The general solution of the differential equation $x\frac{dy}{dx}+2y = x^2, x≠ 0$ is :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $y=\frac{x^2}{4}+\frac{c}{x^2}$(c is constant of integration)
$x\frac{dy}{dx}+2y = x^2$
$\frac{1}{x}(x\frac{dy}{dx}+2y)=\frac{1}{x}(x^2)⇒\frac{dy}{dx}+\frac{2}{x}y=x$
so $I.F.=e^{\frac{2}{x}}dx=e^{2\log|x|}=x^2$
so multiplying eq. with $x^2$ and integrating
$⇒∫x^2\frac{dy}{dx}+2xydx=∫x^3dx$
$⇒x^2y=\frac{x^4}{4}+C⇒y=\frac{x^2}{4}+\frac{C}{x^2}$