The general solution of the differential equation $$(x - y^2)dx + y(5x + y^2)dy = 0$$ is

Given differential equation,

$$(x-y^2)\,dx+y(5x+y^2)\,dy=0$$

Put

$$y^2=v$$

Then,

$$dv=2y\,dy$$

$$y\,dy=\frac{dv}{2}$$

Substituting,

$$(x-v)\,dx+(5x+v)\frac{dv}{2}=0$$

Multiply throughout by $$2$$:

$$2(x-v)\,dx+(5x+v)\,dv=0$$

Now this is a homogeneous equation.

Put

$$v=kx$$

Then,

$$dv=k\,dx+x\,dk$$

Substituting,

$$2(x-kx)\,dx+(5x+kx)(k\,dx+x\,dk)=0$$

$$2(1-k)x\,dx+(5+k)x(k\,dx+x\,dk)=0$$

Divide throughout by $$x$$:

$$2(1-k)\,dx+(5+k)(k\,dx+x\,dk)=0$$

$$\left(2-2k+5k+k^2\right)dx+(5+k)x\,dk=0$$

$$\left(k^2+3k+2\right)dx+(5+k)x\,dk=0$$

Factor:

$$\left(k+1\right)\left(k+2\right)dx+(5+k)x\,dk=0$$

Therefore,

$$\frac{dx}{x}+\frac{k+5}{(k+1)(k+2)}\,dk=0$$

Using partial fractions,

$$\frac{k+5}{(k+1)(k+2)}=\frac4{k+1}-\frac3{k+2}$$

Hence,

$$\int\frac{dx}{x}+\int\left(\frac4{k+1}-\frac3{k+2}\right)dk=0$$

Integrating,

$$\ln|x|+4\ln|k+1|-3\ln|k+2|=\ln C$$

$$\ln\left|\frac{x(k+1)^4}{(k+2)^3}\right|=\ln C$$

Therefore,

$$\frac{x(k+1)^4}{(k+2)^3}=C$$

Now substitute

$$k=\frac{v}{x}=\frac{y^2}{x}$$

$$\frac{x\cdot\frac{(y^2+x)^4}{x^4}}{\frac{(y^2+2x)^3}{x^3}}=C$$

Simplifying,

$$\frac{(y^2+x)^4}{(y^2+2x)^3}=C$$

Hence, the general solution is

$$\boxed{(y^2+x)^4=C|y^2+2x|^3}$$