homogeneous equation.

# Find the Complementary Function of Higher-order Differential Equations

In the context of linear differential equations, the complementary function is a solution to the homogeneous equation associated with the differential equation. The complementary function is also called the homogeneous solution, and it provides a general solution to the differential equation.

To find the complementary function of a high-order differential equation, the first step is to find the characteristic equation associated with the differential equation. The characteristic equation is obtained by setting the coefficient of each derivative to zero and replacing the derivatives with corresponding powers of a variable, usually denoted by “r”. For example, for a third-order differential equation of the form:

ay”’ + by” + cy’ + dy = 0

the characteristic equation would be:

ar3 + br2 + c*r + d = 0

Once the characteristic equation is obtained, the roots of the equation are found, either by factoring the equation or using the quadratic formula for higher-order equations. If the roots are real and distinct, then the complementary function is of the form:

yc(x) = c1e(r1x) + c2e(r2x) + c3e(r3x)

where c1, c2, and c3 are arbitrary constants that are determined by applying initial or boundary conditions to the differential equation.

If the roots are real and repeated, then the complementary function is of the form:

yc(x) = c1e(r1x) + c2xe(r1x) + c3x2e(r1x)

where c1, c2, and c3 are arbitrary constants.

If the roots are complex conjugates, then the complementary function is of the form:

yc(x) = e(ax)(c1cos(bx) + c2sin(bx))

where a is the real part of the complex root and b is the imaginary part, and c1 and c2 are arbitrary constants.

For example, consider the third-order differential equation:

y”’ – 6y” + 11y’ – 6y = 0

The characteristic equation is:

r3 – 6r2 + 11r – 6 = 0

which can be factored as:

(r – 1)(r – 2)(r – 3) = 0

Therefore, the roots are r1 = 1, r2 = 2, and r3 = 3. The complementary function is:

yc(x) = c1ex + c2e(2x) + c3*e(3x)

where c1, c2, and c3 are constants that can be determined by applying initial or boundary conditions to the differential equation.

# Describe Particular Integral

In differential equations, a particular solution is the solution to the differential equation that satisfies the equation and all of its initial or boundary conditions. A particular integral, sometimes called a particular solution, is a solution to a non-homogeneous linear differential equation.

A non-homogeneous linear differential equation is one in which the non-zero function appears on the right-hand side of the equation. For example, a non-homogeneous linear differential equation of the second order looks like this:

y” + p(x) y’ + q(x) y = f(x)

where f(x) is a non-zero function.

The general solution of the differential equation is a combination of the complementary function and the particular integral. The complementary function is the solution to the corresponding homogeneous differential equation, which is obtained by setting f(x) = 0.

The particular integral is any solution to the non-homogeneous differential equation that is not a solution to the corresponding homogeneous differential equation. There are several methods for finding the particular integral, including the method of undetermined coefficients, variation of parameters, and the annihilator method.

The method of undetermined coefficients is used to find the particular integral when the function f(x) is a polynomial, an exponential function, a trigonometric function, or a combination of these functions. The method involves guessing a form for the particular integral and then finding the coefficients that satisfy the non-homogeneous differential equation.

For example, consider the non-homogeneous linear differential equation:

y” – 4y’ + 3y = 2ex

The corresponding homogeneous differential equation is:

y” – 4y’ + 3y = 0

which has the complementary function:

yc(x) = c1 ex + c2 e(3x)

To find the particular integral using the method of undetermined coefficients, we guess a form for the particular integral based on the function on the right-hand side of the equation. Since f(x) = 2ex, we guess that the particular integral has the form:

yp(x) = A ex

where A is a constant to be determined. We substitute this form into the non-homogeneous differential equation and obtain:

y”p – 4y’p + 3yp = 2ex

Taking derivatives and substituting the form of yp(x) yields:

A ex – 4A ex + 3A ex = 2ex

Solving for A gives:

A = 1

Therefore, the particular integral is:

yp(x) = ex

The general solution to the non-homogeneous linear differential equation is then:

y(x) = yc(x) + yp(x) = c1 ex + c2 e(3x) + ex

where c1 and c2 are constants determined by the initial or boundary conditions.

# Find Particular Integral of eax

In differential equations, the particular integral is a solution to the nonhomogeneous equation that is specific to a given nonhomogeneous term. A nonhomogeneous equation is a differential equation that has a forcing function or non-zero terms on the right-hand side. A particular integral is added to the complementary function to produce a general solution to the nonhomogeneous equation.

When the nonhomogeneous term of a differential equation takes the form of an exponential function, the particular integral can be found using the method of undetermined coefficients. In this method, we assume that the particular integral has the same form as the nonhomogeneous term and determine the coefficients through substitution into the differential equation.

For example, let’s consider the differential equation:

y” – 2y’ + y = e(3x)

The associated homogeneous equation is y” – 2y’ + y = 0, which has the complementary function yc = c1ex + c2xex. To find the particular integral, we assume that it has the form yp = Ae(3x), where A is a constant to be determined.

Substituting this into the differential equation, we get:

9Ae(3x) – 6Ae(3x) + Ae(3x) = e(3x)

Simplifying, we get:

4Ae(3x) = e(3x)

Thus, A = 1/4, and the particular integral is yp = (1/4)e(3x). The general solution to the differential equation is then given by the sum of the complementary function and particular integral:

y = c1ex + c2xex + (1/4)e(3x)

# Find Particular Integral of xn

The particular integral of an nth degree polynomial function xn is a function that satisfies a differential equation of the form:

y(n) + p1 y(n-1) + p2 y(n-2) + … + pn-1 y’ + pn y = f(x)

where f(x) is a function of x and p1, p2, …, pn-1, pn are constants.

