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

4 min read Jun 07, 2024

Linear Differential Equation: Solving (x^2-1)dy/dx+2y=(x+1)^2

In this article, we will discuss how to solve a linear differential equation of the form (x^2-1)dy/dx+2y=(x+1)^2. This type of equation is commonly encountered in mathematics and physics, and its solution requires a good understanding of differential equations and integration.

The Given Equation

The given equation is:

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

Step 1: Rearrange the Equation

To solve this equation, we need to rearrange it in the standard form of a linear differential equation, which is:

dy/dx + P(x)y = Q(x)

Rearranging the given equation, we get:

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

Now, we can identify P(x) and Q(x) as:

P(x) = 2/(x^2-1) Q(x) = ((x+1)^2)/(x^2-1)

Step 2: Find the Integrating Factor

To solve the differential equation, we need to find the integrating factor, which is given by:

μ(x) = e^(∫P(x)dx)

In this case, we have:

μ(x) = e^(∫(2/(x^2-1))dx)

To evaluate this integral, we can use the substitution u = x^2-1, which gives:

μ(x) = e^(2∫(1/u)du) = e^(2ln|u|) = |u|^2 = (x^2-1)^2

Step 3: Solve the Differential Equation

Now, we can multiply both sides of the equation by the integrating factor μ(x):

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

Simplifying and rearranging, we get:

d((x^2-1)^2y)/dx = (x+1)^2(x^2-1)

Integrating both sides, we get:

(x^2-1)^2y = ∫(x+1)^2(x^2-1)dx + C

where C is the constant of integration.

Step 4: Simplify and Solve

Simplifying the integral, we get:

(x^2-1)^2y = (x^4 + 2x^3 - x^2 - 2x + 1)/3 + C

Dividing both sides by (x^2-1)^2, we get the final solution:

y = (x^4 + 2x^3 - x^2 - 2x + 1)/3(x^2-1)^2 + C/(x^2-1)^2

And that's it! We have solved the linear differential equation (x^2-1)dy/dx+2y=(x+1)^2.