Question

Find the general solution of the following differential equation. Primes denote derivatives with respect to x.
x(5x+4y)y' +y(15x+4y) = 0
Solution should be implicit general solution in the form ​F(x,y)​=C, where C is an arbitrary constant.
P.s. this is supposed to be done using v = y/x substitution of homogenous differential equations.

Answer 1

The given differential equation

$x(5x+4y)y' + y(15x+4y) = 0$

is indeed homogeneous, since we can write

$\implies y' = -\dfrac{y(15x+4y)}{x(5x+4y)} = -\dfrac{\frac yx \left(15 + 4\frac yx\right)}{5 + 4\frac yx}$

Let $y = xv$, so that $y' = xv' + v$. Then $v=\frac yx$ and the equation transforms to

$xv' + v = -\dfrac{v(15+4v)}{5+4v}$

Rewrite a bit and separate the variables:

$xv' = -\dfrac{v(15+4v)}{5+4v} - v$

$xv' = -\dfrac{v(15+4v) + v(5+4v)}{5+4v}$

$xv' = -\dfrac{v(20+8v)}{5+4v}$

$\dfrac{5+4v}{v(20+8v)} \, dv = -\dfrac{dx}x$

Integrate both sides. On the left, take the partial fraction decomposition,

$\dfrac{5+4v}{v(20+8v} = \dfrac av + \dfrac b{20+8v}$

$\implies 5+4v = a(20+8v) + bv = 20a + (8a+b)v$

$\implies \begin{cases}20a=5 \\ 8a+b=4 \end{cases} \implies a=4,b=-28$

Proceed with the integrals.

$\displaystyle \int \left(\frac4v - \frac{28}{20+8v}\right) \, dv = - \int \frac{dx}x$

$4 \ln|v| - \dfrac{14}4 \ln|20+8v| = -\ln|x| + C$

Get the solution back in terms of $y$ and $x$.

$4 \ln\left|\dfrac yx\right| - \dfrac{14}4 \ln\left|20+8\dfrac yx\right| = -\ln|x| + C$

$16 \ln\left|\dfrac yx\right| - 14 \ln\left|20+8\dfrac yx\right| = -4\ln|x| + C$

$16 (\ln|y| - \ln|x|) - 14 (\ln|20x+8y| - \ln|x|) = -4\ln|x| + C$

$\boxed{16 \ln|y| - 14 \ln|20x+8y| = -2 \ln|x| + C}$