Homogeneous differential equation proof Show that a straight line through the origin intersects all integral curves of a homogeneous equation at the same angle.
I tried like this in homogeneous equation $y'=f(x,y)$ that is $f(tx,ty)=f(x,y)$.
Afterwards I do not know what to do.
 A: Consider the homogeneous DE $y'=f(y/x)$, and the straight line $y=mx$ passing through the origin. We will show that angle between $y=mx$ and the tangent line to the integral curve of $y'=f(y/x)$ remains constant. Slope of the tangent to each integral curve of $y'=f(y/x)$ is $f(y/x)$. At the point $(x,y)$ of intersection with $y=mx$, it becomes $f(m)$. So the angle $\theta$ between $y=mx$ and the tangent line to the integral curve is given by
$$\tan\theta=\frac{m-f(m)}{1+mf(m)},$$
which is a constant.
A: The angle between a line $L$ and a curve $\Gamma$ is the angle between $L$ and the tangent line to $\Gamma$ at the point of intersection. Therefore, you should prove that the tangent lines at all points of the form $(\lambda x, \lambda y)$ are parallel, as $\lambda$ varies.
Writing the equation as $m\,dx+n\,dy=0$ is a good idea. This form tells you that $(m(x,y),n(x,y))$ is a normal vector at the solution curve at $(x,y)$. Therefore, for any $\lambda>0$, $(m(\lambda x,\lambda y),n(\lambda x,\lambda y))$ is a normal vector at the solution curve at $(\lambda x,\lambda y)$. But $(m(\lambda x,\lambda y),n(\lambda x,\lambda y))$ is just a multiple of $(m(x,y),n(x,y))$. Thus, the normal vectors at $(x,y)$ and at  $(\lambda x,\lambda y)$ are parallel. So are the tangent lines. 
