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

The solution for y^{''}+9y=e^t,y(0)=0 is y=-1/10 cos(3t)+c_{2}sin(3t)+1/10 e^t

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

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

y = - \frac{1}{10} cos (3 t) + c_{2} sin (3 t) + \frac{1}{10} e^{t}

Solution steps

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

Solve linear ODE:

y = - \frac{1}{10} cos (3 t) + c_{2} sin (3 t) + \frac{1}{10} e^{t}

y = - \frac{1}{10} cos (3 t) + c_{2} sin (3 t) + \frac{1}{10} e^{t}

y^{^{''}} + y^{^{'}} + y = (x + x^{2}) e^{x}

y^{^{''}} - 2 y^{^{'}} = 2 x + 9 - e^{- 2 x}

y^{^{'''}} + y^{^{'}} = 4 x

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} = \frac{e ^{x}}{x ^{2} + 1}

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

What is the solution for y^{''}+9y=e^t,y(0)=0 ?
The solution for y^{''}+9y=e^t,y(0)=0 is y=-1/10 cos(3t)+c_{2}sin(3t)+1/10 e^t