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I know how to solve an exact equation like

$$M(x,y) + N(x,y)y=0 $$

We just check $$\frac{\partial M}{\partial y} =\frac{\partial N}{\partial x} $$

If so, then it's just a little bit of algebra, taking anti-derivative and solve, such that $f = f(x,y) = c$, where we can obtain $f$ by just taking integal w.r.t. $x$ etc.

But now I'm stuck with integrating factors. I can do it if the integrating factor $\mu=\mu(x)$ depends only on one variable. But now I want to do another exercise, which uses a function of the form $\mu(x+y)$.

How can I solve $$(7x^3+3x^2y+4y) +(4x^3+x+5y)y'=0 $$ I'm given a hint that I should use a function of the form $\mu(x+y)$.

I guess I should just multiply by $(x+y)^m$ for an arbitrary $m$, but I don't know how to do the algebra after I multiplied...

I guess something like this :

$$\frac{\partial (M \mu)}{\partial y}= m(x+y)^{m-1}(7x^3+3x^2y+4y) + (x+y)^m(3x^2+4) =$$

$$\frac{\partial (N\mu)}{\partial x} = m(x+y)^{m-1}(4x^3+x+5y) + (x+y)^m(12x^2+1) $$

How solve for $\mu$ ?

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your integrating factor doesn't need to be of the form $(x+y)^m$. It could also be something like $e^{(x+y)^2}$ or something like that. –  Slugger Jan 26 '13 at 14:43
    
You are doing the same course ? ;-). Thanks, I didn't realize that –  MSKfdaswplwq Jan 26 '13 at 14:44
    
How about changing to $w=x+y$ and writing the equation in terms of $x,w$. Now $\mu=\mu(w)$. –  Maesumi Jan 26 '13 at 15:00
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2 Answers

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Hint: Let you have an OE like $$M(x,y)dx+N(x,y)dy=0$$ and let $\mu=\mu(z(x,y))$ be an integrating factor for it. If $$\frac{M_y-N_x}{Nz_x-Mz_y}$$ be a function respect to $z$ then $$\mu=\exp\left(\int\frac{M_y-N_x}{Nz_x-Mz_y}dz\right)$$. Here $z=x+y$, so $z_x=1,z_y=1$ and therefore your integrating factor is $\mu=\exp\left(\int\frac{M_y-N_x}{N-M}dz\right)$. You can easily find $\mu$ by doing $$(\mu M)_y=(\mu N)_x$$ as well. For thsi problem $$\mu=\exp\left(\int\frac{3x^2+4-12x^2-1}{4x^3+x+5y-7x^3-3x^2y-4y}dz\right)=\exp\left(\int\frac{3(1-3x^2)}{(x+y)(1-3x^2)}dz\right)$$ which if $3x^2-1\neq0$ then $$\mu=\exp\left(\int\frac{3}{x+y}dz\right)=\exp\left(\int\frac{3}{z}dz\right)$$ I think you can do the rest.

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Your integrating factor is $$ \mu(x+y)=(x+y)^3. $$

To find it you use the usual strategy expressing $$ \mu_yM-\mu_xN=\mu(N_x-M_y).\tag{1} $$ After some simplification you can find that $$ \frac{N_x-M_y}{M-N}=\frac{3}{x+y}. $$ Hence, if $\mu(x,y)=\mu(x+y)$ $(1)$ simplifies to $$ \mu'_w=\frac{3\mu}{w},\quad w=x+y, $$ from which you find your integrating factor.

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