What is the solution for y^{''}+9y=18-18e^{-t},y(0)=0,y^'(0)=0 ?

The solution for y^{''}+9y=18-18e^{-t},y(0)=0,y^'(0)=0 is y=-1/5 cos(3t)-3/5 sin(3t)+2-9/5 e^{-t}

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y^{''}+9y=18-18e^{-t},y(0)=0,y^'(0)=0

y^{^{''}} + 9 y = 18 - 18 e^{- t}, y (0) = 0, y^{^{'}} (0) = 0

y = - \frac{1}{5} cos (3 t) - \frac{3}{5} sin (3 t) + 2 - \frac{9}{5} e^{- t}

Solution steps

y^{''} (t) + 9 y = 18 - 18 e^{- t}

Solve linear ODE:

y = - \frac{1}{5} cos (3 t) - \frac{3}{5} sin (3 t) + 2 - \frac{9}{5} e^{- t}

y = - \frac{1}{5} cos (3 t) - \frac{3}{5} sin (3 t) + 2 - \frac{9}{5} e^{- t}

\int 2 csc (y) (cot (y) - csc (y)) d y

x^{3} + 12 x^{2} + 8

\frac{d}{d x} ((x - 2) e^{x})

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

\frac{d}{d x} (\frac{x ^{2}}{2 + x})

What is the solution for y^{''}+9y=18-18e^{-t},y(0)=0,y^'(0)=0 ?
The solution for y^{''}+9y=18-18e^{-t},y(0)=0,y^'(0)=0 is y=-1/5 cos(3t)-3/5 sin(3t)+2-9/5 e^{-t}