the following problem is this: Find the General Solution of the $DE:xy'-y=y^3cos(19x)$
I know that Bernoulli Sub would work here, but I want to see if Homogeneous would work too.
My Work: Divide both sides by x
$y'- \frac yx = \frac yxy^2cos(19x)$
Then move $\frac yx$ over to the right side of the equation
$y' = \frac yx + \frac yxy^2cos(19x)$
Let's apply the homogeneous substitution
let $v = \frac yx$
then $y=vx$
and $y'=v+v'x$
Sub these values into the DE to get
$v+v'x=v+v\cdot v^2x^2cos(19x)$
subtract v from both sides
$v'x=v^3x^2cos(19x)$
Divide both sides by x
$v'=v^3xcos(19x)$
$\cfrac {dv}{dx} = v^3xcos(19x)$
Seperate and Solve
$\int\cfrac {dv}{v^3} = \int xcos(19x)dx$
$\cfrac {-1}{2v^2} = \cfrac {19xsin(19x)+cos(19x)}{361}+C$
Take Reciprocal
$-2v^2= \cfrac{361}{19xsin(19x)+cos(19x)+C}$
Divide both sides by -2
$v^2= \cfrac{-180.5}{19xsin(19x)+cos(19x)+C}$
Take the square root of both sides
$v= \sqrt{\cfrac{-180.5}{19xsin(19x)+cos(19x)+C}}$, $v = -\sqrt{\cfrac{-180.5}{19xsin(19x)+cos(19x)+C}}$
back sub y in $\frac yx = \sqrt{\cfrac{-180.5}{19xsin(19x)+cos(19x)+C}}$, $\frac yx = -\sqrt{\cfrac{-180.5}{19xsin(19x)+cos(19x)+C}}$
Solve for y
$y = x \cdot \sqrt{\cfrac{-180.5}{19xsin(19x)+cos(19x)+C}}$, $y = x \cdot -\sqrt{\cfrac{-180.5}{19xsin(19x)+cos(19x)+C}}$
This is the final answer
Did I do this correctly? Is my reasoning fine?
$$xy'-y=y^3\cos(19x)$$ $$\left(\dfrac yx \right)'=\dfrac {y^3 \cos (19x)}{x^2}$$ $$\dfrac {x^3}{y^3}\left(\dfrac yx \right)'= {x \cos (19x)}$$ $$\left(\dfrac xy \right)^3d\left(\dfrac yx \right)= {x \cos (19x)}dx$$ So yes this part of your attempt is correct. $$-\dfrac 12\left(\dfrac xy \right)^2= \dfrac x {19}\sin (19x) -\dfrac 1 {19}\int \sin (19x)dx$$ $$-\dfrac 12\left(\dfrac xy \right)^2= \dfrac x {19}\sin (19x) +\dfrac 1 {19^2} \cos (19x)dx+C$$