To find the particular integral of xn, we use the method of undetermined coefficients. This method involves assuming that the particular integral has the same form as f(x) multiplied by a polynomial of the same degree as xn, with undetermined coefficients.

For example, to find the particular integral of x3 in the differential equation:

y” + 2y’ + y = 2x3

we assume that the particular integral is of the form:

yp = (ax3 + bx2 + cx + d)

where a, b, c, and d are undetermined coefficients.

We then take the first and second derivatives of yp:

yp‘ = 3ax2 + 2bx + c

yp” = 6ax + 2b

Substituting yp, yp‘, and yp” into the differential equation and equating the coefficients of x3, x2, x, and the constant term, we get:

6a = 2 (coefficient of x3 in f(x))

2b + 6a = 0 (coefficient of x2 in f(x))

2c + 2b + 6a = 0 (coefficient of x in f(x))

2d + 2c + 2b + 6a = 2 (constant term in f(x))

Solving this system of equations, we get:

a = 1/6, b = -1/3, c = 1/2, d = 0

Therefore, the particular integral of x3 is:

yp = (1/6)x3 – (1/3)x2 + (1/2)x

We can then add the particular integral to the complementary function (the general solution of the homogeneous equation) to get the general solution of the differential equation.

In summary, to find the particular integral of xn, we assume that the particular integral has the same form as f(x) multiplied by a polynomial of the same degree as xn, with undetermined coefficients. We then substitute this particular integral into the differential equation and solve for the undetermined coefficients.

# Find Particular Integral of sin(ax)

The particular integral of a function sin(ax) is a function that satisfies a differential equation of the form:

y” + p y’ + q y = f(x)

where f(x) is a function of x, p and q are constants, and a is a constant related to the frequency of sin(ax).

To find the particular integral of sin(ax), we use the method of undetermined coefficients. This method involves assuming that the particular integral has the same form as f(x) multiplied by sin(ax) or cos(ax), with undetermined coefficients.

For example, to find the particular integral of sin(2x) in the differential equation:

y” + 2y’ + 5y = 4sin(2x)

we assume that the particular integral is of the form:

yp = A sin(2x) + B cos(2x)

where A and B are undetermined coefficients.

We then take the first and second derivatives of yp:

yp‘ = 2A cos(2x) – 2B sin(2x)

yp” = -4A sin(2x) – 4B cos(2x)

Substituting yp, yp‘, and yp” into the differential equation and equating the coefficients of sin(2x) and cos(2x), we get:

-4A + 2B + 5A = 4 (coefficient of sin(2x) in f(x))

-4B – 2A + 5B = 0 (coefficient of cos(2x) in f(x))

Solving this system of equations, we get:

A = -4/21, B = -8/21

Therefore, the particular integral of sin(2x) is:

yp = (-4/21)sin(2x) – (8/21)cos(2x)

We can then add the particular integral to the complementary function (the general solution of the homogeneous equation) to get the general solution of the differential equation.

In summary, to find the particular integral of sin(ax), we assume that the particular integral has the same form as f(x) multiplied by sin(ax) or cos(ax), with undetermined coefficients. We then substitute this particular integral into the differential equation and solve for the undetermined coefficients.

# Find Particular Integral of cos(ax)

The particular integral of a function cos(ax) is a function that satisfies a differential equation of the form:

y” + p y’ + q y = f(x)

where f(x) is a function of x, p and q are constants, and a is a constant related to the frequency of cos(ax).

To find the particular integral of cos(ax), we use the method of undetermined coefficients. This method involves assuming that the particular integral has the same form as f(x) multiplied by sin(ax) or cos(ax), with undetermined coefficients.

For example, to find the particular integral of cos(2x) in the differential equation:

y” + 2y’ + 5y = 4cos(2x)

we assume that the particular integral is of the form:

yp = A sin(2x) + B cos(2x)

where A and B are undetermined coefficients.

We then take the first and second derivatives of yp:

yp‘ = 2A cos(2x) – 2B sin(2x)

yp” = -4A sin(2x) – 4B cos(2x)

Substituting yp, yp‘, and yp” into the differential equation and equating the coefficients of sin(2x) and cos(2x), we get:

-4B – 2A + 5B = 4 (coefficient of cos(2x) in f(x))

-4A + 2B + 5A = 0 (coefficient of sin(2x) in f(x))

Solving this system of equations, we get:

A = -2/21, B = 4/21

Therefore, the particular integral of cos(2x) is:

yp = (-2/21)sin(2x) + (4/21)cos(2x)

We can then add the particular integral to the complementary function (the general solution of the homogeneous equation) to get the general solution of the differential equation.

In summary, to find the particular integral of cos(ax), we assume that the particular integral has the same form as f(x) multiplied by sin(ax) or cos(ax), with undetermined coefficients. We then substitute this particular integral into the differential equation and solve for the undetermined coefficients.

# Find Particular Integral of eax f(x)

The particular integral of a function e(ax)f(x) is a function that satisfies a differential equation of the form:

y” + p y’ + q y = e(ax)f(x)

where f(x) is a function of x, p and q are constants, and a is a constant.

To find the particular integral of e(ax)f(x), we use the method of undetermined coefficients. This method involves assuming that the particular integral has the same form as f(x) multiplied by e(ax), with an undetermined coefficient.

