# First Order Ordinary Differential Equations

### Contents

**Describe the Order and Degree of a Differential Equation**2**Determine the Order and Degree of a Differential Equation**3**Explain the formulation of Differential Equations**5**Describe the General and Particular Solution of Differential Equations**6**Explain the variable Separable Method to find the Solutions of Differential Equations**9**Describe the Homogeneous Differential Equations**11**Find the Solutions of Homogeneous Differential Equations**13**Describe the Linear Differential Equations**14**Find the solutions of Linear Differential Equations**17**Describe the Exact Differential Equations**18**Find the solutions of Exact Differential Equations**20**Describe the Differential Equation reducible to the variable Separable form**21**Describe the Differential Equation reducible to the Homogeneous Form**23**Find the solutions of Variable Separable and Homogeneous Differential Equations**25**Describe the Differential Equation reducible to the Exact Form**28**Find the solutions of Exact Differential Equations**30**Describe the Differential Equation reducible to the Linear form (Bernoulli Equation)**32**Find the solutions of Linear Differential Equations**34**Describe Orthogonal Trajectory of Cartesian Curve**37**Describe Orthogonal Trajectory of Polar Curve**38**Calculate the orthogonal Trajectory of Polar Curve**40**Describe differential equation solvable for p(dy/dx)**41**Describe differential equation solvable for y**43**Describe differential equation solvable for X**44**Explain Clairaut’s Equation**45

Describe the Order and Degree of a Differential Equation

First, let’s define what a differential equation is. A differential equation is an equation that relates a function and its derivatives. Differential equations are commonly used in physics, engineering, and other sciences to model real-world phenomena.

The order of a differential equation is the highest order of derivative in the equation. For example, the differential equation

y” + 3y’ – 4y = 0

has a second order, since the highest derivatives in the equation is the second derivative y”. Similarly, the differential equation

y’ + 2y = 5

has a first order, since the highest derivative is the first derivative y’.

The degree of a differential equation is the power to which the highest derivative is raised. For example, the differential equation

y” + 3y’ – 4y = 0

has a degree of 2, since the highest derivative y” is raised to the power of 2. Similarly, the differential equation

y’ + 2y = 5

has a degree of 1, since the highest derivative y’ is raised to the power of 1.

Let’s look at another example. The differential equation

y”’ – 2y” + y’ – 3y = 0

has a third order, since the highest derivative is the third derivative y”’. The degree of the differential equation is 1, since the highest derivative y”’ is raised to the power of 1.

In general, it’s important to understand the order and degree of a differential equation in order to determine its properties and to find a solution. For example, the order and degree can help you determine how many initial or boundary conditions are needed to solve the differential equation, and whether the differential equation is linear or nonlinear.

In conclusion, the order and degree of a differential equation are important properties that help us understand and solve differential equations. By being able to describe these properties, we can better understand the behavior of the functions and the physical systems they model.

Determine the Order and Degree of a Differential Equation

A differential equation is an equation that involves derivatives of a function, which relates the rate of change of a quantity to its current value. The order of a differential equation refers to the highest order derivative present in the equation. For example, consider the following differential equations:

- y” + 4y’ + 3y = 0
- y”’ + y’ = sin(x)
- y’ = x
^{2}– 3x

In equation 1, the highest derivative is the second derivative, y”. Therefore, this differential equation is of second order.

In equation 2, the highest derivative is the third derivative, y”’, making it a third order differential equation.

In equation 3, the highest derivative is the first derivative, y’, making it a first order differential equation.

The degree of a differential equation is the highest power to which the highest order derivative is raised. For example, let’s consider the following differential equation:

- y” + 2y’ + y = sin(x)

The highest order derivative is the second derivative, y”. The degree of this differential equation is 1, because the highest order derivative is raised to the power of 1.

In general, it is important to determine the order and degree of a differential equation in order to identify the appropriate method for solving it. For example, first-order differential equations can be solved using separation of variables, while higher-order differential equations often require more complex methods such as integration by parts, Laplace transforms, or numerical methods.

In conclusion, being able to determine the order and degree of a differential equation is an important skill in solving differential equations. It helps to identify the appropriate methods to use in solving the equation and allows us to better understand the properties and behavior of the system being modelled.

Explain the formulation of Differential Equations

This Learning Outcome requires you to explain the formulation of differential equations. Differential equations are mathematical equations that involve the derivatives of an unknown function. They are used to describe the behavior of physical systems and processes that change over time, such as the motion of objects, the flow of fluids, and the growth of populations.

The formulation of a differential equation involves two key steps: identifying the unknown function and formulating the relationship between the function and its derivatives.

- Identifying the Unknown Function:

The first step in formulating a differential equation is to identify the unknown function that we are trying to model. This function may represent a physical quantity such as position, velocity, temperature, or concentration. Let’s consider some examples:

- Suppose we want to model the motion of a particle in one dimension. We can choose the unknown function to be the particle’s position as a function of time, denoted by x(t).
- If we want to model the temperature of a rod that is heated from one end, we can choose the unknown function to be the temperature as a function of position along the rod, denoted by T(x).
- If we want to model the growth of a population, we can choose the unknown function to be the population size as a function of time, denoted by N(t).

