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I need to find the solution to the following differential equation: $$y'+a(x)y=R(x)$$
With $a(x)$ and $R(x)$ smooth functions.

I can re-order this to the following:

$$1.dy+(a(x)y-R(x))dx=0$$

$Q:=1$
$P:=(a(x)y-R(x))$

First thing I need to do is finding out if the equation is exact by checking if $\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}$ is true. I didn't add the calculation because it's pretty easy.

I now need to find an integrating factor. If the following function:
$$g(x):= \frac{1}{Q}(\frac{\partial P}{\partial y}-\frac{\partial Q}{\partial x})=a(x)$$ Is not dependent of y, then $e^{\int g(x)dx} = e^{\int a(x)dx}$ is a valid integrating factor.

This means that
$$(a(x)y-R(x)).e^{\int a(x)dx}dx+e^{\int a(x)dx}dy=0$$
should be an exact differential equation.

Applying the same formulas as earlier (with the new $P$ and $Q$) verifies that since the result to both partial derivatives is $a(x)e^{\int a(x)dx}$

Now to find the solution I need to integrate $P$ for $x$ and $Q$ for $y$ but here's where I get in trouble:

$\int P\ dx =\int (a(x).y-R(x))e^{\int a(x)dx}dx$

How do I integrate this? Is it even possible?

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$$ \frac{d}{dx}\left(e^{\int a(x)dx}y\right) = e^{\int a(x)dx}R(x) \\ e^{\int a(x)dx}y(x)=\int e^{\int a(x)dx}R(x)dx +C \\ y(x) = e^{-\int a(x)dx}\int e^{\int a(x)dx}R(x)dx+Ce^{-\int a(x)dx} $$

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  • $\begingroup$ I'm not sure how you got to the first line. I understand how you got the left part and the right part but why are they equal? $\endgroup$ – Joshua Jun 4 '15 at 16:29
  • $\begingroup$ @Joshua : differentiate the left side of the first line and cancel the common exponential factors from both sides, and you'll see that they're equal. $\endgroup$ – Disintegrating By Parts Jun 4 '15 at 17:24

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