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I have the following differential equation ;

$$\frac{dx\left(t\right)}{dt}=ay\left(t\right)-bx\left(t\right)$$

where $a$ and $b$ are positive constant terms. $t$ indicates time.

I am trying to solve in the following manner ; First, I put $bx(t)$ in the LHS and multiply two sides by $e^{bt}$ ;

$$\frac{dx\left(t\right)}{dt}e^{bt}+bx\left(t\right)e^{bt}=ay\left(t\right)e^{bt}$$

After, I write ;

$$\frac{d\left(x\left(t\right)e^{bt}\right)}{dt}=ay\left(t\right)e^{bt}$$

My objective is to solve this integral between time $0$ and $\infty$ ;

$$\int_{0}^{\infty}\frac{d\left(x\left(t\right)e^{bt}\right)}{dt}=\int_{0}^{\infty}ay\left(t\right)e^{bt}$$

which yields

$$\left[x\left(t\right)e^{bt}\right]_{0}^{\infty}=\int_{0}^{\infty}ay\left(t\right)e^{bt}$$

The problem is that the left hand side explodes and goes to infity.

How can I solve this differential equation in a correct way ?

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    $\begingroup$ You cannot tell if the LHS explodes at $\infty$. This depends on the right side! If it is finite, then the left side obviously cannot be infinite. This would mean that $x(\infty)=0$. But first you might try integrating from $t_0$ to $t$ to obtain $x(t)$ explicitly, then ask what happens in the limit. $\endgroup$ – krvolok May 4 '15 at 11:13
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    $\begingroup$ Two observations: 1. You only multiply the LHS by $e^{bt}$ and do not change the RHS. 2. Why do you integrate till $\infty?$ $\endgroup$ – gammatester May 4 '15 at 11:14
  • $\begingroup$ @gammatester It was a typo, you are right. I integrate till $\infty$ to because normally I have a maximization program between $0$ and $\infty$ (I did not put the whole program) $\endgroup$ – optimal control May 4 '15 at 11:20
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Actually it yields result as

$x(t)=abe^{-bt} (\int^t y(s)e^{bs}\,\mathrm d s) +C e^{-bt}$

And this is less likely but is gonna depend on what your function $y(t)$ is but in most of the cases the exponential might just make it move to $0$. Probably when $b>0$. Here, I haven't applied the limits because you can use them later to replace the constant C.

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  • $\begingroup$ Could you replace dummy variables from $t$ to something else maybe? This might look confusing. :) $\endgroup$ – krvolok May 4 '15 at 11:29
  • $\begingroup$ Eh , do it , I have to go work, can't do at the moment :(. No problem! $\endgroup$ – Mann May 4 '15 at 11:31

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