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

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

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y^{''}-y=2,y(0)=2,y^'(0)=0

y^{^{''}} - y = 2, y (0) = 2, y^{^{'}} (0) = 0

y = 2 e^{t} + 2 e^{- t} - 2

Solution steps

y^{''} (t) - y = 2

Solve linear ODE:

y = 2 e^{t} + 2 e^{- t} - 2

y = 2 e^{t} + 2 e^{- t} - 2

y^{^{''}} - 2 y^{^{'}} + 5 y = sin (x)

y^{^{''}} + 4 y = 4 sec^{2} (2 t)

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

y^{^{''}} - y = sinh (x)

4 y^{^{''}} - y = e^{\frac{x}{2}} + 9

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