DO4

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Find the general solution of (D^{2}-1)y=2x+e^{{2x}}\,.

The auxiliary equation is m^{2}-1=0\, with roots m=-1,1\,.

The complimentary solution is y_{c}=Ae^{x}+Be^{{-x}}\,.

Guess the particular solution based on the inhomogenous term of the DE: y_{p}=c_{1}x+c_{2}e^{{2x}}\,.

(D^{2}-1)y_{p}=4c_{2}e^{{2x}}-c_{1}x-c_{2}e^{{2x}}=2x+e^{{2x}}\,.

Equate coefficients.

3c_{2}=1,\,\,\,-c_{1}=2\,

So c_{2}=1/3,\,\,\,c_{1}=-2\,.

y_{p}(x)=-2x+{\frac  {1}{3}}e^{{2x}}\,

The general solution is y=y_{c}+y_{p}\,.

y(x)=Ae^{x}+Be^{{-x}}-2x+{\frac  {1}{3}}e^{{2x}}\,


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