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

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

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

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

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}

y^{^{''}} + 5 y^{^{'}} + 6 y = 10 sin (x)

(D^{2} + 2 D + 2) y = x^{2} + x + 1

y^{^{''}} + y^{^{'}} = sin (3 x)

y^{^{''}} + y = 3 x^{2} + 4

y^{^{''}} + 4 y^{^{'}} + 5 y = 2 e^{- 4 x}, y (0) = 2, y^{^{'}} (0) = 3

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