- Formulating the Relationship between the Function and its Derivatives:

The second step in formulating a differential equation is to express the relationship between the unknown function and its derivatives. This relationship can be written as an algebraic equation, where the derivatives of the unknown function appear as variables. The order and degree of the differential equation depend on the highest order derivative of the unknown function that appears in the equation. Let’s consider some examples:

- The equation for the motion of a particle under constant acceleration is given by x”(t) = a, where a is the constant acceleration. This is a second-order differential equation, since the highest order derivative that appears is the second derivative x”(t).
- The equation for the temperature distribution in a rod is given by d²T/dx² = -k T(x), where k is the thermal conductivity of the rod. This is a second-order differential equation, since the highest order derivative that appears is the second derivative d²T/dx².
- The equation for the growth of a population in the absence of external factors is given by dN/dt = rN, where r is the growth rate. This is a first-order differential equation, since the highest order derivative that appears is the first derivative dN/dt.

In conclusion, the formulation of a differential equation involves identifying the unknown function that we want to model and formulating the relationship between the function and its derivatives. The order and degree of the differential equation depend on the highest order derivative of the unknown function that appears in the equation. Differential equations are used to model physical systems and processes that change over time, and are essential in many areas of science and engineering.

Describe the General and Particular Solution of Differential Equations

This Learning Outcome requires you to describe the general and particular solution of a differential equation. Differential equations are mathematical equations that involve the derivatives of an unknown function, and are used to describe the behavior of physical systems and processes that change over time.

- General Solution of a Differential Equation:

The general solution of a differential equation is the most general form of the solution that satisfies the equation. It includes all possible solutions that can be obtained by varying the constants of integration. For example, consider the following differential equation:

dy/dx = x

The general solution of this differential equation can be found by integrating both sides with respect to x:

y = x^{2}/2 + C

where C is a constant of integration. The general solution includes all possible solutions of the differential equation that can be obtained by varying the value of C.

- Particular Solution of a Differential Equation:

A particular solution of a differential equation is a specific solution that satisfies the equation for a given set of initial or boundary conditions. For example, consider the following differential equation:

dy/dx + y = e^{x}

To find a particular solution, we need to solve the differential equation with a specific initial or boundary condition. For example, if we have the initial condition y(0) = 1, we can use this to find a particular solution:

dy/dx + y = e^{x }y(0) = 1

The solution of the differential equation with this initial condition is:

y = e^{x} – 1

This is a particular solution that satisfies the differential equation for the given initial condition.

- Relationship between General and Particular Solutions:

The general solution of a differential equation includes all possible solutions that can be obtained by varying the constants of integration. A particular solution is a specific solution that satisfies the equation for a given set of initial or boundary conditions. Any particular solution can be obtained from the general solution by specifying the appropriate values of the constants of integration.

For example, consider the differential equation dy/dx = x. The general solution is y = x^{2}/2 + C, where C is a constant of integration. If we have the initial condition y(0) = 1, we can find the particular solution by substituting x = 0 and y = 1 into the general solution:

1 = 0^{2}/2 + C C = 1

Therefore, the particular solution with the given initial condition is y = x^{2}/2 + 1.

In conclusion, the general solution of a differential equation includes all possible solutions that can be obtained by varying the constants of integration. A particular solution is a specific solution that satisfies the equation for a given set of initial or boundary conditions. Any particular solution can be obtained from the general solution by specifying the appropriate values of the constants of integration. The general and particular solutions are both important in solving differential equations and understanding the behavior of physical systems and processes that change over time.

Explain the variable Separable Method to find the Solutions of Differential Equations

The Variable Separable Method is a powerful technique used to solve first-order ordinary differential equations. This method involves separating the variables of a differential equation, that is, expressing the dependent and independent variables on different sides of the equation. Once the variables are separated, the resulting equation can be easily integrated to obtain the solution.

Example:

Let’s consider the following first-order differential equation:

dy/dx = x^{2} y

We can solve this equation using the Variable Separable Method as follows:

Step 1: Separate the variables

dy/y = x^{2} dx

Step 2: Integrate both sides

∫ dy/y = dx

ln|y| = (1/3)x^{3} + C

Step 3: Solve for y

|y| = e^{(1/3 x³ + C)}

y = ± e^{(1/3 x³ + C)}

This is the general solution to the differential equation. Note that there is a constant of integration C that needs to be determined based on any initial or boundary conditions.

Explanation of the Method:

The Variable Separable Method involves rearranging a differential equation into a form where the dependent variable and independent variable are separated on different sides of the equation. Once the variables are separated, they can be integrated using the appropriate integration techniques.

The method can be summarised in the following steps:

- Separate the variables by moving all the terms that contain the dependent variable to one side and all the terms that contain the independent variable to the other side.
- Integrate both sides of the resulting equation with respect to their respective variables.
- Add a constant of integration, which is usually denoted by C. This constant is determined by any initial or boundary conditions.
- Solve for the dependent variable to obtain the general solution.

The Variable Separable Method can be used to solve a wide range of first-order differential equations. It is particularly useful when the differential equation is of the form dy/dx = f(x)g(y), where f(x) and g(y) are functions of the independent and dependent variables, respectively.

Conclusion:

In summary, the Variable Separable Method is a powerful technique for solving first-order ordinary differential equations. This method involves separating the variables of a differential equation and then integrating both sides of the equation to obtain the solution. The method is particularly useful when the differential equation is of the form dy/dx = f(x)g(y).

Describe the Homogeneous Differential Equations

A homogeneous differential equation is a type of ordinary differential equation where all the terms can be expressed as a function of the dependent variable and its derivatives. These equations are important in many areas of science and engineering, as they arise in the modeling of physical phenomena such as fluid flow, heat transfer, and electromagnetic waves.