For example, to find the particular integral of e(2x)sin(x) in the differential equation:

y” + 2y’ + 5y = e(2x)sin(x)

we assume that the particular integral is of the form:

yp = Ae(2x)sin(x) + Be(2x)cos(x)

where A and B are undetermined coefficients.

We then take the first and second derivatives of yp:

yp‘ = (2A + B)e(2x)cos(x) + (2B – A)e(2x)sin(x)

yp” = (4A + 4B)e(2x)sin(x)

Substituting yp, yp‘, and yp” into the differential equation and equating the coefficients of e(2x)sin(x) and e(2x)cos(x), we get:

(4A + 2B) + (2A + 4B) + 5A = 0 (coefficient of e(2x)cos(x) in f(x))

(-2A + 2B) + (4A – 2B) + 5B = 1 (coefficient of e(2x)sin(x) in f(x))

Solving this system of equations, we get:

A = 3/34, B = 1/34

Therefore, the particular integral of e(2x)sin(x) is:

yp = (3/34)e(2x)sin(x) + (1/34)e(2x)cos(x)

We can then add the particular integral to the complementary function (the general solution of the homogeneous equation) to get the general solution of the differential equation.

In summary, to find the particular integral of e(ax)f(x), we assume that the particular integral has the same form as f(x) multiplied by e(ax), with an undetermined coefficient. We then substitute this particular integral into the differential equation and solve for the undetermined coefficient.

# Find Particular Integral of sec(ax)

The particular integral of a function sec(ax) is a function that satisfies a differential equation of the form:

y” + p y’ + q y = sec(ax)

where p and q are constants, and a is a constant.

To find the particular integral of sec(ax), we use the method of undetermined coefficients. This method involves assuming that the particular integral has the same form as a linear combination of sec(ax) and tan(ax), with undetermined coefficients.

For example, to find the particular integral of sec(2x) in the differential equation:

y” + 2y’ + 5y = sec(2x)

we assume that the particular integral is of the form:

yp = A sec(2x) + B tan(2x)

where A and B are undetermined coefficients.

We then take the first and second derivatives of yp:

yp‘ = 2A sec(2x) tan(2x) + 2B sec2(2x)

yp” = 2A (sec2(2x) + 2 tan2(2x)) + 8B sec(2x) tan(2x)

Substituting yp, yp‘, and yp” into the differential equation and equating the coefficients of sec(2x) and tan(2x), we get:

2A + 2B = 0 (coefficient of sec(2x) in f(x))

2A = 1 (coefficient of tan(2x) in f(x))

Solving this system of equations, we get:

A = 1/2, B = -1/2

Therefore, the particular integral of sec(2x) is:

yp = (1/2) sec(2x) – (1/2) tan(2x)

We can then add the particular integral to the complementary function (the general solution of the homogeneous equation) to get the general solution of the differential equation.

In summary, to find the particular integral of sec(ax), we assume that the particular integral has the same form as a linear combination of sec(ax) and tan(ax), with undetermined coefficients. We then substitute this particular integral into the differential equation and solve for the undetermined coefficients.

# Describe Homogeneous Linear Differential Equations

A homogeneous linear differential equation is a differential equation of the form:

an (x)y(n) + a{n-1}(x)y(n-1) + … + a1(x)y’ + a0(x)y = 0

where an (x), a{n-1}(x), …, a1(x), a0(x) are continuous functions of x, and y(k) represents the kth derivative of y with respect to x.

This type of differential equation is called homogeneous because all the terms on the left-hand side add up to zero, and the function y(x) is a solution of the equation if and only if c y(x) is a solution for any constant c. In other words, the solutions of the homogeneous linear differential equation form a vector space.

The degree of a homogeneous linear differential equation is the highest derivative in the equation. For example, the equation:

y” + 3xy’ – 4x2 y = 0

is a second-order homogeneous linear differential equation.

One important property of homogeneous linear differential equations is the principle of superposition. This principle states that if y1(x) and y2(x) are solutions of the homogeneous linear differential equation, then any linear combination of y1(x) and y2(x) (i.e., c1 y1(x) + c2 y2(x), where c1 and c2 are constants) is also a solution of the differential equation.

To solve a homogeneous linear differential equation, we can assume that the solution has the form:

y(x) = e(rx)

where r is a constant. Substituting this solution into the differential equation, we get:

an(r)e(rx) + a{n-1}(r)e(rx) + … + a1(r)e(rx) + a0(r)e(rx) = 0

or, equivalently:

p(r)e(rx) = 0

where p(r) is the characteristic polynomial of the differential equation, defined as:

p(r) = an(r) rn + a{n-1}(r) r{n-1} + … + a1(r) r + a0(r)

Since e(rx) is never zero, the equation p(r)e(rx) = 0 is satisfied if and only if p(r) = 0. Therefore, the roots of the characteristic polynomial give us the values of r that correspond to the possible solutions of the homogeneous linear differential equation.

For example, in the differential equation:

y” + 3xy’ – 4x2 y = 0

the characteristic polynomial is:

p(r) = r2 + 3xr – 4x2

which can be factored as:

p(r) = (r-1)(r+4x)

Therefore, the solutions of the differential equation have the form:

y(x) = c1 ex + c2 e(-4x)

where c1 and c2 are constants determined by the initial or boundary conditions.

