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I have this differential equation that needs solving:

$\frac{dx}{dt}+\alpha x=\beta$

Then the solution is supposed to be:

$x=e^{-\alpha t}(C+\beta\int\limits_0^t e^{\alpha y}dy)$

However if I use the simple approach of substituting $z=\beta-\alpha x$, I have

$\frac{dz}{z}=-\alpha\ dt$

Simple integration gives answer which is different from desired solution.

$ln\ z=-\alpha t+C$

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    $\begingroup$ Do you know Laplace transform? $\endgroup$ May 15 '20 at 12:48
  • $\begingroup$ will have to google $\endgroup$
    – user_1_1_1
    May 15 '20 at 12:50
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$$\ln\ z=-\alpha t+C$$ $$ z=Ce^{-\alpha t}$$ $$ \beta-\alpha x=Ce^{-\alpha t}$$ $$ \alpha x=-Ce^{-\alpha t}+\beta$$ $$ x=-\dfrac C {\alpha }e^{-\alpha t}+\dfrac {\beta}{\alpha}$$ $$ x(t)=Ke^{-\alpha t}+\dfrac {\beta}{\alpha}$$ Looks the same for me. $$x=e^{-\alpha t}(C+\beta\int\limits_0^t e^{\alpha y}dy)$$ $$x=e^{-\alpha t}(C+\dfrac {\beta}{\alpha}( e^{\alpha t}-1))$$ $$x=e^{-\alpha t}(C-\dfrac {\beta}{\alpha})+\dfrac {\beta}{\alpha}$$ This can ve rewritten as: $$x(t)=Ke^{-\alpha t}+\dfrac {\beta}{\alpha}$$ It's the same answer. $K$ is just a constant.

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