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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.

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    By the way, I assume $e^2xy$ is a typo for $e^{2x}y$.2011-07-18

4 Answers 4

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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!!!

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    @Gerry is correct, the hint doesn't say that $h = \frac{x}{y} $. In fact, there's no a and b so that $ h = x^a y^b $ turns this into an exact equation so $h$ must be something else.2011-07-18
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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.

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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.

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    @Gerry: Okay I see, I did not even read the question. I just read the answers provided first then comments then the question.2011-07-19
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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 $h(t)=te^t.$