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This problem starts as a 1st order ODE where i need to solve for $\phi(t)$. $$\frac{d\phi(t)}{dt}-D\phi(t)=-f(t)$$ with a final condition $\phi(T)=1$. $D$ is a constant. Using an integrating factor $I(t)=e^{-Dt}$ to solve the ODE for $\phi(t)$, I arrived at the following equation (1): $$\phi(t) = -e^{Dt}\int f(t)e^{-Dt}dt$$How do i go from the above to prove that: $$\phi(t) = e^{-D(T-t)}+\int_t^Te^{-D(\tau-t)}f(\tau) d\tau$$ I've tried to use integration by parts on equation (1) for the expression in the integral but that doesn't seem to converge to the required form.
Thank you.

Gnowl
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  • Does the integral in equation $(1)$ have no born ? Is it a constant that does not depend on $t$ ? – npisinp Apr 06 '14 at 04:07
  • What do you mean by "no born"? only $D$ is a constant, everything else depends on $t$. $T$ is also a constant. – Gnowl Apr 06 '14 at 09:17
  • How $\int f(t)e^{-Dt}dt$ depends on $t$? – npisinp Apr 06 '14 at 11:59
  • By "born" I meant the limit of the integral in equation (1). If we have $\int_{a}^{b} f(t)e^{-Dt}\text{d}t$ and $a$, $b$ do not depends on $t$ then the integral does not depend on $t$. In fact we cannot have $t$ both in the limit of the integral and in the integration variable. – npisinp Apr 06 '14 at 12:32
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    Thanks npisinp, I get your question now.
    – Gnowl Apr 06 '14 at 17:30
  • Thanks npisinp, I get your question now.
    the limits of integration does depend on $t$. The first equation is something i derived. If $f(t)$ was a defined integrable function of $t$, I would've used integration by parts or substitution to solve it. That would give me an answer with a remaining constant $C$, which i would then solve using the final condition.
    However, since $f(t)$ is not defined it is left as is. In the integral of the second equation that i'm trying to get to, the limits of integ. don't depend on $t$, hence the change of variable to $\tau$.     – Gnowl Apr 06 '14 at 17:43
  • Can you give me the original problem? I cannot do it without the limit of the integral in equation $(1)$. – npisinp Apr 06 '14 at 17:48
  • npisinp, I've just updated the post to include the original problem. Hopefully that helps to clarify the problem. thanks. – Gnowl Apr 07 '14 at 04:41
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    One solves equations and proves equalities. – Mariano Suárez-Álvarez Apr 07 '14 at 04:42

1 Answers1

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Hint: use definite integrals all the way through instead of indefinite integrals. This works the appropriate constant terms into the equations from the get go.

$$\frac{d}{dt}\left(e^{-Dt}\phi(t)\right)=-e^{-Dt}f(t)\\ \implies e^{-Dt}\phi(t) - e^{-DT}\phi(T)= -\int_T^t f(\tau)e^{-D\tau}d\tau$$

David H
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