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}

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