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}

\frac{5}{3} y^{^{''}} + 10 y^{^{'}} + 30 y = 300

3 y^{^{''}} - 4 y^{^{'}} + y = x - 7

\frac{d ^{2} y}{d x ^{2}} + y = sec (x)

y^{^{''}} - 6 y^{^{'}} + 9 y = 2 e^{3 x}

y^{^{''}} - y^{^{'}} = e^{x} cos (x)

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