Example:

Let’s consider the following second-order homogeneous differential equation:

y” + 4y’ + 4y = 0

This equation is homogeneous because all the terms can be expressed as a function of y and its derivatives. To solve this equation, we first assume a solution of the form y = e^{(rt)}, where r is a constant. Substituting this into the equation, we obtain:

r^{2 }e^{(rt)} + 4r e^{(rt)} + 4 e^{(rt)} = 0

Dividing through by e^{(rt)}, we get:

r^{2 }+ 4r + 4 = 0

This is a quadratic equation with solutions r = -2, -2. Thus, the general solution to the differential equation is:

y = c_{1} e^{(-2t) }+ c2 t e^{(-2t})

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

Explanation of Homogeneous Differential Equations:

A homogeneous differential equation is of the form:

F(y, y’, y”, …) = 0

where F is a function of the dependent variable y and its derivatives up to some order. The equation is said to be homogeneous if F is a homogeneous function of degree n, which means that:

F(ty, t^{1}y’, t^{2}y”, …) = t^{n} F(y, y’, y”, …)

for any scalar t. In other words, if we replace y with ty, y’ with t^{1}y’, y” with t^{2}y”, and so on, the equation remains unchanged.

Homogeneous differential equations have a special property: if y(t) is a solution to the equation, then so is k*y(t), where k is a constant. This is known as the scaling property of homogeneous equations. This property can be used to simplify the equation by dividing both sides by y^{n}, where n is the highest order of derivative appearing in the equation. This results in a new equation that only involves the derivatives of y in a particular combination, making it easier to solve.

Homogeneous differential equations can be solved using various methods, including the substitution method, the power series method, and the Laplace transform method.

Conclusion:

In summary, homogeneous differential equations are a type of ordinary differential equation where all the terms can be expressed as a function of the dependent variable and its derivatives. These equations have a scaling property that can be used to simplify them, and they can be solved using a variety of methods. Homogeneous differential equations are important in many areas of science and engineering, as they arise in the modeling of physical phenomena.

Find the Solutions of Homogeneous Differential Equations

A homogeneous differential equation is a differential equation in which all the terms can be expressed as a function of the variables and their derivatives. Homogeneous differential equations have a special property that allows the solution to be expressed in terms of a single parameter. The solution to a homogeneous differential equation is found by solving a characteristic equation, which is obtained by assuming that the solution can be expressed in the form of an exponential function.

Consider the following homogeneous differential equation:

y” – 4y’ + 4y = 0

To solve this equation, we assume that the solution has the form:

y = e^{(rt)}

where r is a constant. We substitute this into the differential equation:

y” – 4y’ + 4y = 0

(e^{(rt)})” – 4(e^{(rt)})’ + 4e^{(rt)} = 0

r^{2}e^{(rt)} – 4re^{(rt)} + 4e^{(rt)} = 0

(r^{2} – 4r + 4)e^{(rt)} = 0

The exponential term cannot be zero, so we have:

r^{2} – 4r + 4 = 0

This is a quadratic equation which can be factored as:

(r – 2)^{2} = 0

Therefore, the roots of the characteristic equation are:

r = 2 (multiplicity 2)

This means that the general solution to the differential equation is:

y = c_{1}e^{(2t)} + c^{2}te^{(2t)}

where c_{1} and c_{2} are constants.

So, the solutions to the given homogeneous differential equation are of the form y = c_{1}e^{(2t)} + c_{2}te^{(2t)}.

Describe the Linear Differential Equations

A linear differential equation is a type of differential equation that can be expressed in the form:

a_{n}(x) y^{n} + a_{{n-1}}(x) y^{{n-1}} + … + a_{1}(x) y’ + a_{0}(x) y = f(x)where y is the dependent variable, x is the independent variable, a_{n}(x), a_{{n-1}}(x), …, a_{1}(x), a_{0}(x) are functions of x, and f(x) is a given function of x. The degree of the differential equation is the highest power of y that appears in the equation, which is n.

An example of a linear differential equation is:

y” + 2y’ – 3y = x^{2}

This is a second-order linear differential equation because the highest power of y that appears is y” (second derivative of y).

To solve this equation, we first find the characteristic equation:

r^{2} + 2r – 3 = 0

which can be factored as:

(r + 3)(r – 1) = 0

So the roots are r = -3 and r = 1.

The general solution to the homogeneous equation (y” + 2y’ – 3y = 0) is therefore:

y_{h} = c1 e^{(-3x)} + c2 e^{(x})

To find a particular solution to the non-homogeneous equation (y” + 2y’ – 3y = x^{2}), we use the method of undetermined coefficients. Since the right-hnd side is a polynomial of degree 2, we assume a particular solution of the form:

y_{p} = ax^{2} + bx + c

Taking the derivatives of y_{p}, we get:

y_{p’} = 2ax + b

y_{p}” = 2a

Substituting these into the differential equation, we get:

2a + 4ax + 2b – 3ax^{2 }– 3bx – 3c = x^{2}

Matching coefficients of like terms on both sides, we get:

-3a = 1 (coefficients of x^{2})

4a – 3b = 0 (coefficients of x)

2b – 3c = 0 (coefficients of constant)

Solving for a, b, and c, we get:

a = -1/3, b = -4/9, c = -4/27

So the particular solution to the non-homogeneous equation is:

y_{p} = (-1/3)x^{2} – (4/9)x – (4/27)

Therefore, the general solution to the non-homogeneous equation is:

y = y_{h} + y_{p} = c1 e^{(-3x)} + c2 e^{(x)} – (1/3)x^{2} – (4/9)x – (4/27)

This is the complete solution to the given linear differential equation.

