I'm trying to learn to solve differential equations. Currently in class we are discussing integration factors. Here is one of the problems on the homework -

$$(3xe^y+2y)dx + (x^2e^y+x)dy=0\tag{1}$$

We can quickly see that if we take $\frac{\partial}{\partial y}$ of the left term and $\frac{\partial}{\partial x}$ of the right term and we see that they are not equal, so we need an integrating factor.

An integrating factor of $x$ gets us a bit farther:

$$(3x^2e^y+2xy)dx + (x^3e^y+x^2)dy=0\tag{2}$$

And now when we take $\frac{\partial}{\partial y}$ of the left term and $\frac{\partial}{\partial x}$ of the right term , the two are equal.

From here, I think we would integrate the left term with respect to $x$ and right term with respect to $y$ as follows -

$$\int {(3x^2e^y+2xy)dx} + \int (x^3e^y+x^2)dy = 0 \tag{3}$$ $$2x^3e^y+2x^2y+C=0\tag{4}$$ $$x^3e^y+x^2y=C\tag{5}$$

Am I thinking about this the right way?

Also, in $(5)$, is it fine to move the Constant to the right and absorb the common coefficient of 2 into it to simplify the expression? $(5)$ is the answer in the book and (4) was my answer. Either I have done something wrong, or I am confused in regard to what may be combined into the constant $C$.


$$(3xe^y+2y)dx + (x^2e^y+x)dy=0\tag{1}$$ multiply by X as integrating factor as you wrote $$(3x^2e^y+2yx)dx + (x^3e^y+x^2)dy=0$$ Rearrange terms $$(3x^2e^ydx+x^3e^ydy)+(2yxdx + x^2dy)=0$$ $$d(x^3e^y)+d(x^2y)=0$$ After integration $$x^3e^y+x^2y=K$$ For the constant yes it absorbs the factor 2 and you can have it on the other side. Your result is correct but I wouldn't write the integrals you wrote... $$2x^3e^y+2x^2y+C=0$$ $$\implies x^3e^y+x^2y=-\frac C 2$$ Substitute $$K=-\frac C2 \implies x^3e^y+x^2y=K$$ K is just a constant.

  • $\begingroup$ Thank you, this is helpful. Could you expand on why you wouldn't write the integrals that I wrote? $\endgroup$
    – blueether
    Nov 15 '18 at 7:49
  • $\begingroup$ I used to write them separeately not in an single equation..@blueether How can you justify the step from equation 2 to 3 ?? But if the teacher allows you to do that ...thats it $\endgroup$ Nov 15 '18 at 7:57
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    $\begingroup$ Ah, I see. I've been away from the math for a decade or so and trying to get back into it. I was figuring, since the equation is equal to zero I could move one of the terms to the other side, then integrate both sides with respect to x on the left and y on the right (I simply didn't bother moving anything around). I appreciate your feedback, being an adult and back in school I am far less interested in what the teacher allows and would rather learn to do things the proper way. Style is certainly important. $\endgroup$
    – blueether
    Nov 15 '18 at 8:02
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    $\begingroup$ Well you did a good job except for that step 2 to 3...@blueether note that when you integrate a function of two variable with respect to one variable then the constant of integration is a function of the other variable..For the constant as you can see you made ,no mistake and your answer is the same as that of your book Call the constant C or K these are just names thats all $\endgroup$ Nov 15 '18 at 8:04

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