Question

For the given differential equation, find the general solution: $\frac{dy}{dx} + \frac{y}{x}= x^2$

Answer 1

Answer:

Step-by-step explanation:

Concept:

The solution of

$\frac{dy}{dx}+Py=Q\:is\\\\y.e^{\int{P}dx}=\int{Q.e^{\intPdx}dx}+c$

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

comparing this with

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

we get

$P=\frac{1}{x}\\\\Q= x^2$

Integrating factor

$=e^{\int{P}dx}\\\\=e^{\int{\frac{1}{x}}\:dx}\\\\=e^{logx}\\\\=x$

The solution is

$y.e^{\int{P}dx}=\int{Q.e^{\int{P}dx}dx}+c\\\\y.x=\int{x^2.xdx}+c\\\\xy=\int{x^3dx}+c\\\\xy=\frac{x^4}{4}+c$