In summary, a homogeneous linear differential equation is a differential equation of the form an(x)y(n) + a{n-1}(x)y(n-1) + … + a1(x)y’ + a0(x)y = 0, where all the terms on the left-hand side add up to zero. The solutions of the homogeneous linear differential equation form a vector space, and the principle of superposition holds.

# Find the Solution of Homogeneous Linear Differential Equations

To find the solution of a homogeneous linear differential equation of the form:

an(x)y(n) + a{n-1}(x)y(n-1) + … + a1(x)y’ + a0(x)y = 0

where an(x), a{n-1}(x), …, a1(x), a0(x) are continuous functions of x, and y(k) represents the kth derivative of y with respect to x, we can use the following steps:

Step 1: Find the characteristic polynomial

The characteristic polynomial is obtained by setting the left-hand side of the differential equation to zero, i.e.,

an (x)rn + a{n-1}(x)r{n-1} + … + a1(x)r + a0(x) = 0

This polynomial equation is called the characteristic equation.

Step 2: Find the roots of the characteristic polynomial

The roots of the characteristic polynomial correspond to the values of r that make the left-hand side of the differential equation equal to zero. These values of r are also called the characteristic values or eigenvalues.

Step 3: Find the general solution

The general solution of the differential equation is obtained by combining the solutions of the form y(x) = e(rx), where r is a root of the characteristic polynomial, using the principle of superposition. The general solution has the form:

y(x) = c1 y1(x) + c2 y2(x) + … + cn yn(x)

where y1(x), y2(x), …, yn(x) are linearly independent solutions of the differential equation (i.e., they are not multiples of each other), and c1, c2, …, cn are arbitrary constants.

Step 4: Find the particular solution

The particular solution is obtained by applying the initial or boundary conditions to the general solution. The number of initial or boundary conditions required to determine the values of the constants c1, c2, …, cn depends on the order of the differential equation and the type of conditions.

Let’s consider an example to illustrate the above steps. Suppose we want to solve the second-order homogeneous linear differential equation:

y” + 2y’ – 8y = 0

Step 1: Find the characteristic polynomial

The characteristic polynomial is:

r2 + 2r – 8 = 0

Step 2: Find the roots of the characteristic polynomial

The roots of the characteristic polynomial can be found by using the quadratic formula or factoring the polynomial. In this case, we can factor the polynomial as:

(r + 4)(r – 2) = 0

Therefore, the roots are r = -4 and r = 2.

Step 3: Find the general solution

The general solution is:

y(x) = c1 e(-4x) + c2 e(2x)

Step 4: Find the particular solution

To find the values of c1 and c2, we need to apply the initial or boundary conditions. Suppose we are given that y(0) = 1 and y'(0) = 0. Then, we have:

y(0) = c1 + c2 = 1

y'(0) = -4c1 + 2c2 = 0

Solving these equations, we get c1 = 1/3 and c2 = 2/3.

# Describe Simultaneous Differential Equations

Simultaneous differential equations are a system of two or more differential equations that must be solved together to obtain a solution. These equations are typically used to model physical systems that involve multiple variables that change over time and are interdependent on each other.

The general form of a system of simultaneous differential equations is:

dx1/dt = f1(x1, x2, …, xn, t)

dx2/dt = f2(x1, x2, …, xn, t)

dxn/dt = fn(x1, x2, …, xn, t)

where x1, x2, …, xn are the dependent variables, t is the independent variable, and f1, f2, …, fn are the functions that relate the derivatives of the dependent variables to the independent and dependent variables.

Solving a system of simultaneous differential equations involves finding a set of functions that satisfy all the equations in the system. This can be done using various techniques such as elimination, substitution, and matrix methods.

In practice, many physical systems can be modeled using simultaneous differential equations. For example, a system of two differential equations can be used to model the motion of a mass-spring system, where the position of the mass and the displacement of the spring are dependent variables that change over time and are interdependent on each other.

Here’s an example of a system of simultaneous differential equations:

dx/dt = 2x + y

dy/dt = 3x – y

To solve this system of differential equations, we can use the method of elimination. We can eliminate y from the equations by multiplying the first equation by 3 and the second equation by 2, and subtracting the second equation from the first:

3dx/dt – 2dy/dt = 6x + 3y – (6x – 2y)

d/dt(3x – 2y) = 5y

Integrating both sides with respect to t, we get:

3x – 2y = 5yt + C

where C is the constant of integration.

To solve for y, we can substitute 3x – 2y = 5yt + C into the second equation of the original system:

dy/dt = 3x – y

dy/dt = 3x – (5yt + C)/2

This is a first-order linear differential equation, which can be solved using the integrating factor method. After solving for y, we can substitute the solution into the equation 3x – 2y = 5yt + C to obtain the general solution to the system of differential equations.

# Find the Solution of Simultaneous Differential Equations

Consider the following system of differential equations:

dx/dt = x – 2y

dy/dt = 3x + y

To solve this system, we can use the method of substitution. We can solve for y in terms of x from the first equation and substitute it into the second equation:

dy/dt = 3x + y

y = (x – dx/dt)/2

dy/dt = 3x + (x – dx/dt)/2

Simplifying, we get:

2dy/dt + dx/dt = 5x

This is a first-order linear differential equation in y and x. To solve it, we can use the integrating factor method. The integrating factor is e(∫2dt) = e(2t). Multiplying both sides of the equation by e(2t), we get:

d/dt(e(2t)y) = 5xe(2t)

Integrating both sides with respect to t, we get:

e(2t)y = dt + C

y = (1/2)x + (1/10)e(-2t) + C*e(-2t)

where C is the constant of integration.

