What is the solution for y^{''}=-4y^'-5y+2 ?

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

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y^{''}=-4y^'-5y+2

y^{^{''}} = - 4 y^{^{'}} - 5 y + 2

y = e^{- 2 t} (c_{1} cos (t) + c_{2} sin (t)) + \frac{2}{5}

Solution steps

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

Solve linear ODE:

y = e^{- 2 t} (c_{1} cos (t) + c_{2} sin (t)) + \frac{2}{5}

y = e^{- 2 t} (c_{1} cos (t) + c_{2} sin (t)) + \frac{2}{5}

16 \frac{d ^{4} y}{d x ^{4}} + 24 \frac{d ^{2} y}{d x ^{2}} + 9 = 0

y^{^{''}} - 5 y^{^{'}} + 6 y = 2 sin (4 x)

y^{^{''}} + 7 y^{^{'}} + 10 y = e^{3 x}, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''''}} - y = 15 t + cos (t)

\frac{d ^{2}}{d t ^{2}} (θ (t)) + 4 θ (t) = csc (2) (2 t)

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