Solve $x^2 dy/dx = y^2 − 3xy − 5x^2$ for the case of $ y(1)=0$ and describe what the solution looks like when $|x|$ is very large?

Question

I found the explicit solution of $x^2 dy/dx = y^2 − 3xy − 5x^2$ for the case of $ y(1)=0$ to equal:

(y/x)^2 - 4(y/x) = 5|x|^-6 + 5 (c=5, where c is the constant) ; using the substitution of y=vx [im not 100% sure if this is correct so i would really appreciate if someone would check for me]

EDIT: Here is my working out: part 1 part 2 My problem is what describing how the solution would look like when 'when |x| is very large'

From my knowledge i know that when the constant c>0, y^2 < x^2 but i dont know how that can help.

Thanks!

fyi this is part of my assignment so i'd appreciate if there where no ambiguous answers, im pretty desperate to get this right. Thanks again

Answer 1

then you will have $$x^2(xv'+v)=-5x^2+x^2v^2-3x^2v$$ and then you will get $$\frac{dv(x)}{dx}=\frac{v(x)^2-4v(x)+5}{x}$$

Answer 2

$$x^2y'=y^2-3xy-5x^2$$Let $y=xu$ therefore$$xu'+u=u^2-3u-5\\xu'=u^2-4u-5=(u-5)(u+1)\\\dfrac{u'}{(u-5)(u+1)}=\dfrac{1}{x}\\u'[\dfrac{1}{u-5}-\dfrac{1}{u+1}]=\dfrac{6}{x}\to\ln|u-5|-\ln|u+1|=6\ln |x|+C_1\\|\dfrac{u-5}{u+1}|=C_2x^6$$after applying the initial condition ($y(1)=u(1)=0$) we have $C_2=5$ therefore the final answer is $$|\dfrac{y-5x}{y+x}|=5x^6$$for $|x|\to\infty$ , $|\dfrac{y-5x}{y+x}|=|1-\dfrac{6}{u+1}|\to\infty$ which yields to $u\to -1$ or $y\sim -x$