What is the solution for y^{''}+4y=sin(3t),y(0)=1,y^'(0)=0 ?

The solution for y^{''}+4y=sin(3t),y(0)=1,y^'(0)=0 is y=cos(2t)+3/10 sin(2t)-1/5 sin(3t)

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y^{''}+4y=sin(3t),y(0)=1,y^'(0)=0

y^{^{''}} + 4 y = sin (3 t), y (0) = 1, y^{^{'}} (0) = 0

y = cos (2 t) + \frac{3}{10} sin (2 t) - \frac{1}{5} sin (3 t)

Solution steps

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

Solve linear ODE:

y = cos (2 t) + \frac{3}{10} sin (2 t) - \frac{1}{5} sin (3 t)

y = cos (2 t) + \frac{3}{10} sin (2 t) - \frac{1}{5} sin (3 t)

y^{^{''}} - 2 y^{^{'}} - 3 y = 12 e x - 6

y^{^{''}} - 2 y^{^{'}} - 3 y = 12 e x - 9

y^{^{''}} + 2 y^{^{'}} + 4 y = \frac{x}{e ^{2 x} ( 1 + x ^{2} )}

y^{^{''}} - y^{^{'}} + 64 y = 8 sin (8 t)

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

What is the solution for y^{''}+4y=sin(3t),y(0)=1,y^'(0)=0 ?
The solution for y^{''}+4y=sin(3t),y(0)=1,y^'(0)=0 is y=cos(2t)+3/10 sin(2t)-1/5 sin(3t)