Find the solutions of Linear Differential Equations

Consider the following first-order linear differential equation:

y’ + 3y = 4x

To solve this equation, we can use an integrating factor, which is a function that we multiply both sides of the equation by to make the left-hand side a derivative of a product. In this case, we choose the integrating factor to be:

I(x) = e^{(3x)}

Multiplying both sides of the equation by the integrating factor, we get:

e^{(3x)} y’ + 3e^{(3x)} y = 4xe^{(3x)}

We recognize the left-hand side as the product rule for the derivative of e^{(3x)} y, so we can write:

(e^{(3x)} y)’ = 4xe^{(3x)}

Integrating both sides with respect to x, we get:

e^{(3x)} y = ∫ 4xe^{(3x)} dx = (4/3)xe^{(3x) }– (4/9)e^{(3x)} + C

where C is the constant of integration.

Dividing both sides by e^{(3x)}, we get the general solution to the differential equation:

y = (4/3)x – (4/9) + Ce^{(-3x)}

where C is an arbitrary constant.

So, the solutions to the given linear differential equation are of the form y = (4/3)x – (4/9) + Ce^{(-3x)}.

Describe the Exact Differential Equations

An exact differential equation is a type of differential equation in which the left-hand side of the equation can be written as the total differential of a function, which is a function that can be expressed as the sum of its partial derivatives. In other words, an exact differential equation can be expressed in the form:

M(x, y) dx + N(x, y) dy = 0

where the partial derivatives of M and N with respect to y and x, respectively, are equal:

∂M/∂y = ∂N/∂x

If this condition is satisfied, then the equation is exact and can be solved by finding a function Φ(x, y) such that:

dΦ(x, y) = M(x, y) dx + N(x, y) dy

An example of an exact differential equation is:

(3x^{2 }+ 4y^{3}) dx + (12xy^{2}) dy = 0

To check if this equation is exact, we need to calculate the partial derivatives of M and N with respect to y and x, respectively:

∂M/∂y = 12y^{2}

∂N/∂x = 12y^{2}

Since ∂M/∂y = ∂N/∂x, the equation is exact.

To find a function Φ(x, y), we integrate M(x, y) with respect to x, treating y as a constant:

Φ(x, y) = ∫ (3x^{2} + 4y^{3}) dx = x^{3} + 4y^{3}x + C(y)

where C(y) is an arbitrary function of y.

Taking the partial derivative of Φ(x, y) with respect to y and equating it to N(x, y), we get:

∂Φ/∂y = 12y^{2}x + C'(y) = 12xy^{2}

Comparing the coefficients of x and y^{2}, we get:

C'(y) = 0

This means that C(y) is a constant function of y, so we can write:

C(y) = k

where k is a constant.

Substituting this back into Φ(x, y), we get the general solution to the exact differential equation:

x^{3} + 4y^{3}x + k = 0

This is the complete solution to the given exact differential equation.

Find the solutions of Exact Differential Equations

Consider the following exact differential equation:

(2x + 3y) dx + (3x + 5y^{2}) dy = 0

To solve this equation, we need to check if it satisfies the condition for exactness, which is:

∂(2x + 3y)/∂y = ∂(3x + 5y^{2})/∂x

Simplifying this condition, we get:

3 = 3

Since the condition is satisfied, we know that there exists a function Φ(x, y) such that:

dΦ(x, y) = (2x + 3y) dx + (3x + 5y^{2}) dy

To find Φ(x, y), we integrate M(x, y) with respect to x, treating y as a constant:

Φ(x, y) = ∫ (2x + 3y) dx = x^{2} + 3xy + C(y)

where C(y) is an arbitrary function of y.

Taking the partial derivative of Φ(x, y) with respect to y and equating it to N(x, y), we get:

∂Φ/∂y = 3x + C'(y) = 3x + 5y^{2}

Comparing the coefficients of x and y^{2}, we get:

C'(y) = 5y^{2}

Integrating C'(y) with respect to y, we get:

C(y) = (5/3)y^{3} + k

where k is a constant of integration.

Substituting C(y) back into Φ(x, y), we get the general solution to the exact differential equation:

x^{2} + 3xy + (5/3)y^{3} + k = 0

This is the complete solution to the given exact differential equation.

Describe the Differential Equation reducible to the variable Separable form

A differential equation is said to be reducible to the variable separable form if it can be transformed into an equivalent equation of the form:

dy/dx = f(x)g(y)

where f(x) and g(y) are functions of x and y respectively.

To reduce a given differential equation to the variable separable form, we need to rearrange the equation in such a way that we can separate the variables x and y, and then integrate both sides with respect to their respective variables. Here’s an example of a differential equation that can be reduced to the variable separable form:

y’ + xy = x

To reduce this equation to the variable separable form, we can first move the xy term to the right-hand side:

y’ = x – xy

Next, we can factor out y on the right-hand side:

y’ = x(1 – y)

Now, we can separate the variables x and y:

dy/(1 – y) = x dx

We can integrate both sides with respect to their respective variables:

∫ dy/(1 – y) = ∫ x dx

ln|1 – y| = x^{2}/2 + C

where C is the constant of integration.

Finally, we can solve for y:

|1 – y| = e^{(x2/2 + C)}

y = 1 – Ce^{ (x 2/2)}

where C is a constant.

Therefore, the solution to the given differential equation is:

y = 1 – Ce^{ (x 2/2)}

Describe the Differential Equation reducible to the Homogeneous Form

A differential equation is said to be reducible to homogeneous form if it can be transformed into an equivalent equation of the form:

dy/dx = f(y/x)

where f is a function of y/x only.

To reduce a given differential equation to the homogeneous form, we need to make a substitution of the form y = vx, where v is a new function of x. Using the chain rule, we can express dy/dx in terms of v and x:

dy/dx = v + x dv/dx

Substituting y = vx and dy/dx = v + x dv/dx into the original differential equation, we get:

v + x dv/dx = f(v)

Rearranging this equation, we get:

dv/(f(v) – v) = dx/x

Now, we can integrate both sides with respect to their respective variables:

∫ dv/(f(v) – v) = ∫ dx/x

This integration is not always possible in a closed form, but in some cases, it can be done by using integration by partial fractions or other integration techniques.

