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

The solution for y^{''}-y^'-2y=1,y(0)=1,y^'(0)=2 is y=(-1/3+3/2)e^{2t}+1/3 e^{-t}-1/2

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

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

y = (- \frac{1}{3} + \frac{3}{2}) e^{2 t} + \frac{1}{3} e^{- t} - \frac{1}{2}

Solution steps

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

Solve linear ODE:

y = (- \frac{1}{3} + \frac{3}{2}) e^{2 t} + \frac{1}{3} e^{- t} - \frac{1}{2}

y = (- \frac{1}{3} + \frac{3}{2}) e^{2 t} + \frac{1}{3} e^{- t} - \frac{1}{2}

\frac{d ^{2} y}{d x ^{2}} - \frac{d y}{d x} + y = x

y^{^{''}} - y^{^{'}} - 6 y = 32 sin (3 x)

y^{^{''}} - 2 y^{^{'}} + y = 576 e^{9 t}

y^{^{''}} + 6 y^{^{'}} + 9 y = 10 t^{3} e - 3 t + 2 t^{2} e - 3 t

3 y^{^{''}} + 2 y = x^{2} e^{- 3 x}

What is the solution for y^{''}-y^'-2y=1,y(0)=1,y^'(0)=2 ?
The solution for y^{''}-y^'-2y=1,y(0)=1,y^'(0)=2 is y=(-1/3+3/2)e^{2t}+1/3 e^{-t}-1/2