What is the solution for y^{''}-y=e^{-t},y(0)=1,y^'(0)=0 ?

The solution for y^{''}-y=e^{-t},y(0)=1,y^'(0)=0 is y=(1-1/4)e^t+1/4 e^{-t}-(e^{-t}t)/2

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y^{''}-y=e^{-t},y(0)=1,y^'(0)=0

y^{^{''}} - y = e^{- t}, y (0) = 1, y^{^{'}} (0) = 0

y = (1 - \frac{1}{4}) e^{t} + \frac{1}{4} e^{- t} - \frac{e ^{- t} t}{2}

Solution steps

y^{''} (t) - y = e^{- t}

Solve linear ODE:

y = (1 - \frac{1}{4}) e^{t} + \frac{1}{4} e^{- t} - \frac{e ^{- t} t}{2}

y = (1 - \frac{1}{4}) e^{t} + \frac{1}{4} e^{- t} - \frac{e ^{- t} t}{2}

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f (x) = e^{5 x} cos (π x), at x = 1

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What is the solution for y^{''}-y=e^{-t},y(0)=1,y^'(0)=0 ?
The solution for y^{''}-y=e^{-t},y(0)=1,y^'(0)=0 is y=(1-1/4)e^t+1/4 e^{-t}-(e^{-t}t)/2