<s>[INST] For this non-linear differential equation: $ xy' = y + ax \sqrt{x^2+y^2} $, where $ x>0 $ and $ a>0$ is a constant This is a 3 part question (I seem to have found (a), but I am not confident about (b) and thus (c) as well):
(a) if y is a solution, show that v:= $yx^{-1}$ satisfies the differential equation:
$v'=a  \sqrt{1+v^2}$ 
My solution is as follows:
$ v=\dfrac{y}{x}$
$v'=\dfrac{d}{dx}(\dfrac{y}{x})$
$ v'= \dfrac{y'}{x}-\dfrac{y}{x^2}$
Now substituting y', I have:
$v' = \dfrac{1}{x}(\dfrac{y}{x}+a\sqrt{x^2+y^2})-\dfrac{y}{x^2} $
Distributing the $\dfrac{1}{x}$, factoring out $x^2$ from the square root, and then substituting for v I get:
$v' = \dfrac{y}{x^2}+\dfrac{ax\sqrt{1+v^2})}{x}-\dfrac{y}{x^2} $
Cancelling gives:
$v'=a  \sqrt{1+v^2}$ 
Does this sufficiently answer this question? (Note: This is a 2 point question)
(b) Use the differential equation in part (a) to get the general solution to:
$ xy' = y + ax \sqrt{x^2+y^2} $
This is a 3 point question and where I am having a little trouble, my strategy has been as follows:
Since $ v = \dfrac{y}{x}$, then:
$\int v'\,dv = a\int \sqrt{1+v^2}\,dv = \dfrac{y}{x}$
Substituting $v = sinhx$, (I am not sure if I can do this, or have used this correctly) I get:
$v= a\int coshx\, dx = asinhx+c=\dfrac{y}{x}$
Therefore, $y = axsinhx+c$
Now since $y'= \dfrac{y}{x}+a\sqrt{x^2 + y^2}$
substituting $y = axsinhx+c$, and pulling the constant out of the square root and simply adding it to the end, I have just observed and tried out that when $a=1$, 
$y = (1)xsinhx+c$ seems to be a valid solution. 
But I just guessed $a$, so I need a bit of guidance to give a solid answer to this.
(c) Find the particular solution of $ xy' = y + ax \sqrt{x^2+y^2} $ with $y(1)=0$
Perhaps trivially:
$0 = (1)sinh(1) + c$, gives $c=-sinh(1)$
Thus my final answer is: $y = (1)xsinhx-sinh(1)$
Any help/guidance on this would be greatly appreciated!
 [/INST] After getting differential eqn $$\frac{dv}{dx}=a\sqrt{1+v^2}$$ $$\implies \int\frac{dv}{\sqrt{1+v^2}} =\int a dx$$ $$\implies \ln(v+\sqrt{1+v^2})=ax+c$$ 
Putting $v=\frac{y}{x}$ gives 
$$\ln(\frac{y+\sqrt{x^2+y^2}}{x})=ax+c$$
Since $y(1)=0\implies \ln(1)=a+c\implies c=-a$ (as $\ln(1 )=0$)
Thus, particular solution is $$ \ln(\frac{y+\sqrt{x^2+y^2}}{x})=a(x-1)$$
EDIT:
If you want $y$ explicitly, then,
$$\frac{y+\sqrt{x^2+y^2}}{x}=e^{a(x-1)}\implies {y+\sqrt{x^2+y^2}}=xe^{a(x-1)}$$
$$\sqrt{x^2+y^2}=xe^{a(x-1)}-y$$
Squaring both sides,
$$x^2+y^2=x^2e^{2a(x-1)}+y^2-2xye^{a(x-1)}$$
$$\implies 2xye^{a(x-1)}=x^2(e^{2a(x-1)}-1)$$
$$\implies y=x(\frac{e^{a(x-1)}-e^{-a(x-1)}}{2})$$
But squaring usually changes the domain, so be careful. 
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