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 September 19th, 2019, 11:44 AM #1 Senior Member   Joined: Dec 2015 From: Earth Posts: 826 Thanks: 113 Math Focus: Elementary Math Non-order DE Solve equation : $\displaystyle y''=y+e^{2x}$ .
 September 19th, 2019, 01:14 PM #2 Senior Member   Joined: Jun 2019 From: USA Posts: 380 Thanks: 205 $y = \Sigma c_n e^{\lambda_n x}$ $\lambda = 2$ 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 Thanks from topsquark and idontknow
 September 19th, 2019, 08:28 PM #3 Global Moderator   Joined: Dec 2006 Posts: 21,110 Thanks: 2326 $e^{-x}y'' - e^{-x}y = e^x \\ e^{-x}y' + e^{-x}y = e^x + 2\text{A} \\ e^xy' + e^xy = e^{3x} + 2\text{A}e^{2x} \\ e^xy = \frac13e^{3x} + \text{A}e^{2x} + \text{B} \\ y = \frac13e^{2x} + \text{A}e^x + \text{B}e^{-x}$ Thanks from topsquark, v8archie, idontknow and 1 others
 September 21st, 2019, 05:48 AM #4 Senior Member   Joined: Dec 2015 From: Earth Posts: 826 Thanks: 113 Math Focus: Elementary Math Looks similiar to Wronskian . $\displaystyle dW(y,e^{-x} )=e^{x}dx$. Thanks from topsquark
September 21st, 2019, 05:28 PM   #5
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Quote:
 Originally Posted by skipjack $e^{-x}y'' - e^{-x}y = e^x \\ e^{-x}y' + e^{-x}y = e^x + 2\text{A} \\ e^xy' + e^xy = e^{3x} + 2\text{A}e^{2x} \\ e^xy = \frac13e^{3x} + \text{A}e^{2x} + \text{B} \\ y = \frac13e^{2x} + \text{A}e^x + \text{B}e^{-x}$
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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