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

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

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y^{''}-y=1,y(0)=0

y^{^{''}} - y = 1, y (0) = 0

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

Solution steps

y^{''} (t) - y = 1

Solve linear ODE:

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

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

y^{^{''}} + 4 y^{^{'}} + 6 y = 1, + e^{-} t y (0) = 0, y^{^{'}} (0) = 0

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

y^{^{'''}} + y^{^{''}} + y^{^{'}} + y = e^{- t} + 14 t

y^{^{''}} - 64 y = 16

\frac{d ^{2} y}{d x ^{2}} + 3 \frac{d y}{d x} + 2 y = sin (e^{x})

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