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

The solution for y^{''}-y^'-2y=1+t,y(0)=2,y^'(0)=0 is y=(-4/3+9/4)e^{2t}+4/3 e^{-t}-t/2-1/4

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

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

y = (- \frac{4}{3} + \frac{9}{4}) e^{2 t} + \frac{4}{3} e^{- t} - \frac{t}{2} - \frac{1}{4}

Solution steps

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

Solve linear ODE:

y = (- \frac{4}{3} + \frac{9}{4}) e^{2 t} + \frac{4}{3} e^{- t} - \frac{t}{2} - \frac{1}{4}

y = (- \frac{4}{3} + \frac{9}{4}) e^{2 t} + \frac{4}{3} e^{- t} - \frac{t}{2} - \frac{1}{4}

What is the solution for y^{''}-y^'-2y=1+t,y(0)=2,y^'(0)=0 ?
The solution for y^{''}-y^'-2y=1+t,y(0)=2,y^'(0)=0 is y=(-4/3+9/4)e^{2t}+4/3 e^{-t}-t/2-1/4