Answer
$y=\sqrt \frac{25(\ln x)^2+C}{2x^2}$
Work Step by Step
We are given:
$$\frac{dy}{dx}+\frac{y}{x}=\frac{25\ln x}{2x^3y}$$
Dividing both sides by $y^2$:
$$y\frac{dy}{dx}+\frac{1}{x}y^2=\frac{25\ln x}{2x^3}$$
Assume that $u=y^2\rightarrow \frac{du}{dx}=2y\frac{dy}{dx}$
The equation becomes:
$$\frac{1}{2}\frac{du}{dx}+\frac{1}{x}u=\frac{25\ln x}{2x^3}$$
$$\frac{du}{dx}+\frac{2}{x}u=\frac{25 \ln x}{x^3}$$
Itegrating factor:
$$I=e^{\int \frac{2}{x}dx}=e^{2\ln x}=x^2$$
The equation becomes:
$$\frac{d}{dx} (ux^2)=25x^{-1}\ln x$$
Integrating both sides:
$$ux^2=\frac{25}{2}(\ln x)^2+C$$
$$y^2=\frac{\frac{25}{2}(\ln x)^2+C}{x^2}$$
The general solution is $y=\sqrt \frac{25(\ln x)^2+C}{2x^2}$