Substituting this solution for y back into the first equation, we get:

dx/dt = x – 2((1/2)x + (1/10)e(-2t) + Ce(-2t))

dx/dt = -x + (4/5)e(-2t) + 2Ce(-2t)

This is a first-order linear differential equation in x. To solve it, we can again use the integrating factor method. The integrating factor is e(∫-1dt) = e(-t). Multiplying both sides of the equation by e(-t), we get:

d/dt(e(-t)x) = (4/5)e(-3t) + 2Ce(-3t)

Integrating both sides with respect to t, we get:

e(-t)x = (-2/15)e(-3t) + C1e(-t) + C2

where C1 and C2 are constants of integration.

Therefore, the general solution to the system of differential equations is:

x = (-2/15) + C1e(t) + C2e(t)

y = (1/2)x + (1/10)e(-2t) + C*e(-2t)

where C, C1, and C2 are constants of integration.

# Describe the Wronskian for Linearly Dependent or Linearly Independent solutions

The Wronskian is a determinant-based method that is used to determine whether a set of solutions to a homogeneous linear differential equation are linearly dependent or linearly independent. It is defined as:

W(y1, y2, …, yn) = |y1 y2 … yn|

|y1‘ y2‘ … yn‘|

|y1” y2” … yn”|

| … … … |

|y1(n-1) y2(n-1) … yn(n-1)|

where y1, y2, …, yn are n functions that are solutions to a homogeneous linear differential equation of nth order.

If the Wronskian is non-zero for any value of x, then the set of solutions is said to be linearly independent. If the Wronskian is zero for all values of x, then the set of solutions is said to be linearly dependent.

For example, consider the following homogeneous linear differential equation:

y” – 4y’ + 4y = 0

The characteristic equation of this differential equation is r2 – 4r + 4 = 0, which has a double root at r = 2. Therefore, the general solution is:

y(x) = (c1 + c2 x) e(2x)

To check whether this set of solutions is linearly independent, we can calculate its Wronskian:

W(y1, y2) = |e(2x) (1 + 0x) e(2x)|

|2e(2x) 2e(2x) |

W(y1, y2) = 2e(4x) – 2e(4x) = 0

Since the Wronskian is zero for all values of x, the set of solutions is linearly dependent. Therefore, we need to find another solution that is linearly independent of the first one.

# Describe method of Variation of Parameters

The method of variation of parameters is used to find a particular solution to a non-homogeneous linear differential equation of the form:

y” + p(x)y’ + q(x)y = f(x)

To apply this method, we first find the general solution to the corresponding homogeneous equation:

y” + p(x)y’ + q(x)y = 0

Let the two linearly independent solutions be y1(x) and y2(x). Then the general solution to the homogeneous equation is:

y(x) = c1 y1(x) + c2 y2(x)

To find a particular solution to the non-homogeneous equation, we assume that the solution has the form:

y(x) = u1(x) y1(x) + u2(x) y2(x)

where u1(x) and u2(x) are functions to be determined.

Taking the first and second derivatives of y(x) with respect to x, we get:

y'(x) = u1‘(x) y1(x) + u2‘(x) y2(x) + u1(x) y1‘(x) + u2(x) y2‘(x)

y”(x) = u1”(x) y1(x) + u2”(x) y2(x) + 2u1‘(x) y1‘(x) + 2u2′(x) y2′(x) + u1(x) y1”(x) + u2(x) y2”(x)

Substituting these expressions into the non-homogeneous equation, we get:

u1”(x) y1(x) + u2”(x) y2(x) + 2u1′(x) y1‘(x) + 2u2′(x) y2′(x) + u1(x) y1”(x) + u2(x) y2”(x) + p(x) (u1′(x) y1(x) + u2′(x) y2(x) + u1(x) y1′(x) + u2(x) y2′(x)) + q(x) (u1(x) y1(x) + u2(x) y2(x)) = f(x)

Since y1(x) and y2(x) are solutions to the homogeneous equation, we have:

y1”(x) + p(x) y1′(x) + q(x) y1(x) = 0

y2”(x) + p(x) y2′(x) + q(x) y2(x) = 0

Therefore, the above equation reduces to:

u1”(x) y1(x) + u2”(x) y2(x) + p(x) (u1′(x) y1(x) + u2′(x) y2(x)) = f(x)

We can simplify this equation by defining the Wronskian as:

W(y1, y2) = |y1 y2|

|y1′ y2′|

Then, we have:

u1”(x) y1(x) + u2”(x) y2(x) = f(x) / W(y1, y2)

To find u1(x) and u2(x), we integrate this equation with respect to x:

u1′(x) = – y2(x) (f(x) / W(y1, y2))

u2′(x) = y1(x) (f(x) / W(y1, y2))

# Describe method of Variation of Parameters

The method of variation of parameters is a technique used to solve non-homogeneous linear differential equations of the form y”(x) + p(x)y'(x) + q(x)y(x) = g(x), where g(x) is a known function and p(x) and q(x) are continuous functions. This method involves assuming a particular solution in the form of a linear combination of two linearly independent functions and finding the coefficients using the boundary conditions.

Example:

Consider the differential equation y”(x) – y(x) = x2. To solve this equation using the method of variation of parameters, we first need to find the complementary function (CF) by solving the homogeneous equation y”(x) – y(x) = 0. The characteristic equation is r2 – 1 = 0, which gives the solutions r = ±1. Therefore, the CF is yc(x) = c1ex + c2e(-x).

