I was studying "continuous and discrete signals and systems" by Samir S. Soliman where I encountered with this first order differential equation: $$ \frac{dy(t)}{dt} + \frac{R_1R_2}{L(R_1+R_2)}y(t) = \frac{R_2}{L(R_1+R_2)}x(t) $$ They mention that in next line " To compute an explicit expression for y(t) in terms of x(t), we must solve the differential equation for an arbitrary input x(t) applied for $\ t >= t_o $ . The complete solution is of the form "

$$ y(t) = y(t_o) exp[-\frac{R_1R_2}{L(R_1+R_2)}(t-t_o)] + \frac{R_2}{L(R_1+R_2)}\int_{t_o}^{t}exp[-\frac{R_1R_2}{L(R_1+R_2)}(t-\tau)]x(\tau) d\tau ;t >= t_o $$

But there is no explanation how they get this result in detail. How did the solve this? How did they get this limit or$\ \tau $ . Please explain this total solution in detail (step by step approach to solve this kind of problems is appreciated ) .

I have introductory knowledge on differential equation solving but never encountered this kind of problem.

  • 1
    $\begingroup$ This is called variation of parameters. There is a Wikipedia article about it to get you started. And I have no doubt that searching for "variation of parameters" here will provide dozens of examples, and explanations. $\endgroup$ – Chappers Jun 16 '15 at 19:12
  • $\begingroup$ $\tau$ is just a dummy variable for the integral. It takes the place of $t$ since the integral limits involve $t$. Another common choice of integration variable is $s$. $\endgroup$ – Zach Jun 16 '15 at 19:18
  • $\begingroup$ Thnx.. I got the basic idea. But can't figure out where they get $(t-t_o) $ in first place . shouldn't it be only t ? $\endgroup$ – Anklon Jun 16 '15 at 19:44
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    $\begingroup$ mathworld.wolfram.com/IntegratingFactor.html A far simpler method than variation of parameters. $\endgroup$ – grdgfgr Jun 16 '15 at 19:58

General Solution (using the method of integrating factor):

Let $\alpha = \frac{R_1R_2}{L(R_1+R_2)}$ and $\beta=\frac{R_2}{L(R_1+R_2)}$. The linear inhomogeneous differential equation becomes:

$$\frac{dy(t)}{dt}+\alpha y(t) = \beta x(t)$$

Let $\mu(t) = e^{\int \alpha dt} = e^{\alpha t}$ and multiply both sides by $\mu(t)$: $$e^{\alpha t} \frac{dy(t)}{dt}+\alpha e^{\alpha t} y(t) = \beta e^{\alpha t} x(t)$$

Substitute $\alpha e^{\alpha t} = \frac{d}{dt}(e^{\alpha t})$: $$e^{\alpha t} \frac{dy(t)}{dt}+\frac{d}{dt}(e^{\alpha t}) y(t) = \beta e^{\alpha t} x(t)$$ Apply the reverse product rule $g \frac{df}{dt}+f \frac{dg}{dt} = \frac{d}{dt}(f g)$ to the left-hand side: $$\frac{d}{dt}(e^{\alpha t} y(t)) = \beta e^{\alpha t} x(t)$$ Integrate both sides with respect to $t$: $$\int \frac{d}{dt}(e^{\alpha t} y(t)) dt = \int \beta e^{\alpha t} x(t) dt$$ Evaluate the integrals: $$e^{\alpha t} y(t) = \beta \int e^{\alpha t} x(t) dt+C$$ where $C$ is an arbitrary constant.

Divide both sides by $\mu(t) = e^{\alpha t}$ to get the general solution: $$y(t) = \beta e^{-\alpha t} \int e^{\alpha t} x(t) dt+C e^{-\alpha t}$$

I will leave the rest for you. Give it a shot.


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