What is the solution for 8y^{''}+4y^'+5y=1,y(0)=1 ?

The solution for 8y^{''}+4y^'+5y=1,y(0)=1 is y=e^{-t/4}(4/5 cos((3t)/4)+c_{2}sin((3t)/4))+1/5

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8y^{''}+4y^'+5y=1,y(0)=1

8 y^{^{''}} + 4 y^{^{'}} + 5 y = 1, y (0) = 1

y = e^{- \frac{t}{4}} (\frac{4}{5} cos (\frac{3 t}{4}) + c_{2} sin (\frac{3 t}{4})) + \frac{1}{5}

Solution steps

8 y^{''} (t) + 4 y^{'} (t) + 5 y = 1

Solve linear ODE:

y = e^{- \frac{t}{4}} (\frac{4}{5} cos (\frac{3 t}{4}) + c_{2} sin (\frac{3 t}{4})) + \frac{1}{5}

y = e^{- \frac{t}{4}} (\frac{4}{5} cos (\frac{3 t}{4}) + c_{2} sin (\frac{3 t}{4})) + \frac{1}{5}

y^{^{'''}} - 4 y^{^{'}} = cot (2 x)

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

\frac{d ^{2} y}{d x ^{2}} - \frac{d y}{d x} - 2 y = e^{3 x}

y^{^{''}} + 3 y^{^{'}} - 10 y = e^{x} cos (3 x)

y^{^{''}} - 2 y^{^{'}} - 3 y = 4 e^{t}

What is the solution for 8y^{''}+4y^'+5y=1,y(0)=1 ?
The solution for 8y^{''}+4y^'+5y=1,y(0)=1 is y=e^{-t/4}(4/5 cos((3t)/4)+c_{2}sin((3t)/4))+1/5