Question

The solution of the differential equation $\frac{d y}{d x}=\frac{y^{2} + x ln x}{2 x y}$ is (where, $c$ is the constant of integration)

• A. $2x^{2}=y\left(ln x\right)^{2}+2cy$
• B. $2y^{2}=x\left(ln x\right)^{2}+2cx$
• C. $x^{2}=y\left(ln x\right)^{2}+c$
• D. $2y^{2}=\frac{x}{y}\left(ln x\right)^{2}+cx$

Answer 1

Answer: (B)

$2xy\frac{d y}{d x}=y^{2}+xln x$
$2y\frac{d y}{d x}-\frac{y^{2}}{x}=ln x$
Put, $y^{2}=t\Rightarrow 2y\frac{d y}{d x}=\frac{d t}{d x}$
$\frac{d t}{d x}-\frac{t}{x}=ln x$
I.F. $=e^{\int \frac{- 1}{x} d x}=e^{- l n x}=\frac{1}{x}$
The solution is
$\frac{t}{x}=\int \frac{ln x}{x} d x$
$\frac{t}{x}=\frac{\left(ln x\right)^{2}}{2}+c$
$\frac{y^{2}}{x}=\frac{\left(ln x\right)^{2}}{2}+c$
$2y^{2}=x\left(ln x\right)^{2}+2cx$