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I'm trying to solve: $$y''-5y'+6y=12e^{2x}$$

The roots for the homogeneous equation are $r=2$ and $r=3$. Homogeneous solution is $Ae^{2x}+Be^{3x}$

Via method of undetermined coefficients I got the particular solution as: $$y_p=-12xe^{2x}$$

Now with variation of parameters I get the same answer along with a second term: $$y_p=-12xe^{2x}-12e^{2x}$$

Why is there a second term with Variation of Parameters?

Here's my working out:

Let:
$y_1$ = $e^{2x}$, $y_2$ = $e^{3x}$

Find the Wronskian: W= $$ \left[ \begin{array}{ c c } e^{2x} & e^{3x} \\ 2e^{2x} & 3e^{3x} \end{array} \right] $$ $$W=e^{5x}$$

Variation of Parameters formula:

$$y_p=-y_1\int \frac{y_2 g(x)}{W} dx +y_2\int \frac{y_1 g(x)}{W} dx$$

Let: $g(x)=12e^{2x}$

Plug in our values: $$y_p=-e^{2x}\int \frac{(e^{3x}) (12e^{2x})}{e^{5x}} dx +e^{3x}\int \frac{(e^{2x}) (12e^{2x})}{e^{5x}} dx$$

$$y_p=-e^{2x}\int \frac{12e^{5x}}{e^{5x}} dx +e^{3x}\int \frac{12e^{4x}}{e^{5x}} dx$$

$$y_p=-e^{2x}\int 12 \ dx +e^{3x}\int \frac{12}{e^{x}} dx$$

$$y_p=-12xe^{2x} +12e^{3x}\int e^{-x} \ dx$$ $$y_p=-12xe^{2x} +12e^{3x}(-e^{-x})$$ $$y_p=-12xe^{2x} -12e^{2x}$$ What is that second term doing here? Why isn't this WORKING!?

Please help me figure this out or I will cry manly tears from my tear ducts and it may wet my Stewart Calculus Textbook (5th Edition) thus reducing its selling value from 5 dollars to 4 when I sell it to the pirates my university's book store.

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2 Answers 2

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It doesn't matter that that second term is there. Why? I'll show you.

$y=\color{green}{y_c}+\color{red}{y_p}=\color{green}{Ae^{2x}+Be^{3x}}\color{red}{-12xe^{2x}-12e^{2x}}=\underbrace{(A-12)}_{\text{another arbitrary constant}}\cdot e^{2x}+Be^{3x}-12xe^{2x}.$

If we re-label $A-12$ as $C$, we get:

$$\boxed{y=Ce^{2x}+Be^{3x}-12xe^{2x}}.$$

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    $\begingroup$ Ahhhhhhhhhhhhhhhh... I get it now. It gets a little confusing when the constants keep the same name without a label change. I understand that it is for convenience instead of having $c_1$ morph to $c_2$ to $c_3$ to $c_4$ etc. every time it is subjected to a change. $\endgroup$
    – Ursa Major
    Jun 17, 2014 at 13:27
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I haven't checked your work, but there's nothing wrong with the second solution. The extra term is a solution of the homogeneous equation: By starting with any particular solution and adding a solution of the homogeneous equation you get another particular solution. (P.S. Nice touch of humor!)

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