Proving the Solution to a First Order Linear Differential Equation

[name]

Hello all,

I'm trying to prove to myself that the following solution to the DE shown works. I can't start using it until i prove to my self it works (it's this psycological thing i have were i can't use anything unless i know where it comes from).

Here is the Equation and it's solution
http://img142.imageshack.us/img142/3437/defa7.png

and here is me trying to prove to my self it works...
http://img137.imageshack.us/img137/1831/desolhv6.png Am I doing anything wrong? Or can anyone please show me a proof which shows that this is a solution to the Differential Equation?

Thanks in advance edit: I don't think this is a homework question, as you know, I am just trying to prove it to my self.

 

[HallsofIvy]

Let me see if I can put into Latex what you have so others won't have to wait for those to load:

Your first reference asserts that the solution to the first order, linear, differential equation
\frac{dx}{dt}= ax(t)+ f(t)
with x(0) and f(t) given is
x(t)= e^{at}x(0)+ \int_0^t e^{a(t-s)}ds[/itex]<br /> <br /> 2) You method of solution is: an integrating factor for the problem is e^-at so<br /> e^{-at}\frac{dx}{dt}= ae^{-at}x(t)+ e^{-at}f(t)<br /> e^{-at}\frac{dx}{dt}- ae^{-at}x(t)= e^{-at}f(t)<br /> \frac{de^{-at}x}{dt}= e^{-at}f(t)<br /> <br /> Yes, so far this is completely correct. You then integrate to get<br /> e^{-at}x(t)= \int e^{-at}f(t)dt+ C<br /> so <br /> x(t)= e^{at}\int e^{-at}f(t)dt+ Ce^{at}<br /> and try to determine C by setting t= 0<br /> x(0)= e^{a0}\int e^{-a(0)}f(0)dt+ Ce^{a0}<br /> <b>That&#039;s</b> your mistake! You are treating the &quot;t&quot; inside the integral as if it were the same as the &quot;t&quot; outside. It&#039;s not- it&#039;s a &quot;dummy&quot; variable.<br /> Remember that \int_0^1 t^2dt= 3. You can&#039;t &quot;set&quot; t equal to 0 and declare that \int_0^1 0^2 dt= 3!<br /> <br /> Go back and use a different variable in your integral:<br /> e^{-at}x(t)= \int^t e^{-as}f(s)ds<br /> Notice the single &quot;t&quot; as a limit on the integral. That tells people we mean for the final result of the integral to be in the variable t. Also notice there is no &quot;C&quot;. Strictly speaking, that is included in the indefinite integral. A better technique, which you should learn, is to write that indefinite integral as a definite integral with a variable limit:<br /> e^{-at}x(t)= \int_{0}^t e^{-as}f(s)ds+ C<br /> I now have &quot;+ C&quot; because choosing a lower limit is the same as choosing a specific constant for the indefinite integral which we don&#039;t want to do yet.<br /> I took the lower limit as 0 because we know x(0). The upper limit is the variable t. Of course, if t= 0, that integral is from 0 to 0 and so is 0 no matter what is being integrated:<br /> e^{-a(0)}x(0)= x(0)= \int_0^0 e^{-as}f(s)ds+ C= C<br /> so<br /> e^{-at}x(t)= \int_0^t e^{-as}f(s)ds+ x(0)<br /> Now multiply by e^at to get<br /> x(t)= e^{at}\int_0^t e^{-as}f(s)ds+ x(0)e^{at}<br /> x(t)= \int_0^t e^{a(t-s)}f(s)ds+ x(0)e^{at}<br /> as claimed. (Of course we can take that e^at inside the integral as if it were a constant because it does not depend on the variable of integration, s.)<br /> <br /> (You don&#039;t <b>think</b> this is a homework question? Don&#039;t you <b>know</b> for sure?<img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f642.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":smile:" title="Smile :smile:" data-smilie="1"data-shortname=":smile:" />

 

[name]

Let me see if I can put into Latex what you have so others won't have to wait for those to load:

...

(You don't think this is a homework question? Don't you know for sure?

HallsofIvy,

Thank you very much for that. What a disgrace, this is even a fundamental part of first year calculas!. I kind of knew something was wrong in that line - hence those red question marks.

As for the homework part, What i meant to say was that i don't think this should be in the homework section (I wasn't sure what constitutes as "homework" in this forum). And Latex looks powerful, I think i'd better learn it.