Answer
$y=\sqrt 5x^3\ln x-x^3-Cx^{-2}$
Work Step by Step
We are given:
$$\frac{dy}{dx}+\frac{1}{x}y=\frac{25x^2 \ln x}{2y}$$
Multiplying both sides by $y$
$$y\frac{dy}{dx}+\frac{1}{x}y^2=\frac{25x^2 \ln x}{2}$$
Assume $u=y^2 \rightarrow \frac{du}{dx}=2y \frac{dy}{dx}$
The equation becomes:
$$\frac{1}{2}\frac{du}{dx}+\frac{1}{x}du=\frac{25x^2 \ln x}{2}$$
$$\frac{du}{dx}+\frac{2}{x}u=25x^2 \ln x$$
Integrating factor:
$$I=e^{\int \frac{2}{x} dx}=e^{2\ln x dx}=x^2$$
The equation becomes:
$$\frac{d}{dx}(ux^2)=25x^4\ln x$$
Integrating both sides:
$$ux^2=25(\frac{1}{5}x^5\ln x-\frac{1}{25}x^5)+C$$
$$u=5x^3\ln x-x^3-Cx^{-2}$$
Since $u=y^2$
$$y^2=5x^3\ln x-x^3-Cx^{-2}$$
Hence general solution is $y=\sqrt 5x^3\ln x-x^3-Cx^{-2}$
