Solve equation : y''=y+e^{2x} .

D DarnItJimImAnEngineer Jun 2019 493 262 USA Sep 19, 2019 #2 $y = \Sigma c_n e^{\lambda_n x}$ $\lambda = 2$ 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 Reactions: 2 people

$y = \Sigma c_n e^{\lambda_n x}$ $\lambda = 2$ 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

skipjack Forum Staff Dec 2006 21,478 2,470 Sep 19, 2019 #3 $e^{-x}y'' - e^{-x}y = e^x \\ e^{-x}y' + e^{-x}y = e^x + 2\text{A} \\ e^xy' + e^xy = e^{3x} + 2\text{A}e^{2x} \\ e^xy = \frac13e^{3x} + \text{A}e^{2x} + \text{B} \\ y = \frac13e^{2x} + \text{A}e^x + \text{B}e^{-x}$ Reactions: 4 people

$e^{-x}y'' - e^{-x}y = e^x \\ e^{-x}y' + e^{-x}y = e^x + 2\text{A} \\ e^xy' + e^xy = e^{3x} + 2\text{A}e^{2x} \\ e^xy = \frac13e^{3x} + \text{A}e^{2x} + \text{B} \\ y = \frac13e^{2x} + \text{A}e^x + \text{B}e^{-x}$

idontknow Dec 2015 1,082 169 Earth Sep 21, 2019 #4 Looks similiar to Wronskian . \(\displaystyle dW(y,e^{-x} )=e^{x}dx\). Reactions: 1 person

V v8archie Math Team Dec 2013 7,712 2,682 Colombia Sep 21, 2019 #5 skipjack said: $e^{-x}y'' - e^{-x}y = e^x \\ e^{-x}y' + e^{-x}y = e^x + 2\text{A} \\ e^xy' + e^xy = e^{3x} + 2\text{A}e^{2x} \\ e^xy = \frac13e^{3x} + \text{A}e^{2x} + \text{B} \\ y = \frac13e^{2x} + \text{A}e^x + \text{B}e^{-x}$ Click to expand... 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 $y_p=Ae^{2x}$. Reactions: 1 person

skipjack said: $e^{-x}y'' - e^{-x}y = e^x \\ e^{-x}y' + e^{-x}y = e^x + 2\text{A} \\ e^xy' + e^xy = e^{3x} + 2\text{A}e^{2x} \\ e^xy = \frac13e^{3x} + \text{A}e^{2x} + \text{B} \\ y = \frac13e^{2x} + \text{A}e^x + \text{B}e^{-x}$ Click to expand... 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 $y_p=Ae^{2x}$.