What is the solution for (d^2y)/(dt^2)-y=e^t,y(0)=1,y^'(0)=1 ?

The solution for (d^2y)/(dt^2)-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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(d^2y)/(dt^2)-y=e^t,y(0)=1,y^'(0)=1

\frac{d ^{2} y}{d t ^{2}} - 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}

a y^{^{''}} - 2 a y^{^{'}} - 3 a y = x^{2}

y^{^{''}} - 5 y^{^{'}} + 6 y = 10 sin (2 x) + 2 e^{3 x}

y^{^{''}} - 4 y^{^{'}} + 4 y = 1 - 4 e^{t}

y^{^{''}} + 5 y^{^{'}} + 6 y = e^{2 x}

y^{^{''}} + 6 y + 9 = 0

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