What is the solution for y^{''}+4y^'+5y=34e^{2t},y(0)=1,y^'(0)=2 ?

The solution for y^{''}+4y^'+5y=34e^{2t},y(0)=1,y^'(0)=2 is y=e^{-2t}(-cos(t)-4sin(t))+2e^{2t}

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y^{''}+4y^'+5y=34e^{2t},y(0)=1,y^'(0)=2

y^{^{''}} + 4 y^{^{'}} + 5 y = 34 e^{2 t}, y (0) = 1, y^{^{'}} (0) = 2

y = e^{- 2 t} (- cos (t) - 4 sin (t)) + 2 e^{2 t}

Solution steps

y^{''} (t) + 4 y^{'} (t) + 5 y = 34 e^{2 t}

Solve linear ODE:

y = e^{- 2 t} (- cos (t) - 4 sin (t)) + 2 e^{2 t}

y = e^{- 2 t} (- cos (t) - 4 sin (t)) + 2 e^{2 t}

y^{^{''}} = 32 e^{2 t}, y (0) = 3, y^{^{'}} (0) = 8

\int_{a}^{b} \frac{1}{b - a}

16 y^{^{''}} + 24 y^{^{'}} + 9 y = 0, y (2) = 4, y^{^{'}} (- 2) = - 3

- 3 cos (- 3 x)

\frac{d}{d x} (x - 1 - ln (x))

What is the solution for y^{''}+4y^'+5y=34e^{2t},y(0)=1,y^'(0)=2 ?
The solution for y^{''}+4y^'+5y=34e^{2t},y(0)=1,y^'(0)=2 is y=e^{-2t}(-cos(t)-4sin(t))+2e^{2t}