Question

Find the solution of the differential equation $x\frac{dy}{dx}+2y=x^2$ $(x\ne 0)$ given that $y=0$ when $x=1$.

Answer 1

Suppose we have the first order differential equation

$\frac{dy}{dx}+Py=Q$

where$P$ and $Q$ are functions involving x only.

We multiply both sides of the differential equation by the integrating factor$I$ which is defined as

$e^{\int Pdx}$

$x\frac{dy}{dx}+2y=x^2⟹\frac{dy}{dx}+\frac{2y}{x}=x$

Here $P=\frac{2}{x}$

Hence, the integrating factor $I$ becomes $e^{\int \frac{2}{x}dx}=e^{2\ln x}=x^2$

$\frac{dy}{dx}+\frac{2y}{x}=x⟹x^{2}dy+2yxdx=x^{3}dx⟹\int {(x}^{2}dy+2yx)dx=\int x^{3}dx⟹x^{2}y+c=\frac{x^4}{4}$

Given that $y=0,x=1$

$⟹0+c=\frac{1}{4}$

$⟹c=\frac{1}{4}$

The solution of the differential equation is $x^{2}y+\frac{1}{4}=\frac{x^4}{4}$