Question

Solve the differential equation: $\frac{dy}{dx}+\frac{2}{x}y=\sin x$

• A. $x^{2}y=(2+x^2)\cos x+2x\sin x+c.$
• B. $x^{2}y=(2-x^2)\cos x+2x\sin x+c.$
• C. $x^{2}y=(2-x^2)\cos x-2x\sin x+c.$
• D. None of these.

Answer 1

The correct option is B $x^{2}y=(2-x^2)\cos x+2x\sin x+c.$

$\frac{dy}{dx}+\frac{2}{x}y=\sin x$ ...(1)
Here $P=\frac{2}{x}\Rightarrow \int Pdx=\int \frac{2}{x}dx=2\log x=\log x^2$
$\therefore I.F.=e^{\log x^2}=x^2$
Multiplying (1) by $I.F.$ we get
$x^2\frac{dy}{dx}+2xy=x^2\sin x$
Integrating both sides we get
$x^{2}y=\int x^2\sin xdx+c=-x^2\cos x+\int 2x\cos xdx+c=-x^2\cos x+2x\sin x-\int 2\sin xdx+c=-x^2\cos x+2x\sin x+2\cos x+c$

$\therefore x^{2}y=(2-x^2)\cos x+2x\sin x+c$