# The solution of the differential equation $x\large\frac{dy}{dx}$$+2y=x^2 is \begin{array}{1 1} (A)\;y-\large\frac{x^2}{4}=cx^{2} \\ (B)y=\large\frac{x^2}{4}+cx^{-2} \\ (C)\;4y=\large\frac{x^2}{4}+cx^{-2} \\ (D\;y+\large\frac{x^2}{4}=cx^{-2}\end{array} ## 1 Answer Toolbox: • A linear differential equation of the form \large\frac{dy}{dx}$$+Py=Q$ has general solution as $ye^{\int pdx}=\int Q.e^{\int pdx}dx+c$
• Where $e^{\int pdx}$ is the integral factor (I.F)
The solution of the differential equation $x\large\frac{dy}{dx}$$+2y=x^2 is y-\large\frac{x^2}{4}$$=cx^{-2}$
Divide throughout by x
$\large\frac{dy}{dx}+\frac{2y}{x}$$=x This represents linear differential equation of the form \large\frac{dy}{dx}$$+Py=Q$
Where $P=\large\frac{2}{x}$ and $Q=x$
$\int pdx=\int \large\frac{2}{x}dx$
$=2 \log |x|$
$=\log x^2$
Hence $e^{\int pdx}=e^{[\log x^2]}$
$I.F=x^2$
Hence the required solution is
$y.(I.F)=\int Q.(I.F)dx+c$
Where I.F is the integral factor
$y.x^2=\int x.x^2dx+c$
On integrating we get
$yx^2=\large\frac{x^4}{4}+c$
Divide throughout by $x^2$
$y=\large\frac{x^2}{4}+\large\frac{c}{x^2}$