Next, we assume the particular solution (PS) in the form of yp(x) = u1(x)ex + u2(x)e(-x), where u1(x) and u2(x) are functions to be determined. We substitute this into the non-homogeneous equation and simplify to get:

y”p(x) – yp(x) = x2

[u”1(x) + 2u’1(x) + u”2(x) – 2u’2(x)]ex + [u”1(x) – 2u’1(x) + u’‘2(x) + 2u‘2(x)]e(-x) – [u1(x)ex + u2(x)e(-x)] = x2

Now, we equate the coefficients of ex and e(-x) on both sides of the equation, which gives us the following system of equations:

u”1(x) + 2u‘1(x) + u”2(x) – 2u’2(x) = x2

u’‘1(x) – 2u‘1(x) + u”2(x) + 2u’2(x) = 0

We can solve this system of equations to find the values of u1(x) and u2(x). Let’s assume u1(x) = Ax2 + Bx + C and u2(x) = Dx2 + Ex + F. Substituting these values into the above system of equations and solving for the coefficients, we get:

u1(x) = (1/2)x2 – (5/4)x – (1/4)

u2(x) = (1/2)x2 + (5/4)x – (3/4)

Therefore, the particular solution is yp(x) = [(1/2)x2 – (5/4)x – (1/4)]ex + [(1/2)x2 + (5/4)x – (3/4)]e(-x).

# Find the Solution of Differential Equation using method of Variation of Parameters

The method of variation of parameters is a technique used to solve non-homogeneous linear differential equations of the form y”(x) + p(x)y'(x) + q(x)y(x) = g(x), where g(x) is a known function and p(x) and q(x) are continuous functions. This method involves assuming a particular solution in the form of a linear combination of two linearly independent functions and finding the coefficients using the boundary conditions.

Example:

Let’s consider the following differential equation:

y”(x) + 4y'(x) + 4y(x) = x2

Step 1: Find the complementary function (CF)

We first find the complementary function by solving the homogeneous equation y”(x) + 4y'(x) + 4y(x) = 0. The characteristic equation is r2 + 4r + 4 = 0, which gives the repeated root r = -2. Therefore, the CF is yc(x) = c1e(-2x) + c2xe(-2x).

Step 2: Assume the particular solution (PS)

We assume the particular solution in the form of yp(x) = u1(x)e(-2x) + u2(x)xe(-2x), where u1(x) and u2(x) are functions to be determined.

Step 3: Find the coefficients

We substitute the PS into the non-homogeneous equation and simplify to get:

[u”1(x) + 2u’1(x) + u”2(x) – 2u’2(x)]e(-2x) + [u”1(x) – 2u‘1(x) + u”2(x) + 2u’2(x)]xe(-2x) = x2

Now, we equate the coefficients of e(-2x) and xe(-2x) on both sides of the equation, which gives us the following system of equations:

u”1(x) + 2u’1(x) + u”2(x) – 2u‘2(x) = 0

u”1(x) – 2u’1(x) + u’2(x) + 2u’2(x) = x2We can solve this system of equations to find the values of u1(x) and u2(x). We assume u1(x) = Ax2 + Bx + C and u2(x) = Dx2 + Ex + F. Substituting these values into the above system of equations and solving for the coefficients, we get:

u1(x) = -x2/2 – x/2

u2(x) = x2/2 + x/2

Therefore, the particular solution is:

yp(x) = [-x2/2 – x/2]e(-2x) + [x2/2 + x/2]xe(-2x)

Step 4: Find the general solution

The general solution of the non-homogeneous equation is then given by y(x) = yc(x) + yp(x) = c1e(-2x) + c2xe(-2x) + [-x2/2 – x/2]e(-2x) + [x2/2 + x/2]xe(-2x)

# Find the Complete Solution of the Differential Equation whose one Solution is known

This learning outcome deals with finding the complete solution of a differential equation when one particular solution is already known. This method is called the method of integrating factors.

Method of Integrating Factors:

The method of integrating factors is used to solve first-order linear differential equations of the form y’ + p(x)y = q(x). To use this method, we first need to find a function u(x) such that multiplying both sides of the differential equation by u(x) will make the left-hand side the derivative of a product. This function u(x) is called the integrating factor. Once we have found the integrating factor, we can then integrate both sides of the differential equation to obtain the complete solution.

Example:

Consider the differential equation y’ + 2y = 4x. We can find the complete solution of this differential equation using the method of integrating factors, given that one particular solution is y = 2x – 1.

Step 1: Find the Integrating Factor

To find the integrating factor, we need to find a function u(x) such that multiplying both sides of the differential equation by u(x) will make the left-hand side the derivative of a product. In this case, we can use the integrating factor u(x) = e(∫2dx) = e(2x). Multiplying both sides of the differential equation by e(2x), we get:

e(2x)y’ + 2e(2x)y = 4xe(2x)

Now, the left-hand side is the derivative of the product e(2x)y, so we can write:

(e(2x)y)’ = 4xe(2x)

Step 2: Integrate Both Sides

Integrating both sides of the differential equation, we get:

e(2x)y =dx = 2xe (2x) – dx = 2xe (2x) – e (2x) + C

where C is the constant of integration.

Step 3: Solve for y

To find the complete solution, we need to solve for y. Dividing both sides by e(2x), we get:

y = 2x – 1 + Ce(-2x)

where C is the constant of integration. This is the complete solution of the differential equation. Note that the particular solution y = 2x – 1 is a special case of the complete solution when C = 0.

