Find the general solution of the given differential equation. x²y’ + x(x + 2)y = e^x

Solution:

The given differential equation is

x²y’ + x(x + 2)y = e^x

Let us multiply be ex on both sides

x²e^xy’ + (x² + 2x)e^xy = e^2x

We know that

(x²e^x)’ = 2xe^x + x²e^x = (x² + 2x)e^x

Let us collapse the LHS and write it as the derivative of the product.

(x²e^xy)’ = e^2x

$\\x^{2}e^{x}y=\int e^{2x}dx \\ \\x^{2}e^{x}y=\frac{e^{2x}}{2}+C \\ \\y=\frac{e^{x}}{2x^{2}}+\frac{C}{x^{2}e^{x}}$

Depending on the initial value (0, ∞) or (-∞, 0) would be the largest interval.

When x → ∞ $\frac{C}{x^{2}e^{x}}$ term approaches zero, so it is the only transient term.

Therefore, the general solution is $y=\frac{e^{x}}{2x^{2}}+\frac{C}{x^{2}e^{x}}$.

Find the general solution of the given differential equation. x²y’ + x(x + 2)y = e^x

Summary:

The general solution of the given differential equation x²y’ + x(x + 2)y = e^x is $y=\frac{e^{x}}{2x^{2}}+\frac{C}{x^{2}e^{x}}$.