What is the solution for y^{''}-4y=4t-8e^{-2t},y(0)=0,y^'(0)=5 ?

The solution for y^{''}-4y=4t-8e^{-2t},y(0)=0,y^'(0)=5 is y=e^{2t}-e^{-2t}-t+2e^{-2t}t

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y^{''}-4y=4t-8e^{-2t},y(0)=0,y^'(0)=5

y^{^{''}} - 4 y = 4 t - 8 e^{- 2 t}, y (0) = 0, y^{^{'}} (0) = 5

y = e^{2 t} - e^{- 2 t} - t + 2 e^{- 2 t} t

Solution steps

y^{''} (t) - 4 y = 4 t - 8 e^{- 2 t}

Solve linear ODE:

y = e^{2 t} - e^{- 2 t} - t + 2 e^{- 2 t} t

y = e^{2 t} - e^{- 2 t} - t + 2 e^{- 2 t} t

y^{^{''}} + 4 y = x - x^{2}

y^{^{''}} - 6 y^{^{'}} + 5 y = e^{- 3 x}

y^{^{''}} - y^{^{'}} = x^{2} - 4

y^{^{''}} + y = \frac{1}{x ^{2}}

\frac{d ^{2} y}{d t ^{2}} + 2 \cdot \frac{d y}{d t} + 2 y = 11 t^{3}

What is the solution for y^{''}-4y=4t-8e^{-2t},y(0)=0,y^'(0)=5 ?
The solution for y^{''}-4y=4t-8e^{-2t},y(0)=0,y^'(0)=5 is y=e^{2t}-e^{-2t}-t+2e^{-2t}t