Differential Equations - Linear Differential Equation & It'S Solution

Differential Equations - Linear Differential Equation & its Solution

In this lecture, we will learn about linear differential equations and how to find their solutions.
A linear differential equation can be written in the form: $$\frac{{dy}}{{dx}} + P(x)y = Q(x)$$
Here, the functions P(x) and Q(x) are given.
The goal is to find the function y(x) that satisfies this equation.
Let’s consider an example to understand this better.

Find the solution to the differential equation: $$\frac{{dy}}{{dx}} + 2xy = x$$
Solution:
To solve this equation, we need to find an integrating factor, which is given by the formula: $$I(x) = e^{\int P(x) dx}$$
In our case, P(x) = 2x, so the integrating factor becomes: $$I(x) = e^{\int 2x dx} = e^{x^2}$$
Multiply the given differential equation by the integrating factor: $$e^{x^2} \frac{{dy}}{{dx}} + 2xe^{x^2}y = xe^{x^2}$$
Notice that the left-hand side can be simplified using the product rule of differentiation. Can you give it a try?

Find the solution to the differential equation: $$\frac{{dy}}{{dx}} - \frac{{2x}}{{x^2 + 1}}y = \cos(x)$$
Solution:
To solve this equation, we need to find an integrating factor, which is given by the formula: $$I(x) = e^{\int P(x) dx}$$
In our case, P(x) = -\frac{{2x}}{{x^2 + 1}}, so the integrating factor becomes: $$I(x) = e^{\int -\frac{{2x}}{{x^2 + 1}} dx}$$
This integral can be evaluated using a substitution. Can you give it a try? (10)

Multiply the given differential equation by the integrating factor: $$e^{\int -\frac{{2x}}{{x^2 + 1}} dx} \frac{{dy}}{{dx}} - \frac{{2x}}{{x^2 + 1}}e^{\int -\frac{{2x}}{{x^2 + 1}} dx}y = \cos(x)e^{\int -\frac{{2x}}{{x^2 + 1}} dx}$$
Simplify the left-hand side using the product rule of differentiation: $$\frac{{d}}{{dx}}\left(ye^{\int -\frac{{2x}}{{x^2 + 1}} dx}\right) = \cos(x)e^{\int -\frac{{2x}}{{x^2 + 1}} dx}$$

Integrate both sides of the equation: $$\int \frac{{d}}{{dx}}\left(ye^{\int -\frac{{2x}}{{x^2 + 1}} dx}\right) dx= \int \cos(x)e^{\int -\frac{{2x}}{{x^2 + 1}} dx}dx$$
Apply the fundamental theorem of calculus to evaluate the integrals on both sides.

Solve the integral on the left-hand side: $$ye^{\int -\frac{{2x}}{{x^2 + 1}} dx} = \int \cos(x)e^{\int -\frac{{2x}}{{x^2 + 1}} dx}dx + C$$
Here, C is the constant of integration.

Divide both sides of the equation by the integrating factor: $$y = e^{-\int -\frac{{2x}}{{x^2 + 1}} dx} \left(\int \cos(x)e^{\int -\frac{{2x}}{{x^2 + 1}} dx}dx + C\right)$$
Simplify the exponential term using the properties of exponents.

Evaluate the integral on the right-hand side: $$y = e^{-\int -\frac{{2x}}{{x^2 + 1}} dx} + \sin(x) + C$$
Substitute the value of the integral back into the equation.

Now, let’s consider an initial value problem to find the particular solution to the differential equation.
An initial value problem involves finding the solution that satisfies certain conditions at a specific point.
For example, let’s say we have the following initial condition: $$y(0) = 2$$
We can substitute this value into the general solution to find the particular solution.

Substituting the initial condition into the general solution: $$2 = e^{-\int -\frac{{2(0)}}{{(0)^2 + 1}} dx} + \sin(0) + C$$ $$2 = e^0 + 0 + C$$ $$2 = 1 + C$$

Solving for C: $$C = 2 - 1 = 1$$
Therefore, the particular solution to the given initial value problem is: $$y = e^{-\int -\frac{{2x}}{{x^2 + 1}} dx} + \sin(x) + 1$$

In summary, we learned about linear differential equations and how to find their solutions.
We used the concept of an integrating factor to simplify the equation and solve for the unknown function.
By integrating both sides of the equation and applying the fundamental theorem of calculus, we found the general solution.
We also discussed how to solve initial value problems and find the particular solution that satisfies certain conditions.
Differential equations are an important topic in mathematics and have wide applications in various fields.

Example 3:
Find the solution to the differential equation: $$\frac{{dy}}{{dx}} + \frac{{2y}}{{x}} = 3x$$
Solution:
To solve this equation, we need to find an integrating factor, which is given by the formula: $$I(x) = e^{\int P(x) dx}$$
In our case, P(x) = \frac{{2}}{{x}}, so the integrating factor becomes: $$I(x) = e^{\int \frac{{2}}{{x}} dx}$$
Evaluate the integral to find the value of the integrating factor.

Multiply the given differential equation by the integrating factor: $$e^{\int \frac{{2}}{{x}} dx}\frac{{dy}}{{dx}} + \frac{{2}}{{x}}e^{\int \frac{{2}}{{x}} dx}y = 3xe^{\int \frac{{2}}{{x}} dx}$$
Simplify the left-hand side using the product rule of differentiation.

Integrate both sides of the equation: $$\int \frac{{d}}{{dx}}\left(ye^{\int \frac{{2}}{{x}} dx}\right) dx= \int 3xe^{\int \frac{{2}}{{x}} dx}dx$$
Apply the fundamental theorem of calculus to evaluate the integrals.

Solve the integral on the left-hand side: $$ye^{\int \frac{{2}}{{x}} dx} = \int 3xe^{\int \frac{{2}}{{x}} dx}dx + C$$
Here, C is the constant of integration.

Divide both sides of the equation by the integrating factor: $$y = \frac{{1}}{{e^{\int \frac{{2}}{{x}} dx}}} \left(\int 3xe^{\int \frac{{2}}{{x}} dx}dx + C\right)$$
Simplify the exponential term.

Evaluate the integral on the right-hand side: $$y = \frac{{1}}{{e^{\int \frac{{2}}{{x}} dx}}} \left(\int 3x \cdot x^2 dx + C\right)$$
Simplify the multiplication.

Solve the integral on the right-hand side: $$y = \frac{{1}}{{e^{\int \frac{{2}}{{x}} dx}}} \left(x^4 + C\right)$$
Substitute the value of the integral back into the equation.

Final Solution: $$y = \frac{{1}}{{e^{\int \frac{{2}}{{x}} dx}}} \left(x^4 + C\right)$$
This is the general solution to the given differential equation.

Now, let’s consider an initial value problem to find the particular solution to the differential equation.
An initial value problem involves finding the solution that satisfies certain conditions at a specific point.
For example, let’s say we have the following initial condition: $$y(1) = 5$$
We can substitute this value into the general solution to find the particular solution.

Substituting the initial condition into the general solution: $$5 = \frac{{1}}{{e^{\int \frac{{2}}{{x}} dx}}} \left(1^4 + C\right)$$ $$5 = \frac{{1}}{{e^{\int \frac{{2}}{{x}} dx}}} (1 + C)$$