After integrating both sides, we get:

ln|f(v) – v| = ln|x| + C

where C is the constant of integration.

Solving for v, we get:

v = ±√(Cx + v^{2})

where we take the positive or negative square root depending on the sign of f(v) – v.

Finally, we can substitute y = vx back into the equation to obtain the general solution to the original differential equation.

Here’s an example of a differential equation that can be reduced to the homogeneous form:

xy’ = y + x^{2}

To reduce this equation to the homogeneous form, we can make the substitution y = vx, which gives:

x(dx/dt)v + x(v + xdv/dt) = vx + x^{2}

Simplifying this equation, we get:

x^{2} dv/dx + xv = v + x

Now, we can divide both sides by x^{ 2}:

dv/dx + v/x = (1 + x^{ 2})/x^{ 2}

This equation is now in the form of a first-order linear differential equation, which can be solved using the integrating factor method or any other suitable method.

After solving for v, we can substitute y = vx back into the equation to obtain the general solution to the original differential equation.

Find the solutions of Variable Separable and Homogeneous Differential Equations

Examples of solving variable separable and homogeneous differential equations.

- Variable Separable Differential Equation:

Consider the differential equation:

dy/dx = 2x/(y^{ 2+1})

This is a variable separable differential equation, as we can write it in the form:

(y^{2+1})dy = 2xdx

Integrating both sides, we get:

∫(y^{2+1})dy = ∫2xdx

Solving the integrals, we get:

y^{3}/3 + y = x^{2} + C

where C is the constant of integration.

Therefore, the general solution of the given differential equation is:

y^{3} + 3y = 3x^{2} + C

- Homogeneous Differential Equation:

Consider the differential equation:

x^{2}dy/dx = y(x^{2} – y^{2})

This is a homogeneous differential equation, as we can write it in the form:

dy/dx = (y/x2)(x^{4} – y^{2})

To solve this equation, we can make the substitution y = vx, which gives:

dy/dx + v = xdv/dx

Substituting this in the original equation, we get:

xdv/dx = v(x^{2} – v^{2})

Dividing both sides by v(x^{2} – v^{2}), we get:

dv/v = (x dx)/x^{3}

Integrating both sides, we get:

ln|v| = -1/(2x^{2}) + C

where C is the constant of integration.

Solving for v, we get:

v = ±(Cx^{(-1/2)})

where C is a constant.

Substituting y = vx, we get:

y = ±x√(Cx^{(-1/2)})

Therefore, the general solution of the given differential equation is:

y = ±Cx^{(-1/2)}x

where C is the constant of integration.

Describe the Differential Equation reducible to the Exact Form

Exact differential equations are a special type of first-order differential equations where the total differential of the equation can be expressed as the product of two functions of the independent variable x and the dependent variable y. A differential equation is said to be exact if it can be written in the form:

M(x, y)dx + N(x, y)dy = 0

where M(x, y) and N(x, y) are functions of both x and y, and the partial derivative of M with respect to y is equal to the partial derivative of N with respect to x, that is:

∂M/∂y = ∂N/∂x

If this condition is satisfied, then the equation is exact, and it can be solved by finding a function ϕ(x,y) such that the partial derivative of ϕ with respect to x is equal to M, and the partial derivative of ϕ with respect to y is equal to N. That is:

∂ϕ/∂x = M, ∂ϕ/∂y = N

The solution to the exact differential equation is then given by the equation:

ϕ(x,y) = C

where C is the constant of integration.

Example:

Consider the following differential equation:

(3x^{2}y + 2xy^{2})dx + (x^{3} + 3x^{2}y)dy = 0

To check if this equation is exact, we need to calculate the partial derivative of the first term with respect to y and the partial derivative of the second term with respect to x:

∂M/∂y = 3x^{2} + 4xy

∂N/∂x = 3x^{2 }+ 6xy

Since ∂M/∂y is not equal to ∂N/∂x, the equation is not exact.

To convert this equation into an exact form, we can multiply the whole equation by an integrating factor. The integrating factor is given by:

I(x) = e^{(∫(∂M/∂y – ∂N/∂x)}/N(x,y) dx)

where ∂M/∂y – ∂N/∂x = xy/(x^{3} + 3x^{2}y). Therefore, we get:

I(x) = e^{(∫xy/(x3 + 3x2y)} dx)

Solving this integral gives:

I(x) = e^{(ln|x| – ln|x+3y|)}

I(x) = |x|/|x+3y|

Now, we multiply the given differential equation by the integrating factor I(x):

(3x^{2}y + 2xy2)/|x+3y| dx + (x^{3} + 3x^{2}y)/|x+3y| dy = 0

After multiplying and simplifying, we find that the equation becomes exact. We can then find the solution by integrating both sides of the equation with respect to x, which gives:

x^{3}y + x^{2}y^{2} + f(y) = C

where f(y) is the constant of integration with respect to y.

Find the solutions of Exact Differential Equations

An exact differential equation is a type of first-order differential equation that can be written in the form:

M(x, y) dx + N(x, y) dy = 0

where M and N are functions of x and y. An exact differential equation can be solved by finding a function ϕ(x, y) such that:

dϕ/dx = M(x, y) and dϕ/dy = N(x, y)

The solution to the exact differential equation is then given by the equation:

ϕ(x, y) = C

where C is a constant of integration.

