User:Egm4313.s12.team11.sheider/R.6.6: Lecture 10 p.10-9

Problem Statement
Given: For the following differential equation: $$ y_{p}''+4y_{p}'+3y_{p} = 2 e^{2x} cos(3x) \!$$

With a particular solution of the form: $$ y_{p} = x e^{-2x}(Mcos(3x) + Nsin(3x)) \!$$

Verify that this solution has a final expression of $$ y_{p} = 6 e^{-2x} (Ncos(3x) - Msin(3x)) \!$$ as follows:

1) Simplify the term $$ y_{p}'' \!$$

2) Simplify the 2nd term $$4y_{p}'\!$$ and combine with the simplified first term

3) Finally add the 3rd term $$ 13y_{p}\!$$

4) Find the final expression for $$ y_{p}(x) \!$$

Solution
1) To find the derivative and simplify $$ y_{p}'' \!$$ we will use www.wolframalpha.com.

The "Simplify" command will be exploited as well as the notation for taking a derivative in mathematica. The following was entered into www.wolframalpha.com:

Simplify[D[D[x*exp(-2x)*(Mcos(3x) + Nsin(3x)),x],x]]

The answer that was yielded is as follows:

$$ y_{p}'' = -6Me^{-2x}sin(3x) + 12Me^{-2x}x sin(3x) - 4Me^{-2x}cos(3x) - 5Me^{-2x} x cos(3x)\!$$

$$ - 4Ne^{-2x}sin(3x) -5Ne^{-2x} x sin(3x) + 6Ne^{-2x}cos(3x) - 12Ne^{-2x} x cos(3x) \!$$

2) In the same fashion, $$4y_{p}'\!$$ was evaluated and simplified using www.wolframalpha.com

The following was entered into www.wolframalpha.com:

simplify[4*[D[x*exp(-2x)*(Mcos(3x) + Nsin(3x)),x]]]

The answer that was yielded is as follows:

$$ 4y_{p}' = -12Me^{-2x} x sin(3x) + 4M e^{-2x}cos(3x) - 8Me^{-2x} x cos(3x) + 4Ne^{-2x}sin(3x) - 8Ne^{-2x} x sin(3x) + 12Ne^{-2x} x cos(3x) \!$$

This term can be added to the previous, and then simplified yet again using www.wolframalpha.com

The following was entered into www.wolframalpha.com

Simplify[D[D[x*exp(-2x)*(Mcos(3x) + Nsin(3x)),x],x]+ 4*[D[x*exp(-2x)*(Mcos(3x) + Nsin(3x)),x]]

The answer that was yielded is as follows:

$$ y_{p}'' + 4y_{p}' = -6Me^{-2x}sin(3x) - 13Me^{-2x} x cos(3x) -13Ne^{-2x} x sin(3x) + 6Ne^{-2x}cos(3x) \!$$

3) The final term $$ 13y_{p} \!$$ is now added to this result, as follows:

$$ y_{p}'' + 4y_{p}' + 13y_{p} = -6Me^{-2x}sin(3x) - 13Me^{-2x} x cos(3x) -13Ne^{-2x} x sin(3x) + 6Ne^{-2x}cos(3x) + 13Me^{-2x} x cos(3x) +13Ne^{-2x} x sin(3x) \!$$

$$ y_{p}'' + 4y_{p}' + 13y_{p} = -6Me^{-2x}sin(3x) + (13-13)Me^{-2x} x cos(3x) + (13-13)Ne^{-2x} x sin(3x) + 6Ne^{-2x}cos(3x) \!$$

$$ y_{p}'' + 4y_{p}' + 13y_{p} = -6Me^{-2x}sin(3x) + 6Ne^{-2x}cos(3x) \!$$

$$ y_{p}'' + 4y_{p}' + 13y_{p} = 6e^{-2x}(Ncos(3x)-Msin(3x) ) \!$$

4) As verified above, the final term is as follows:

$$ y_{p}(x) = 6e^{-2x}(Ncos(3x)-Msin(3x) ) \!$$

--Egm4313.s12.team11.sheider (talk) 22:30, 10 April 2012 (UTC)