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

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

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

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

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}

y^{^{''}} - 4 y^{^{'}} + 13 y = 3 e^{2 x} - 5 e^{3 x}

y^{^{''}} - y = cosh (3 x)

y^{^{''}} - 4 y^{^{'}} + 4 y = - 5 e^{2 x}

y^{^{''}} - 2 y^{^{'}} + y = 10 cos (x) + 8 sin (x)

y^{^{''}} + 10 y^{^{'}} + 16 y = 949 sin (3 x)

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