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Differential Equations Ordinary and Partial Differential Equations Math Forum


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September 19th, 2019, 11:44 AM   #1
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Non-order DE

Solve equation : $\displaystyle y''=y+e^{2x}$ .
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September 19th, 2019, 01:14 PM   #2
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$y = \Sigma c_n e^{\lambda_n x}$
$\lambda = 2$ or $\lambda^2 - 1 = 0$
$\rightarrow y=\frac{1}{3} e^{2x} + c_1 e^x + c_2 e^{-x}$
$c_1, ~c_2$ from initial/boundary conditions
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September 19th, 2019, 08:28 PM   #3
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$e^{-x}y'' - e^{-x}y = e^x \\
e^{-x}y' + e^{-x}y = e^x + 2\text{A} \\
e^xy' + e^xy = e^{3x} + 2\text{A}e^{2x} \\
e^xy = \frac13e^{3x} + \text{A}e^{2x} + \text{B} \\
y = \frac13e^{2x} + \text{A}e^x + \text{B}e^{-x}$
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September 21st, 2019, 05:48 AM   #4
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Looks similiar to Wronskian . $\displaystyle dW(y,e^{-x} )=e^{x}dx$.
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September 21st, 2019, 05:28 PM   #5
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Quote:
Originally Posted by skipjack View Post
$e^{-x}y'' - e^{-x}y = e^x \\
e^{-x}y' + e^{-x}y = e^x + 2\text{A} \\
e^xy' + e^xy = e^{3x} + 2\text{A}e^{2x} \\
e^xy = \frac13e^{3x} + \text{A}e^{2x} + \text{B} \\
y = \frac13e^{2x} + \text{A}e^x + \text{B}e^{-x}$
Very nice.

The more standard approach would be to solve the characteristic polynomial for $y''-y=0$ ($r=\pm1$) and them find a particular solution for the original equation, probably using the method of undetermined coefficients with $y_p=Ae^{2x}$.
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