Answer
The solution of the given differential equation
$$
2 y e^{y^{2}} y^{\prime}=2 x+3 \sqrt{x}
$$
is
$$
y =\pm \sqrt {\ln \left(x^{2}+2 x^{3 / 2}+C\right)}
$$
where $C$ is any nonzero constant.
Work Step by Step
$$
2 y e^{y^{2}} y^{\prime}=2 x+3 \sqrt{x}
$$
We write the equation in terms of differentials and integrate both sides:
$$
2 y e^{y^{2}} d y=(2 x+3 \sqrt{x}) d x
$$
$$
\int 2 y e^{y^{2}} d y=\int(2 x+3 \sqrt{x}) d x
$$
$\quad\quad\quad\quad \Rightarrow$
$$
e^{y^{2}}=x^{2}+2 x^{3 / 2}+C \Rightarrow y^{2}=\ln \left(x^{2}+2 x^{3 / 2}+C\right)
$$
$\quad\quad\quad\quad \Rightarrow$
$$
y =\pm \sqrt {\ln \left(x^{2}+2 x^{3 / 2}+C\right)}
$$
where $C$ is any nonzero constant.
So the solution of the differential equation
$$
2 y e^{y^{2}} y^{\prime}=2 x+3 \sqrt{x}
$$
is
$$
y =\pm \sqrt {\ln \left(x^{2}+2 x^{3 / 2}+C\right)}
$$
where $C$ is any nonzero constant.
