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We have this:

$$y'=\frac{y}{x}+e^{-\frac{y}{x}}$$

The way to solve this equations (as I have learned) is to make the var. change $y=xu$, so you get a linear equation: $$\frac{du}{f(1,u)-u}=\frac{dx}{x}$$ Being $f(x,y)=y'$

Now, when you have an equation with a no homogenous term $b(x)$, you can remove it, find a solution for the new equation, make the constant of that solution a function of $x$, substitute that solution in the initial equation and find what that function is. You couldn't do it with the above equation because the $e^{-\frac{y}{x}}$ is not a function of $x$ only. But if you do it, you will get to the solution $$y=x\log|\log|x|+k|,\;\;k\in\mathbb{R}$$ And this equation is a solution of the above, so the method works, why? Is it correct? I did it in an exam and I don't know if I did it right. Obviously the final solution is correct, I didn't bother to recheck the problem because I pluged it into the equation and it worked, but I don't know why this works.

Thanks in advance.


ADDED Example asked by Babak Sorouh $y'+2y=3x$ We remove the $3x$ and solve, the solution is $ce^{-2x}$, we make the $c$ a function of $x$ and substitute in the first one: $c'e^{-2x}=3x\Longrightarrow c(x)=3/2xe^{2x}-3/4e^{2x}+k;\;\;k\in\mathbb{R}$ Substitute the $c(x)$ in the solution for the equation without the $3x$, and you got it: $3/2x-3/4+ke^{-2x}$

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may i ask you to give me an example including that $b(x)$? Thanks –  B. S. Feb 8 '13 at 15:05
    
@BabakSorouh Done –  MyUserIsThis Feb 8 '13 at 15:14
    
Looks good to me. –  Ron Gordon Feb 8 '13 at 16:01
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1 Answer 1

I don't think this will work for all cases. What you have described is essentially the method for exact equations, but with the addition of integrating factors. (If you have Boyce/DiPrima, it's on pg 99.) There are condition(s) for the method of exact equations to work.

An exact equation is of the form: $$M(x, y) + N(x, y)y' = 0$$ ...where $M_y = N_x$.

So, if you have the DE (using your example): $$y' + 2y - 3x = 0$$ This implies: $$M(x,y) = 2y-3x$$ $$N(x,y) = 1$$

So, this is not an exact equation ($0\ne2$). However, if $\frac{M_y - N_x}{N}$ is a function of $x$ only, there exists a factor to make the equation exact. (I am leaving out the derivation of this.) For this DE, $\frac{M_y - N_x}{N} = \frac{2 - 0}{1} = 2$, which satisfies the criteria.

Thus, we find our integrating factor as follows: $$\frac{d\mu}{dx} = \frac{M_y - N_x}{N}\mu$$ $$\frac{d\mu}{dx} = 2\mu$$ $$\int\frac{d\mu}{\mu} = \int 2dx$$ $$\ln|\mu| = 2x$$ $$\mu = e^{2x}$$

So, multiplying through the original DE: $$e^{2x}y' + (2y - 3x)e^{2x} = 0$$ Now, $M(x, y) = (2y - 3x)e^{2x}$, and $N(x, y) = e^{2x}$. This equation is exact, as $M_y = N_x = 2e^{2x}$

Now solve using method for exact equations as usual. (If needed I can demonstrate. I don't think it adds much to the answer, though...)

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And why is the method working on my first equation? It is not an exact equation... and no conditions exist to find integrating factors (at least not trivially). My only question is why is the method working on the FIRST equation. The second was an example asked by an user in which you first remove the $b(x)$ function. Thanks for your help btw. –  MyUserIsThis Feb 8 '13 at 16:43
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