# Find the Solution of Differential Equation by using Removal of First Derivative method

The removal of the first derivative method is a technique used to solve first-order ordinary differential equations. This method involves making a substitution that transforms the original equation into a separable form, i.e., one in which the variables can be separated and the equation can be integrated.

Method of Removal of First Derivative:

The method of removal of the first derivative involves making a substitution of the form y = f(x)u, where f(x) is a function of x and u is a function of y. We then differentiate both sides of the equation with respect to x and use the product rule to eliminate the first derivative of y. This gives us a new differential equation that is separable.

Example:

Consider the differential equation y’ – 2xy = x. To solve this equation using the method of removal of first derivative, we first make the substitution y = f(x)u, where f(x) is a function of x and u is a function of y. Let us choose f(x) = e(x^2), which will simplify the derivative of y. Substituting y = f(x)u into the differential equation, we get:

f'(x)u + f(x)u’ – 2xf(x)u = x

Now, we differentiate both sides of the equation with respect to x and use the product rule to eliminate the first derivative of y:

f”(x)u + 2f'(x)u’ + f(x)u” – 2f(x)u – 2xf'(x)u = 1

Multiplying both sides by e(x^2), we get:

(e(x^2)f”(x) + 2xe(x2)f'(x))u + e(x^2)f(x)u” = e(x^2)

Note that the left-hand side of the equation is now a function of x and u, while the right-hand side is only a function of x. Therefore, we can separate the variables by dividing both sides by e(x^2)f(x)u’, giving us:

[e(x^2)f”(x) + 2xe(x^2)f'(x)]/e(x^2)f(x) = 1/u + 2x/e(x^2)f(x)

The left-hand side is a function of x, while the right-hand side is a function of y. Therefore, we can integrate both sides with respect to their respective variables to obtain the solution:

∫ [e(x^2)f”(x) + 2xe(x^2)f'(x)]/e(x^2)f(x) dx = ∫ 1/u + 2x/e(x^2)f(x) dy

The integral on the left-hand side can be simplified using the substitution v = f'(x)/f(x), which gives:

∫ [e(x^2)f”(x) + 2xe(x^2)f'(x)]/e(x^2)f(x) dx = ∫ (v + 2x)e(x^2) dx

Integrating both sides, we get:

ln|f(x)| = -e(-x^2) + Ce(x^2)

where C is the constant of integration. Solving for u, we get:

u = Ce(∫2x/e(x^2)f(x)dx)

# Find the Solution of a Differential Equation by Changing the Independent Variable

Changing the independent variable is a technique used to solve ordinary differential equations. This technique involves making a substitution of the form y = f(x)z, where f(x) is a function of x and z is a function of a new independent variable t. We then differentiate both sides of the equation with respect to x and use the chain rule to transform the differential equation into a new form that is easier to solve.

Method of Changing the Independent Variable:

The method of changing the independent variable involves making a substitution of the form y = f(x)z, where f(x) is a function of x and z is a function of a new independent variable t. We then differentiate both sides of the equation with respect to x and use the chain rule to transform the differential equation into a new form that is easier to solve. We choose f(x) such that it cancels out some of the terms in the original equation, and we choose z such that it simplifies the remaining terms.

Example:

Consider the differential equation y’ + 2y = x. To solve this equation using the method of changing the independent variable, we first make the substitution y = e(-x^2)z, where z is a function of a new independent variable t. We then differentiate both sides of the equation with respect to x:

y’ = -2xe(-x^2)z + e(-x^2)z’

Substituting the values of y and y’ into the original differential equation, we get:

-2xe(-x^2)z + e(-x^2)z’ + 2e(-x^2)z = x

Multiplying both sides of the equation by e(x^2), we get:

-2xz + e(x^2)z’ + 2z = xe(x^2)

We now choose t such that it satisfies the equation:

t = dx

This gives us:

dt/dx = e(x^2)

We can now substitute t into the equation above, and differentiate with respect to t using the chain rule to obtain:

dz/dt = x

Substituting these values back into the differential equation, we get:

-2tz + dz/dt + 2z = t

This is now a separable differential equation, which we can solve by separating the variables:

(2z – t)dz = (2t – 1)dt

Integrating both sides, we get:

z2 – tz = t2 – t/2 + C

where C is the constant of integration. Substituting back the value of y, we get:

y = e(-x^2)(z2 – tz)

This is the general solution of the differential equation.

# Apply Differential Equations in Electrical Circuits

Differential equations are used to model a variety of physical systems, including electrical circuits. Electrical circuits are made up of components such as resistors, capacitors, and inductors, and these components can be described by differential equations. By modelling an electrical circuit as a system of differential equations, we can determine the behavior of the circuit over time, and use this information to design and optimize the circuit.

Method of Applying Differential Equations in Electrical Circuits:

The method of applying differential equations in electrical circuits involves modelling the circuit as a system of differential equations. This is done by using Kirchhoff’s laws, which describe the behavior of electrical circuits. Kirchhoff’s laws state that the sum of the voltages around any closed loop in a circuit is equal to zero, and the sum of the currents at any node in a circuit is equal to zero. By applying these laws to each component in the circuit, we can derive a set of differential equations that describe the behavior of the circuit.

