The second question asks: What are its solution curves?
These questions were extracted from Ordinary Differential equations: Garrett Birkhoff.
What I did: I differentiated the first equation (of the ellipse) and got the differential equation shown, but it was too simple, so I am not quite sure that that was the expected way to prove it.
Also, I tried to do it the other way around (finding the first equation from the differential equation given). However, I could only get to the normal form of the differential equation, and I can not figure out which method should I use:
$y'=\dfrac{-5x-3y}{3x+5y}$
On the other hand, what are its solution curves? I could not understand what this question expected me to do, so I supposed it meant finding the solution curves of the orthogonal trajectories, so I just replaced the "inverse" of the slope given (normal form of differential equation) $y'=\frac{3x+5y}{5x+3y}$ in the ODE given at first, $(5x+3y)+(3x+5y)y'=0$.
I really appreciate your answer
Note that $$(5x+3y)+(3x+5y)y'=0 \quad \iff \quad (5x+3y)dx+(3x+5y)dy=0$$ is an exact equation because: $$ \frac{\partial}{\partial y}(5x+3y)=\frac{\partial}{\partial x}(xx+5y)=3 $$ So the solution is $f(x,y)=C$ where $f(x,y)$ is a function such that $$ \frac{\partial f(x,y)}{\partial x}=(5x+3y) \quad \mbox{and}\quad \frac{\partial f(x,y)}{\partial y}=(3x+5y) $$
that is exactly $f(x,y)=5x^2+6xy+5y^2$. And the solution curves (or integral curves) are the ellipse of equation $5x^2+6xy+5y^2=C$