Answer
See below
Work Step by Step
Let the equation be:
$\frac{dy}{dx}=\frac{4x+y}{x-4y}\\\rightarrow (x-4y)dy=(4x+y)dx\\\rightarrow (x-4y)dy-(4x+y)dx=0$
We will transform the variables into polar coordinates:
$x=r\cos \theta\\y=r\sin \theta\\dx=\frac{\partial x}{\partial r}dr+\frac{\partial x}{\partial \theta}d\theta=\cos \theta dr-r\sin \theta d\theta\\dy=\frac{\partial y}{\partial r}dr+\frac{\partial y}{\partial \theta}d\theta=\sin \theta dr+r\cos \theta d\theta$
Therefore, we have:
$(r\cos \theta y-4r\sin \theta)(\sin \theta dr+r\cos \theta d\theta)-(4r\cos \theta +r\sin \theta)(\cos \theta dr-r\sin \theta d\theta)=0\\ \rightarrow r\sin \theta\cos \theta dr-4r\sin^2 \theta dr +r^2\cos^2\theta d\theta -4r^2\sin \theta\cos \theta d\theta-(4r\cos^2dr-4r^2\sin \theta \cos \theta d\theta+r\sin \theta\cos \theta dr-r^2\sin^2 \theta d\theta)=0\\ \rightarrow
r\sin \theta \cos \theta dr-4r\sin^2 \theta dr +r^2\cos^2\theta d\theta -4r^2\sin \theta\cos \theta d\theta-4r\cos^2dr+4r^2\sin \theta \cos \theta d\theta-r\sin \theta\cos \theta dr +r^2\sin^2 \theta d\theta)=0\\ \rightarrow r(\sin \theta \cos \theta-4\sin^2 \theta -4\cos^2-\sin \theta\cos \theta) dr-r^2(\cos^2 \theta-4\sin \theta \cos \theta + 4\sin \theta\cos \theta-\sin^2 \theta)d\theta=0\\\rightarrow r(-4\sin^2\theta-4\cos^2\theta)dr+r^2(\cos^2\theta+\sin^2\theta)d\theta=0\\
\rightarrow -4r(\sin^2\theta+\cos^2\theta)dr+r^2d\theta=0\\\rightarrow -4rdr+r^2d\theta=0\\
\rightarrow 4rdr=r^2d\theta\\
\rightarrow \frac{dr}{r}=\frac{1}{4}d\theta$
Integrate both sides of the last equation:
$\int \frac{dr}{r}=\int \frac{1}{4}d\theta\\
\rightarrow \ln |r|=\frac{1}{4}\theta+ c_1$
Since $c=e^{c_1}$, thus $r=ce^{\theta /4}$
