What is the solution for y^{''}-y^'-2y=1 ?

The solution for y^{''}-y^'-2y=1 is y=c_{1}e^{2t}+c_{2}e^{-t}-1/2

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y^{''}-y^'-2y=1

y^{^{''}} - y^{^{'}} - 2 y = 1

y = c_{1} e^{2 t} + c_{2} e^{- t} - \frac{1}{2}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{- t} - \frac{1}{2}

y = c_{1} e^{2 t} + c_{2} e^{- t} - \frac{1}{2}

v^{^{''}} - 16 v = 2 e^{4 x}

y^{^{''}} + 2 y^{^{'}} + y = t^{5} e^{- t}

y^{^{''}} + 49 y = 17 sec^{2} (7 t)

y^{^{''}} + 4 y = (5 x^{2} + 4 x - 3) e^{x}

(\frac{1}{2}) y^{^{''}} + y^{^{'}} + 5 y = sin (3 t)

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