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I have the following differential equation ;

$$\frac{\partial z}{\partial t}+\alpha z\left(t\right)=y\left(t\right)$$

I tried to find the general solution by multiplying two sides by $e^{\alpha t}$ but it did not work and I can not figure out how to find the general solution.

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  • $\begingroup$ What did you obtain after multiplying by $\exp(\alpha t)$? Where do you get stuck? $\endgroup$ – TZakrevskiy May 21 '15 at 11:14
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    $\begingroup$ don't you have to specify what $y(t)$ is first? $\endgroup$ – David Quinn May 21 '15 at 11:14
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I like to do it like this:

Substitute $z=uv$

$u'v+uv'+a uv=y$

If I had no unprimed function, then I could simply integrate the expression to find the answer. So I will focus on the second and the third term. I will make it zero. Since I choose $u$ and $v$ I can do this:

$$uv'+a uv=0$$

$$\frac{v'}{v}=-a$$

$$v=e^{-at}$$

OR if you want to be more general:

$$v=e^{-\int a dt}$$

This will work if a is a function of t

Substitute this back to the original equation, remember the middle two terms are zero:

$$u'e^{-at}=y$$

$$u'=ye^{at}$$

$$u=\int ye^{at}dt+C$$

Combine $z=uv$

$$z=e^{-at}(\int ye^{at}dt+C)$$

Or using the more general expression:

$$z=e^{-\int a dt}(\int ye^{\int a dt}dt+C)$$

This formula will work for ALL first order linear differential equations (nonconstant coefficients and nonhomogenous). Just divide by the leading coefficient.

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