# Tricky Separable Differential Equation

$y' + ay +b = 0$ ($a$ not zero) is supposed to be separable and has solution

$y = ce^{-ax} - \frac ba$

Here is my start to this problem:

$\frac{dy}{dx} + ay = -b$ is as far as I can go with this. How should I go about separating $x$ and

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$\dfrac{dy}{dx} +ay+b = 0$

$\dfrac{dy}{dx} = -ay-b$

$\dfrac{dy}{ay+b} = -dx$

Now, can you solve this?

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Yes that makes perfect sense and yes I can solve this. The integral on left is a natural logarithm. Thank you so much. – Jules Manson Jun 26 '14 at 4:52

You want to rewrite the equation in the form $f(y)dy=g(x)dx$. There is no need to isolate the constant, as it can be considered a function of $y$ (a constant function). Hint: $g(x)=1$

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Of course now I see it. Thank you for your help. – Jules Manson Nov 25 '14 at 17:29

We can take the differential equation $$y' + ay +b = 0$$ and write $$y'+ay = -b$$ then multiply by $e^{ax}$: $$e^{ax}y'+ae^{ax}y = -5 e^{ax}$$

Now the term on the left becomes $$(ye^{ax})' = -5 e^{ax}.$$ Next antidifferentiate and solve for $y$.

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I like this alternative way to solve this. – Jules Manson Jun 26 '14 at 7:25
Thank you :) It is a favorite approach of mine. – Joel Jun 26 '14 at 13:13