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' = -(y(15x+4y))/(x(5x+4y)) = -(\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 = -(v(15+4v))/(5+4v)`

Rewrite a bit and separate the variables:

`xv' = -(v(15+4v))/(5+4v) - v`

`xv' = -(v(15+4v) + v(5+4v))/(5+4v)`

`xv' = -(v(20+8v))/(5+4v)`

`(5+4v)/(v(20+8v)) \, dv = -\frac{dx}x`

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

`(5+4v)/(v(20+8v) = \frac av + \frac 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 - (28)/(20+8v)\right) \, dv = - \int \frac{dx}x`

`4 \ln|v| - \frac{14}4 \ln|20+8v| = -\ln|x| + C`

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

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

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

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

`\boxed - 14 \ln`