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

**Solution:**

The given differential equation is

x^{2}y’ + x(x + 2)y = e^{x}

Let us multiply be ex on both sides

x^{2}e^{x}y’ + (x^{2} + 2x)e^{x}y = e^{2x}

We know that

(x^{2}e^{x})’ = 2xe^{x} + x^{2}e^{x} = (x^{2} + 2x)e^{x}

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

(x^{2}e^{x}y)’ = 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. x^{2}y’ + x(x + 2)y = e^{x}

**Summary:**

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

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