What is the solution for y^{''}+9y=10e^{-t},y(0)=0,y^'(0)=0 ?

The solution for y^{''}+9y=10e^{-t},y(0)=0,y^'(0)=0 is y=-cos(3t)+1/3 sin(3t)+e^{-t}

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y^{''}+9y=10e^{-t},y(0)=0,y^'(0)=0

y^{^{''}} + 9 y = 10 e^{- t}, y (0) = 0, y^{^{'}} (0) = 0

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

Solution steps

y^{''} (t) + 9 y = 10 e^{- t}

Solve linear ODE:

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

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

y^{^{'''}} - y^{^{'}} = 5 sin (t)

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

y^{^{''}} + 3 y^{^{'}} + 33 y = 9 cos (5 x)

2 y^{^{''}} + 3 y^{^{'}} - 2 y = 14 x^{2} - 4 x - 11

4 y^{^{''}} - y = \frac{e x}{2 + 7}

What is the solution for y^{''}+9y=10e^{-t},y(0)=0,y^'(0)=0 ?
The solution for y^{''}+9y=10e^{-t},y(0)=0,y^'(0)=0 is y=-cos(3t)+1/3 sin(3t)+e^{-t}