Question

x(dy)/(dx)=x^(2) + 5y

Answer 1

$$xy'=x^2+5y$$

$$y'-\frac{5}{x}y=x$$

We put

$$y=uv\\ y'=u'v+uv'$$

Then

$$u'v+uv'-\frac{5}{x}uv=x$$

$$u(v'-\frac{5}{x}v)+u'v=x$$

Let

$$v'-\frac{5}{x}v=0$$

Then

$$v'=\frac{5}{x}v$$

$$\ln v=5\ln x$$

$$v=x^5$$

Thus

$$x^5u'=x$$

$$u'=x^{-4}$$

$$u=-1/3x^{-3}+C$$

Finally

$$y=uv=(-1/3x^{-3}+C)x^5$$

$$y=-1/3x^2+Cx^5$$