What is the solution for y^{''}+4y=e^{3t},y(0)=0,y^'(0)=0 ?

The solution for y^{''}+4y=e^{3t},y(0)=0,y^'(0)=0 is y=-1/13 cos(2t)-3/26 sin(2t)+1/13 e^{3t}

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y^{''}+4y=e^{3t},y(0)=0,y^'(0)=0

y^{^{''}} + 4 y = e^{3 t}, y (0) = 0, y^{^{'}} (0) = 0

y = - \frac{1}{13} cos (2 t) - \frac{3}{26} sin (2 t) + \frac{1}{13} e^{3 t}

Solution steps

y^{''} (t) + 4 y = e^{3 t}

Solve linear ODE:

y = - \frac{1}{13} cos (2 t) - \frac{3}{26} sin (2 t) + \frac{1}{13} e^{3 t}

y = - \frac{1}{13} cos (2 t) - \frac{3}{26} sin (2 t) + \frac{1}{13} e^{3 t}

What is the solution for y^{''}+4y=e^{3t},y(0)=0,y^'(0)=0 ?
The solution for y^{''}+4y=e^{3t},y(0)=0,y^'(0)=0 is y=-1/13 cos(2t)-3/26 sin(2t)+1/13 e^{3t}