To solve an exact differential equation, we can use the following steps:

Step 1: Check if the equation is exact. To do this, we check whether:

∂M/∂y = ∂N/∂x

If this condition is satisfied, the equation is exact, and we can proceed to the next step. If not, we can attempt to make the equation exact by using an integrating factor.

Step 2: Find ϕ(x, y) by integrating M(x, y) with respect to x and integrating N(x, y) with respect to y. We can then equate the two expressions to find ϕ(x, y).

Step 3: Solve for the constant of integration C by substituting the initial conditions.

Example:

Consider the following differential equation:

(2xy^{3} + 3x^{2}y^{2})dx + (x^{3}y^{2} + 3xy^{2})dy = 0

Step 1: Check if the equation is exact. We have:

∂M/∂y = 6xy^{2} + 6xy^{2} = 12xy^{2}

∂N/∂x = 3x^{2}y^{2} + 3y^{2} = 3y^{2}(x^{2} + 1)

Since ∂M/∂y is not equal to ∂N/∂x, the equation is not exact.

Step 2: To make the equation exact, we need to find an integrating factor. The integrating factor can be found by dividing one side of the equation by M(x, y) and the other side by N(x, y). We then take the partial derivative of the quotient with respect to y and equate it to the partial derivative of the quotient with respect to x. This gives:

(3x^{2}y^{2} + 2xy^{3})/(x^{3}y^{2} + 3xy^{2}) = g(y)/h(x)

where g(y) and h(x) are functions of y and x, respectively. We can then find the integrating factor by taking the exponential of the integral of the function g(y)/h(x) with respect to y. This gives:

I(x, y) = e ^{(∫g(y)/h(x) dy)}

Solving the integral and simplifying, we get:

I(x, y) = (x^{3}y^{2} + 3xy^{2})^{(-2)}

Multiplying the differential equation by the integrating factor, we get:

(2xy^{3} + 3^{x2}y^{2})/(x^{3}y^{2} + 3xy^{2})^{2} dx + (x^{3}y^{2} + 3xy^{2})^{(-1)} dy = 0

The equation is now exact, and we can find ϕ(x, y) by integrating M(x, y) with respect to x and N(x, y) with respect to y:

ϕ(x, y) = x

Describe the Differential Equation reducible to the Linear form (Bernoulli Equation)

A Bernoulli differential equation is a type of nonlinear differential equation that can be transformed into a linear differential equation through a suitable substitution. It has the following form:

dy/dx + P(x)y = Q(x)y^{n}

where P(x) and Q(x) are functions of x, y is the dependent variable and n is a constant other than 1.

This type of equation can be reduced to a linear differential equation by making the substitution:

z = y^{(1-n)} / (1-n)

Taking the derivative of z with respect to x and substituting in the original equation, we get:

dz/dx + (1-n)P(x)z = (1-n)Q(x)

This equation is linear and can be solved using methods such as integrating factor or separation of variables.

Example:

Solve the Bernoulli differential equation:

y’ – y/x = 2xy^{2}

We notice that the equation is in the Bernoulli form, with n = 2. To transform it into a linear equation, we make the substitution:

z = y^{(1-2)}/(1-2) = -y^{-1}

Taking the derivative of z with respect to x, we get:

dz/dx = (1/n)y^{(1-n)}y’

Substituting y’ and z in the original equation, we get:

(1/(-z)) * dz/dx – (-z)/x = 2x(-z)^{-2}

Simplifying, we get:

dz/dx + (1/x)z = -2

This is a linear differential equation that can be solved using integrating factor or separation of variables. The solution is:

z = Ce^{(-ln(x2)) }– 2x

Substituting back y and simplifying, we get the general solution:

y = (C/x^{2}) – 2x^{2}

Find the solutions of Linear Differential Equations

A linear differential equation is an equation of the form:

a_{n}(x)y^{ (n)} + a_{n-1}(x)y^{ (n-1)} + … + a_{1}(x)y’ + a_{0}(x)y = f(x)

where a_{i}(x) and f(x) are functions of x and y is the dependent variable. If f(x) is zero, the equation is called homogeneous, otherwise it is non-homogeneous.

To solve a linear differential equation, the general approach is to find a solution to the associated homogeneous equation, called the complementary function, and then add a particular solution to the non-homogeneous equation. The sum of these two solutions gives the general solution to the original equation.

To find the complementary function, we assume a solution of the form:

y_{c} = e^{ (rx)}

where r is a constant. Substituting this into the homogeneous equation, we get the characteristic equation:

a_{n}(r)^{ n} + a_{n-1}(r)^{ (n-1)} + … + a_{1}(r)r + a_{0}(r) = 0

Solving this equation gives the roots r_{1}, r_{2}, …, r_{n}. The general solution to the complementary function is then:

y_{c} = C_{1}e^{ (r1x)} + C_{2}e^{ (r2x)} + … + C_{n}e^{ (rnx)}

where C_{1}, C_{2}, …, C_{n} are constants determined by the initial or boundary conditions.

To find a particular solution to the non-homogeneous equation, we use a method called the method of undetermined coefficients or variation of parameters, depending on the form of f(x).

Example:

Solve the following linear differential equation:

y” + 3y’ + 2y = 2x + 3

The associated homogeneous equation is:

y” + 3y’ + 2y = 0

The characteristic equation is:

r^{2 }+ 3r + 2 = 0

The roots are:

r_{1} = -1 and K_{2} = -2

The complementary function is:

y_{c} = C_{1}e^{ (-x)} + C_{2}e^{ (-2x)}

To find a particular solution, we notice that the right-hand side of the non-homogeneous equation is of the form ax + b, which suggests a particular solution of the form:

y_{p} = Ax + B

Substituting this into the non-homogeneous equation, we get:

2A = 2x + 3

This gives A = 1 and B = 1. Therefore, the particular solution is:

y_{p} = x + 1

The general solution is then:

y = y_{c} + y_{p}

y = C_{1}e^{ (-x)} + C_{2}e^{ (-2x)} + x + 1

The constants C_{1} and C_{2} are determined by the initial or boundary conditions.

