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=(-3/4+11/4)e^t+3/4 e^{-2t}-t/2-3/4

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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{3}{4} + \frac{11}{4}) e^{t} + \frac{3}{4} e^{- 2 t} - \frac{t}{2} - \frac{3}{4}

Solution steps

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

Solve linear ODE:

y = (- \frac{3}{4} + \frac{11}{4}) e^{t} + \frac{3}{4} e^{- 2 t} - \frac{t}{2} - \frac{3}{4}

y = (- \frac{3}{4} + \frac{11}{4}) e^{t} + \frac{3}{4} e^{- 2 t} - \frac{t}{2} - \frac{3}{4}

y^{^{''}} + y = 2 tan (x)

\frac{1}{2} x^{^{''}} + \frac{1}{2} x^{^{'}} - 6 x = 10 cos (3 t)

y^{^{''}} + 4 y^{^{'}} + 4 y = 4 x + 5

y^{^{''}} + 4 y^{^{'}} + 4 y = 4 x + 6

y^{^{''}} - 2 y^{^{'}} - 8 y = - 16 t + 16 t^{2}

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=(-3/4+11/4)e^t+3/4 e^{-2t}-t/2-3/4