What is the solution for y^{''}+9y=12cos(3t),y(0)=-1,y^'(0)=1 ?

The solution for y^{''}+9y=12cos(3t),y(0)=-1,y^'(0)=1 is y=-cos(3t)+1/3 sin(3t)+2tsin(3t)

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y^{''}+9y=12cos(3t),y(0)=-1,y^'(0)=1

y^{^{''}} + 9 y = 12 cos (3 t), y (0) = - 1, y^{^{'}} (0) = 1

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

Solution steps

y^{''} (t) + 9 y = 12 cos (3 t)

Solve linear ODE:

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

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

4 x^{^{''}} + 324 x = 8 sin (t)

y^{^{''}} + 5 y^{^{'}} + 4 y = - 2 e^{3 t}

y^{^{''}} + 4 y = 8 e^{2 t}

y^{^{''}} - 4 y^{^{'}} + 3 y = e^{(3 x)}

y^{^{''}} + 3 y^{^{'}} - 4 y = 7 e^{x}

What is the solution for y^{''}+9y=12cos(3t),y(0)=-1,y^'(0)=1 ?
The solution for y^{''}+9y=12cos(3t),y(0)=-1,y^'(0)=1 is y=-cos(3t)+1/3 sin(3t)+2tsin(3t)