Describe Orthogonal Trajectory of Cartesian Curve

In mathematics, the orthogonal trajectory of a cartesian curve is another curve which intersects the first curve at a right angle. The term orthogonal trajectory comes from the fact that the two curves are perpendicular or orthogonal at every point where they intersect.

The orthogonal trajectory of a given cartesian curve can be found by solving a differential equation. Specifically, if the cartesian curve is given by the equation f(x, y) = 0, then the orthogonal trajectory can be found by solving the differential equation

y’ = -f_{y}(x,y)/f_{x}(x,y),

where f_{x} and f_{y }are the partial derivatives of f with respect to x and y, respectively.

For example, consider the cartesian curve given by the equation x^{2} + y^{2} = 1, which is a circle centered at the origin with radius 1. To find its orthogonal trajectory, we need to solve the differential equation

y’ = -(2x)/(2y) = -x/y.

To do this, we can separate variables and integrate:

y dy = -x dx

Integrating both sides, we get

y^{2}/2 + x^{2}/2 = C

where C is a constant. This is the equation of the orthogonal trajectory, which is a family of hyperbolas centered at the origin. Each hyperbola in the family intersects the circle at a right angle.

The concept of orthogonal trajectories has applications in physics, engineering, and other areas of science. For example, in electricity and magnetism, the lines of electric and magnetic fields are orthogonal to each other, and the orthogonal trajectories of these lines can be used to study the behavior of electromagnetic waves.

Describe Orthogonal Trajectory of Polar Curve

In mathematics, the orthogonal trajectory of a polar curve is a curve that intersects the given polar curve at right angles. The term orthogonal means “perpendicular” or “at right angles”. The method of finding the orthogonal trajectory of a polar curve is the same as that of the cartesian curve, except that the polar coordinates are used instead of the cartesian coordinates.

To find the orthogonal trajectory of a polar curve, we need to express the given polar curve in terms of the cartesian coordinates x and y, and then apply the method of finding the orthogonal trajectory of a cartesian curve.

For example, consider the polar curve r = a cos(θ), where a is a positive constant. To find its orthogonal trajectory, we first need to express it in terms of the cartesian coordinates x and y:

x = r cos(θ) = a cos²(θ)

y = r sin(θ) = a cos(θ) sin(θ)

Then, we find the partial derivatives of x and y with respect to θ:

dx/dθ = -2a cos(θ) sin(θ)

dy/dθ = a (cos²(θ) – sin²(θ))

Next, we form the differential equation for the orthogonal trajectory by taking the negative reciprocal of the derivative of y with respect to x:

dy/dx = -dx/dy = (2a cos(θ) sin(θ)) / (a(cos²(θ) – sin²(θ)))

This simplifies to:

dy/dx = -2 cot(θ)

We can now solve this differential equation to find the orthogonal trajectory of the given polar curve. The solution is:

y = C x² – a²

where C is a constant. This is the equation of a family of parabolas, which are the orthogonal trajectories of the given polar curve.

The concept of orthogonal trajectories has applications in physics, engineering, and other areas of science. For example, in electromagnetic theory, the lines of electric and magnetic fields are orthogonal to each other, and the orthogonal trajectories of these lines can be used to study the behavior of electromagnetic waves.

Calculate the orthogonal Trajectory of Polar Curve

To calculate the orthogonal trajectory of a polar curve, we follow the same steps as in finding the orthogonal trajectory of a cartesian curve. First, we express the polar curve in terms of the cartesian coordinates x and y, and then find the differential equation for the orthogonal trajectory by taking the negative reciprocal of the derivative of y with respect to x. Finally, we solve this differential equation to obtain the equation of the orthogonal trajectory.

As an example, let’s consider the polar curve r = a sin(2θ), where a is a positive constant. To find its orthogonal trajectory, we need to express it in terms of the cartesian coordinates x and y. Using the conversion formulas for polar coordinates, we get:

x = r cos(θ) = a sin(2θ) cos(θ) = (a/2) sin(3θ)

y = r sin(θ) = a sin(2θ) sin(θ) = (a/2) cos(θ)

Next, we find the derivative of y with respect to x:

dy/dx = (dy/dθ) / (dx/dθ) = (-a cos(2θ) cos(θ) – a sin(2θ) sin(θ)) / (a cos(2θ) sin(θ) – a sin(2θ) cos(θ))

= -(cos(2θ) cos(θ) + sin(2θ) sin(θ)) / (cos(2θ) sin(θ) – sin(2θ) cos(θ))

= -cot(3θ)

Taking the negative reciprocal of this derivative, we get the differential equation for the orthogonal trajectory:

dy/dx = 1/cot(3θ) = tan(3θ)

Now we can solve this differential equation to find the equation of the orthogonal trajectory. We integrate both sides with respect to x, treating y as a function of θ:

∫dy = ∫tan(3θ) dx

y = (-1/3) ln|cos(3θ)| + C

where C is a constant of integration. This is the equation of the family of curves that are orthogonal to the given polar curve. The curves are logarithmic spirals that wind around the origin, with the angle of rotation increasing by 60 degrees for each complete turn.

In summary, to calculate the orthogonal trajectory of a polar curve, we need to express it in terms of the cartesian coordinates, find the derivative of y with respect to x, take the negative reciprocal of this derivative to get the differential equation for the orthogonal trajectory, and then solve this differential equation to obtain the equation of the family of curves that are orthogonal to the original curve.