Example:

Consider the following circuit, consisting of a resistor R, a capacitor C, and a voltage source V:

To model this circuit as a system of differential equations, we can use Kirchhoff’s laws. The voltage drop across the resistor is given by Ohm’s law, V = IR, where I is the current through the resistor. The current through the capacitor is given by the rate of change of the voltage across the capacitor, I = C(dV/dt). Kirchhoff’s voltage law states that the sum of the voltages around any closed loop in the circuit is zero. Applying this law to the loop consisting of the resistor and capacitor, we get:

V – IR – (1/C) ∫Idt = 0

Differentiating both sides with respect to time, we get:

-d/dt(IR) – (1/C)I = dV/dt

Simplifying this equation, we get:

RC(dI/dt) + I = (1/R) dV/dt

This is a first-order linear differential equation, which can be solved using techniques such as separation of variables or integrating factors. The solution of this equation gives us the current through the circuit as a function of time. We can then use this information to determine the behavior of the circuit, such as the voltage across the capacitor over time.

In this example, we have modelled an electrical circuit as a system of differential equations, which can be used to determine the behavior of the circuit over time. This is a powerful tool for designing and optimising circuits, and is widely used in electrical engineering and related fields.

# Apply Differential Equations in Simple Harmonic Motion

Simple harmonic motion is a common physical phenomenon that can be described using differential equations. Simple harmonic motion occurs when a system is subject to a restoring force that is proportional to the displacement of the system from its equilibrium position. Examples of systems that exhibit simple harmonic motion include mass-spring systems and pendulums. By modelling these systems as differential equations, we can analyze their behavior and make predictions about their motion.

Method of Applying Differential Equations in Simple Harmonic Motion:

The method of applying differential equations in simple harmonic motion involves modelling the system as a second-order differential equation. The general form of the differential equation is:

m(d2x/dt2) + kx = 0

where m is the mass of the system, k is the spring constant or equivalent for the system, and x is the displacement of the system from its equilibrium position. This equation can be derived using Newton’s second law, which states that the force on an object is proportional to its acceleration.

The solution to this differential equation is a function of time that describes the motion of the system. For a mass-spring system, the solution is a sinusoidal function that describes the oscillation of the mass around its equilibrium position. For a pendulum, the solution is a combination of sinusoidal and exponential functions that describe the motion of the pendulum.

Example:

Consider a mass-spring system with a mass of 1 kg and a spring constant of 4 N/m. The system is initially displaced 0.1 m from its equilibrium position and released from rest. We want to determine the position of the mass as a function of time.

To model this system as a differential equation, we use the equation:

m(d2x/dt2) + kx = 0

Substituting the values for m and k, we get:

d2x/dt2 + 4x = 0

This is a second-order linear differential equation, which can be solved using techniques such as characteristic equations or Laplace transforms. The general solution to this equation is:

x = A cos(2t) + B sin(2t)

where A and B are constants determined by the initial conditions of the system.

Using the initial conditions given in the problem, we can determine the values of A and B. The initial position of the mass is x(0) = 0.1, so we have:

A cos(0) + B sin(0) = 0.1

which simplifies to:

A = 0.1

The initial velocity of the mass is dx/dt = 0, so we have:

-2A sin(0) + 2B cos(0) = 0

which simplifies to:

B = 0

Substituting these values into the general solution, we get:

x = 0.1 cos(2t)

This equation describes the position of the mass as a function of time. We can use this information to determine the velocity and acceleration of the mass, and to make predictions about its motion.

In this example, we have modelled a mass-spring system as a differential equation, and used the solution to make predictions about its motion. This is a powerful tool for understanding and analyzing systems that exhibit simple harmonic motion, and is widely used in physics and related fields.

# Apply Differential Equations Simple Population Model

Differential equations can be used to model the behavior of populations over time. The simplest population models assume that the population grows at a rate proportional to its size, and that there is no limit to the growth of the population. These models are useful for understanding the basic dynamics of population growth, and for making predictions about future population sizes.

Method of Applying Differential Equations in Simple Population Model:

The method of applying differential equations in simple population models involves modeling the population as a first-order differential equation. The general form of the differential equation is:

dP/dt = rP

where P is the population size, t is time, and r is the growth rate of the population. This equation can be derived using the assumption that the rate of change of the population is proportional to its size. The solution to this differential equation is an exponential function that describes the growth of the population over time.

If we want to include a limiting factor, such as a finite amount of resources, we can modify the differential equation to include a carrying capacity, K, that represents the maximum population size that the environment can support. The modified differential equation is:

dP/dt = rP(1 – P/K)

where the term (1 – P/K) represents the limiting effect of the environment on the growth of the population.

Example:

Consider a population of bacteria that doubles in size every 24 hours. We want to model the growth of this population over time using a differential equation.

To model this system as a differential equation, we use the equation:

dP/dt = rP

where P is the population size and r is the growth rate. In this case, the growth rate is given as doubling every 24 hours, so we have:

r = ln(2)/24

Substituting this value into the differential equation, we get:

dP/dt = (ln(2)/24)P

This is a separable first-order differential equation, which can be solved using separation of variables. The solution to this equation is:

P = P0 * e(ln(2)/24 * t)

where P0 is the initial population size.

Using the initial condition that the population is initially 100 bacteria, we have:

P0 = 100

Substituting this value into the solution, we get:

P = 100 * e(ln(2)/24 * t)

This equation describes the population of bacteria as a function of time. We can use this information to make predictions about the future size of the population.

In this example, we have modelled a population of bacteria as a differential equation, and used the solution to make predictions about its growth over time. This is a useful tool for understanding the basic dynamics of population growth, and can be extended to more complex models that include limiting factors and other variables.