What is the solution for y^{''}+y=e^{-2pi},y(0)=0,y^'(0)=1 ?

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

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y^{''}+y=e^{-2pi},y(0)=0,y^'(0)=1

y^{^{''}} + y = e^{- 2 π}, y (0) = 0, y^{^{'}} (0) = 1

y = - \frac{1}{e ^{2 π}} cos (t) + sin (t) + \frac{1}{e ^{2 π}}

Solution steps

y^{''} (t) + y = e^{- 2 π}

Solve linear ODE:

y = - \frac{1}{e ^{2 π}} cos (t) + sin (t) + \frac{1}{e ^{2 π}}

y = - \frac{1}{e ^{2 π}} cos (t) + sin (t) + \frac{1}{e ^{2 π}}

y^{^{''}} - 6 y^{^{'}} + 8 y = x^{3} - x + 1

y^{^{''}} - y = e^{2 t} + 4 t \cdot e^{2 t} + 8 t^{2} \cdot e^{t}

(d^{2} + 9) y = sec^{3} (x)

9 y^{^{''}} + 4 y = 3 x + 12 e - x, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} - y = 4 cos (t), y (0) = 0, y^{^{'}} (0) = 1

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