(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Greetings,

I am having trouble understanding the solution to this DEQ equation. I am able to solve the equation using the variation of parameters technique, however, when it comes to simplifying the integral I am lost. It appears they are using the Fundamental Theorem of Calculus to simplify the integral, but I am confused as to how this is done?

2. Relevant equations

1) Original DEQ equation:

[itex]

y''-5y'+6y=g(t)

[/itex]

2) Solving the characteristic equation of the homogenous equation yields:

[itex]

y_{H}=c_{1}e^{2t}+c_{2}e^{3t}

[/itex]

3) Solving for [itex]y_{1}, y'_{1}, y_{2}, y'_{2}[/itex], and the Wronskian, and plugging into the theorem of variation of parameters yields:

[itex]

Y(t)=-e^{2t}\int{\frac{e^{3t}g(t)}{e^{5t}}dt}+e^{3t}\int{\frac{e^{2t}g(t)}{e^{5t}}dt}

[/itex]

4) Simplified:

[itex]

Y(t)=-e^{2t}\int{e^{-2t}g(t)dt}+e^{3t}\int{e^{-3t}g(t)dt}

[/itex]

5) BUT, Texbook Solution:

[itex]

Y(t)=\int{[e^{3(t-s)}-e^{2(t-s)}]g(s)ds}

[/itex]

3. The attempt at a solution

So, my problem is how to get from 4) to 5)? I realize they probably applied the FTC, but I am not clear on how to use this to derive the solution in the textbook? The farthest I can get is as below:

[itex]

Y(t)=-e^{2t}\int_{0}^{s}{e^{-2t}g(t)dt}+e^{3t}\int_{0}^{s}{e^{-3t}g(t)dt}

[/itex]

Any help is greatly appreciated.

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# Homework Help: DEQ/Calculus (FTC) Question Simplifying Nonintegrable Function

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