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I have to solve an ODE:$$2(y-3x)dx+x\left(3-\frac{4x}{y} \right) dy=0$$

I am given that I have to use the integrating factor x^a(y^b) where a and b are real numbers in order to turn the problem into a solvable exact ODE. The problem is that I am unsure of how to find these real numbers. From my understanding, an ODE is exact with the integrating factor (I) if the partial derivative of the dx w.r.t y is equal to the partial derivative of the dy w.r.t x (both multiplied with I). I tried doing this but I am unsure of how to proceed with the algebra. So far I have:

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If anyone can help with the algebra to find the values for a and b that would be appreciated

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  • $\begingroup$ Yes I am sure it is the question given $\endgroup$
    – noov101
    Commented May 15, 2020 at 21:47

2 Answers 2

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Multiply by $y$ the DE: $$2y^2dx-6xydx+3xydy-4x^2dy=0$$ The integrating factor is $\mu (x,y)=xy$ $$y^3dx^2-6(xy)^2dx+3(xy)^2dy-2x^3dy^2=0$$ $$y^3dx^2-2y^2dx^3+x^2dy^3-2x^3dy^2=0$$ Rearrange some terms: $$(y^3dx^2+x^2dy^3)-2(y^2dx^3+x^3dy^2)=0$$ $$dx^2y^3-2dx^3y^2=0$$ Integration gives us: $$x^2y^3-2x^3y^2=K$$


To summarize the integrating factor should be $\mu (x,y)=xy^2$.Then the DE becomes exact.

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  • $\begingroup$ This seemed to work, thank you very much $\endgroup$
    – noov101
    Commented May 16, 2020 at 9:53
  • $\begingroup$ You'rer welcomed @noov101 $\endgroup$ Commented May 16, 2020 at 10:37
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In order to make the differential equation exact, there must be a function $F$ for which: $$0=dF= \frac{\partial F}{\partial x}dx+\frac{\partial F}{\partial y}dy$$ So for this problem, we need to find $a$, $b$ and $F$ so that, when we multiply both sides of the expression by $x^ay^b$: $$\begin{align*} \frac{\partial F}{\partial x} &= x^ay^b(2(y-3x)) = 2x^ay^{b+1}-6x^{a+1}y^b\\ \frac{\partial F}{\partial y} &= x^ay^b\left(x\left(3-\frac{4x}{y} \right) \right)= 3x^{a+1}y^b-4x^{a+2}y^{b-1} \end{align*}$$ One way to find $F$ is to integrate one of these two expressions with respect to the variable in the partial derivative, differentiate with respect to the other variable, and then equate it to the other expression. The other way, which you mention, is to take the derivative with respect to the other variable and exploit the fact that the mixed partials must be equal.

We'll take the latter approach. The useful step is that we may first simplify the expressions before we differentiate. Having done so already, we now compute: $$\begin{align*} F_{xy}&= 2(b+1)x^a y^b -6b x^{a+1}y^{b-1} \\ F_{yx}&= 3(a+1)x^ay^b-4(a+2)x^{a+1}y^{b-1} \end{align*}$$ Equating these two expressions, and more specifically noting that their respective coefficients should match, we find some straightforward equations for $a$ and $b$.

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