What is the solution for y^{''}+y-2e^{-t}-1=0 ?

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

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y^{''}+y-2e^{-t}-1=0

y^{^{''}} + y - 2 e^{- t} - 1 = 0

y = c_{1} cos (t) + c_{2} sin (t) + e^{- t} + 1

Solution steps

y^{''} (t) + y - 2 e^{- t} - 1 = 0

Solve linear ODE:

y = c_{1} cos (t) + c_{2} sin (t) + e^{- t} + 1

y = c_{1} cos (t) + c_{2} sin (t) + e^{- t} + 1

\frac{d ^{2}}{d x ^{2}} (θ (x)) + 4 θ (x) = csc^{2} (2 x)

(D^{4} + D^{2}) y = 3 x^{2} + 4 sin (x) - 2 cos (x)

y^{^{''}} + 2 y = - 18 x e^{2 x}

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

y^{^{''}} + y = sin (4 x)

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