Problem solving a couple of ODEs 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.
 A: Concerning the second question. I'll write $y_x$ and $y_t$ for derivatives with respect to $x$ and $t$, respectively. 
We already have $y_x=ty_t$. 
Then $y_{xx}=(ty_t)_x=t_xy_t+t(y_t)_x=t_xy_t+ty_{tt}t_x=ty_t+t^2y_{tt}$. 
The equation becomes $t^2y_{tt}+ty_t-ty_t+t^2y=0$, that is, $t^2y_{tt}+t^2y=0$, which I'm sure you can handle. 
A: Here's a sneaky way to do the first problem: it's actually linear, if you take $x$ as the dependent, and $y$ as the independent, variable. That is, with a little algebra you can rewrite it as $${dx\over dy}-{1\over y}x=-y\cot y$$ 
I expect you can solve first-order linear differential equations. 
A: To find a integrating factor $u(x,y)=h(x/y)$ try this. In a domain $D\subseteq \mathbb{R}^2$, such that $x\neq 0$, $y\neq 0$ and $\cos y\neq 0$, let $M(x,y)$ and $N(x,y)$ as Nana says and $u(x,y)=h(x/y)$, we have
$$\begin{matrix}
M_y= \frac{1}{x}\cos y & u_y = -\frac{x}{y^2}h'\left(\frac{x}{y}\right)\\
N_x= -\frac{y}{x^2}\cos y & u_x=\frac{1}{y}h'\left(\frac{x}{y}\right)
\end{matrix}.$$
Then
$$\begin{align*}
(uM)_y&= \frac{1}{x}h\left(\frac{x}{y}\right)\cos y - \frac{1}{y^2}h'\left(\frac{x}{y}\right)\sin y\\
(uN)_x&= -\frac{y}{x^2}h\left(\frac{x}{y}\right)\cos y + h'\left(\frac{x}{y}\right)\left(\frac{1}{x}\cos y -\frac{1}{y^2}\sin y\right).
\end{align*}$$
If $u$ is an integrating factor
$$\begin{align*}
(uM)_y&= (uN)_x\\
\frac{1}{x}h\left(\frac{x}{y}\right)\cos y&=-\frac{y}{x^2}h\left(\frac{x}{y}\right)\cos y + \frac{1}{x}h'\left(\frac{x}{y}\right)\cos y,
\end{align*}$$
multiplying by $\frac{y}{\cos y}$ in both sides
$$\begin{align*}
\frac{y}{x}h\left(\frac{x}{y}\right) &=-\frac{y^2}{x^2}h\left(\frac{x}{y}\right)+\frac{y}{x}h'\left(\frac{x}{y}\right),
\end{align*}$$
now, put $t=\frac{x}{y}$
$$0=-\left( 1+\frac{1}{t}\right)h(t)+h'\left(t\right).$$
A solution of the last equation is
$$h(t)=te^t.$$
A: I can help you make the first equation exact. First, let $M=\frac{\sin y}{x}$, $N=\frac{y}{x}\cos y-\frac{\sin y}{y}$.  Then $$M_{y}=\frac{\cos y}{x}\,\,\,\,,N_{x}=-\frac{y\cos y}{x^2},$$ and $$M_{y}-N_{x}=\frac{x+y}{x^2}\cos y.$$ So clearly the differential equation is not exact. To find an integrating factor, we use the hint given. That is we try an integrating factor of the form $z=\frac{x}{y}.$ We thus compute the following:  $$z_{x}=\frac{1}{y}\,\,\,,\,z_{y}=-\frac{x}{y^2}.$$ Next we form the differential equation of the form $$\frac{1}{u}\frac{\mathrm{d}u}{\mathrm{d}z}=\frac{M_{y}-N_{x}}{N\cdot z_{x}-M\cdot z{_y}}\qquad\qquad\qquad (*)$$ where the right hand side must depend on only $z$.  
Computing the above expression we find the right hand side to be $\frac{x+y}{x}=1+\frac{y}{x}=1+\frac{1}{z}.$
So (*) becomes $$ \frac{1}{u}\frac{\mathrm{d}u}{\mathrm{d}z}=1+\frac{1}{z}$$ which upon solving yields $$ u=\exp(z+\ln z)=ze^z,$$ which is the integrating factor. Multiplying the original equation by $u$ will surely make the d.e. exact! 
Hope it helps!!!
