Question

Use the Method of Undetermined Coefficients to find the general solution for the differential equation: y"-2y'+2y= e^(x)sinx

Answer should be: y= ce^(x)cosx+ce^(x)sinx-(x/2)e^(x)cosx

Answer 1

$y''\:-2y'\:+2y=e^x\sin \left(x\right)$.................(1)

for homogeneous system find roots

$y''\:-2y'\:+2y=0$

$r_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$a=1,\:b=-2,\:c=2$

$r_{1,\:2}=\frac{-\left(-2\right)\pm \sqrt{\left(-2\right)^2-4\cdot \:1\cdot \:2}}{2\cdot \:1}$

$r_{1,\:2}=\frac{2\pm \sqrt{4-8}}{2}$

$r_{1,\:2}=\frac{2\pm \sqrt{-4}}{2}$

$r_{1,\:2}=\frac{2\pm 2i}{2}$

$r_{1,\:2}=1\pm i$

for complex roots complementary solution is

$y_c=e^{ax}\left(c_1\cos \left(bx\right)+c_2\sin \left(bx\right)\right)$

${\color{Red} y_c= e^x\left(c_1\cos \left(x\right)+c_2\sin \left(x\right)\right)}$

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here  $g\left(x\right)=e^x\sin \left(x\right)$ so assume that particular solution is in the form of

$y=a_0xe^x\sin \left(x\right)+a_1xe^x\cos \left(x\right)$................(2)

take first derivative

$y'= \frac{d}{dx}\left(a_0xe^x\sin \left(x\right)+a_1xe^x\cos \left(x\right)\right)$

$y'= a_0\left(e^x\sin \left(x\right)+x\left(e^x\sin \left(x\right)+e^x\cos \left(x\right)\right)\right)+a_1\left(e^x\cos \left(x\right)+x\left(e^x\cos \left(x\right)-e^x\sin \left(x\right)\right)\right)$

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take second derivative

$y''= \frac{d}{dx}\left( a_0\left(e^x\sin \left(x\right)+x\left(e^x\sin \left(x\right)+e^x\cos \left(x\right)\right)\right)+a_1\left(e^x\cos \left(x\right)+x\left(e^x\cos \left(x\right)-e^x\sin \left(x\right)\right)\right) \right)$

$y''= a_0\left(2e^xx\cos \left(x\right)+2e^x\sin \left(x\right)+2e^x\cos \left(x\right)\right)+a_1\left(-2e^xx\sin \left(x\right)+2e^x\cos \left(x\right)-2e^x\sin \left(x\right)\right)$

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put all value in equation 1

$y''\:-2y'\:+2y=e^x\sin \left(x\right)$

$a_0\left(2e^xx\cos \left(x\right)+2e^x\sin \left(x\right)+2e^x\cos \left(x\right)\right)+a_1\left(-2e^xx\sin \left(x\right)+2e^x\cos \left(x\right)-2e^x\sin \left(x\right)\right)-2\left (a_0\left(e^x\sin \left(x\right)+x\left(e^x\sin \left(x\right)+e^x\cos \left(x\right)\right)\right)+a_1\left(e^x\cos \left(x\right)+x\left(e^x\cos \left(x\right)-e^x\sin \left(x\right)\right)\right)\right )+2\left (a_0xe^x\sin \left(x\right)+a_1xe^x\cos \left(x\right)\right )=e^x\sin \left(x\right)$

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$2a_0e^xx\cos \left(x\right)+2a_0e^x\sin \left(x\right)+2a_0e^x\cos \left(x\right)-2a_1e^xx\sin \left(x\right)+2a_1e^x\cos \left(x\right)-2a_1e^x\sin \left(x\right)-2a_0e^x\sin \left(x\right)-2a_0xe^x\sin \left(x\right)-2a_0xe^x\cos \left(x\right)-2a_1e^x\cos \left(x\right)-2a_1xe^x\cos \left(x\right)+2a_1xe^x\sin \left(x\right)+2a_0xe^x\sin \left(x\right)+2a_1xe^x\cos \left(x\right)=e^x\sin \left(x\right)$

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$2e^x\left(a_0\cos \left(x\right)-a_1\sin \left(x\right)\right)=e^x\sin \left(x\right)$

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$2a_0e^x\cos \left(x\right)-2a_1e^x\sin \left(x\right)=1\cdot \:e^x\sin \left(x\right)$

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compare coefficient both sides

$0=2a_0...........................{\color{Blue} a_0=0}$

$1=-2a_1...........................{\color{Blue} a_1=-\frac{1}{2}}$

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put both constant in equation 2

$y=a_0xe^x\sin \left(x\right)+a_1xe^x\cos \left(x\right)$

$y=0\cdot \:xe^x\sin \left(x\right)+\left(-\frac{1}{2}\right)xe^x\cos \left(x\right)$

$y=0-\frac{e^xx\cos \left(x\right)}{2}$

${\color{Red} y_p=-\frac{e^xx\cos \left(x\right)}{2}}$

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general solution is

$y=y_c+y_p$

$y=y=e^x\left(c_1\cos \left(x\right)+c_2\sin \left(x\right)\right)-\frac{e^xx\cos \left(x\right)}{2}$

${\color{Red} y=c_1e^x\cos \left(x\right)+c_2e^x\sin \left(x\right)- \frac{x}{2}e^x\cos \left(x\right)}$