Answer
$4\ln|{\sqrt y+2}|$ = ${(e^{x^{2}})}+C$
Work Step by Step
$\frac{1}{x}\frac{dy}{dx}$ = $ye^{x^{2}}+2\sqrt{y}e^{x^{2}}$
$\frac{1}{x}\frac{dy}{dx}$ = $e^{x^{2}}(y+2\sqrt {y})$
$\frac{1}{y+2\sqrt {y}}dy$ = $xe^{x^{2}}dx$
$\frac{1}{\sqrt {y}(\sqrt y+2})dy$ = $xe^{x^{2}}dx$
$\int(\frac{1}{\sqrt {y}(\sqrt y+2}))dy$ = $\int(xe^{x^{2}})dx$
$u$ = $\sqrt y+2$
$du$ = $\frac{1}{2\sqrt y}dy$
$2du$ = $\frac{1}{\sqrt y}dy$
$z$ = $x^{2}$
$dz$ = $2xdx$
$\frac{1}{2}dz$ = $xdx$
$\int(\frac{2}{u})du$ = $\int{\frac{1}{2}(e^{z})}dz$
$2\ln|{u}|$ = ${\frac{1}{2}(e^{z})}+C_{1}$
$2\ln|{\sqrt y+2}|$ = ${\frac{1}{2}(e^{x^{2}})}+C_{1}$
$4\ln|{\sqrt y+2}|$ = ${(e^{x^{2}})}+2C_{1}$
$4\ln|{\sqrt y+2}|$ = ${(e^{x^{2}})}+C$
$C$ = $2C_{1}$
