Question

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

• A. $y=\frac{x^2+c}{4x^2}$
• B. $y=\frac{x^2}{4}+c$
• C. $y=\frac{x^4+c}{x^2}$
• D. $y=\frac{x^4+c}{4x^2}$

Answer 1

The correct option is D

$y=\frac{x^4+c}{4x^2}$

Explanation of the correct option.

Compute the required value.

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

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

Compare the equation with standard L.D.E. $\frac{dy}{dx}+Py=Q$.

$P=\frac{2}{x}$ and $Q=x$

Now,

$\begin{array}{rcl}\mathrm{I}.\mathrm{F}.& =& {\mathrm{e}}^{\int \mathrm{P}\mathrm{dx}}\\ & =& {\mathrm{e}}^{\int \frac{2}{\mathrm{x}}\mathrm{dx}}\left[\int \frac{1}{\mathrm{x}}\mathrm{dx}=\ln \left(\mathrm{x}\right)\right]\\ & =& {\mathrm{e}}^{2\ln \left(\mathrm{x}\right)}\\ & =& {\mathrm{e}}^{\ln \left({\mathrm{x}}^2\right)}\\ & =& {\mathrm{x}}^2\end{array}$

Since the solution of L.D.E. is given by $y(\mathrm{I}.\mathrm{F}.)=\int Q\left(\mathrm{I}.\mathrm{F}.\right)dx+c$.

$y.x^2=\int x.x^{2}dx+c_1$

$\Rightarrow$$y.x^2=\frac{x^4}{4}+c_1$

$\Rightarrow$ $y=\frac{x^4+4c_1}{4x^2}$

$\Rightarrow$ $y=\frac{x^4+c}{4x^2}$ $\left[\because 4c_1=c\right]$

Hence option $\left(D\right)$ is the correct option.