First I'm trying to make this equation exact $ \frac{\sin y}{x} dx + (\frac{y}{x} \cos y - \frac{\sin y}{y} ) dy = 0 $
The problem says to use to use an integrating factor $ u(x,y)=h(\frac{x}{y}) $.
To make integrating a little easier I first did a variable change using $v=\frac{x}{y} $, but that didn't help much when I had to derive it.
In general I-ve been having trouble solving most problems that require an integrating factor that involes both variables, unless it's of the form $ u(x,y)=x^a y^b $ with a and b integers to be determined.
I think there's some "simple" way to solve this I'm not seeing, or knew and can't remember. If anyone has any idea, please share.
Now on to the second problem. I need to find a solution for y'' -y' +e^2x y = 0 As a note it says to consider the variable change $ x = \ln t$.
By doing this I get y'' - y' +t y = 0 , which I'm not too sure how to solve.
I gather that something's missing since as that is right now y'= \frac{dy}{dx} , and I need it w.r.t. $t$. So, y'= \frac{dy}{dx} = \frac{dy}{dt} \frac{dt}{dx} = y' \frac{1}{t} y'' = \frac{d^2 y}{dx^2} = \frac{dy'}{dx} = \frac{dy'}{dt} \frac{dt}{dx} = y'' \frac{1}{t^2}
Putting all this together back in the equation I get \frac{y''}{t^2} - \frac{y'}{t} + ty = 0
Now All I can really think about this is adding everything together so \frac{y'' - ty' +t^3 y}{t^2} = 0 y'' - t y' + t^3 y = 0
Again, I'm probably not doing something right here, but if I am, I'm not sure what would be a suitable $y(t)$ to try, since it can't be $t^a$, $e^{at}$ and so on.
Again, any ideas would be more than welcomed.
EDIT: Regarding the second problem, with a little help from Gerry I got (Assuming $\frac{d^2y}{dx^2} = t^2 \frac{d^2y}{dt^2} $ which I think is right) t^2 y'' - t y' + ty = 0 ty'' - y' + y = 0 Either $ \frac{d^2y}{dx^2} = t \frac{d^2y}{dt^2}$ so that I don-t have any $t$ laying around, or there's something else that I'm not getting.