Describe differential equation solvable for p(dy/dx)

In differential calculus, a differential equation is a mathematical equation that relates an unknown function with its derivatives. It expresses the relationship between the function and its derivatives in terms of one or more independent variables. Some differential equations are solvable for the derivative in the form of p(dy/dx), where p is a function of the independent variable. These types of equations are known as “differential equations solvable for p(dy/dx).”

The general form of a differential equation solvable for p(dy/dx) is:

f(y)dy/dx + g(x) = p(dy/dx)

where f(y), g(x), and p(dy/dx) are functions of their respective variables.

To solve this type of differential equation, we can first rearrange it to isolate p(dy/dx) on one side of the equation:

p(dy/dx) = (f(y)dy/dx + g(x)) / 1

Next, we can multiply both sides by dx/dy to get:

p(dy/dx) dx/dy = f(y) + g(x)dy/dx

Then, we can rearrange the terms to get:

(dy/dx) (p(dy/dx) – g(x)) = f(y)

Finally, we can solve for dy/dx to get:

dy/dx = f(y) / (p(dy/dx) – g(x))

As an example, let’s consider the differential equation:

(2y + x)dy/dx + 2y^{2} = x

This equation is solvable for p(dy/dx), where p(dy/dx) = 2y + x.

To solve for the orthogonal trajectory, we can first rearrange the equation to isolate p(dy/dx):

p(dy/dx) = (x – 2y^{2}) / (2y + x)

Next, we can multiply both sides by dx/dy to get:

p(dy/dx) dx/dy = (x – 2y^{2}) / (2y + x)

Then, we can rearrange the terms to get:

(dy/dx) (2y + x) = (x – 2y^{2}) / dx/dy

Finally, we can solve for dy/dx to get:

dy/dx = (x – 2y^{2}) / (2y + x)

This is the equation of the orthogonal trajectory of the curve represented by the original differential equation.

Describe differential equation solvable for y

In differential equations, there are some equations that can be solved for y explicitly. This means that the equation can be written in terms of y only without involving the derivative of y. In particular, a differential equation of the form:

y’ = f(x,y)

is solvable for y if we can find a function F(x,y) such that:

F(x,y) = C

where C is some constant. This means that we can implicitly define y as a function of x by the equation:

F(x,y) = C

To solve for y explicitly, we can solve for y in terms of x and C. The general solution is then given by the family of curves defined by this equation.

For example, consider the differential equation:

y’ = 2xy

We can rewrite this as:

dy/dx = 2xy

Dividing both sides by y, we get:

(1/y) dy/dx = 2x

Integrating both sides with respect to x, we get:

ln|y| = x^{2} + C

where C is an arbitrary constant of integration. Exponentiating both sides, we get:

|y| = e^{(x2+C)} = Ce^{(x2)}

where we have absorbed the positive/negative sign into the constant C. Therefore, the general solution of the differential equation is:

y = Ce^{(x2)}

This is an example of a differential equation solvable for y explicitly.

Describe differential equation solvable for X

A differential equation is an equation that relates a function to its derivatives. A differential equation solvable for X is one where the independent variable X can be isolated on one side of the equation and the dependent variable and its derivatives are on the other side.

One example of a differential equation solvable for X is:

dy/dx = f(x)

This equation relates the derivative of the dependent variable y with respect to the independent variable x to a function f(x). We can solve this equation for X by integrating both sides with respect to x:

∫ dy/dx dx = ∫ f(x) dx

y = ∫ f(x) dx + C

where C is the constant of integration. This gives us the general solution for y in terms of x. To find the particular solution for a given initial condition, we need to use the initial value of y and solve for C.

Another example of a differential equation solvable for X is:

d²y/dx² + p(x) dy/dx + q(x) y = f(x)

This is a second-order linear homogeneous differential equation with variable coefficients, where p(x) and q(x) are functions of x and f(x) is a given function. To solve this equation for X, we can use various methods such as the method of undetermined coefficients, variation of parameters, or the Laplace transform method. The solution will involve finding the roots of the characteristic equation associated with the differential equation and using them to construct the general solution.

Explain Clairaut’s Equation

Clairaut’s equation is a type of first-order differential equation that has the form:

y = xy’ + f(y’)

where y’ denotes the derivative of y with respect to x and f is an arbitrary function of y’. This equation is named after Alexis Clairaut, an 18th-century French mathematician.

Clairaut’s equation is also sometimes written as:

dy/dx = x + f'(dy/dx)

where f’ denotes the derivative of f with respect to y’. This form makes it easier to see that the equation is of the form y = mx + f(m), where m = dy/dx.

One interesting feature of Clairaut’s equation is that it is not necessary to specify an initial condition in order to find a solution. Instead, the equation has a family of solutions, each of which corresponds to a different value of the arbitrary constant in the solution.

To illustrate this, consider the example of the Clairaut’s equation:

y = xy’ – y’^{2}/2

To find a general solution, we first solve for y’:

y’ = (y + c)/x

where c is an arbitrary constant of integration. Next, we substitute this expression for y’ back into the original equation:

y = x(y + c)/x – (y + c)^{2}/(2x^{2})

Simplifying, we obtain:

y^{2} = 2cx

This equation represents a family of parabolas in the xy-plane, each of which corresponds to a different value of the constant c. Therefore, there are infinitely many solutions to the original differential equation, each of which is a member of this family of parabolas.

Clairaut’s equation has many applications in physics and engineering, where it is used to model a wide range of phenomena, including fluid mechanics, elasticity, and quantum mechanics.