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

The solution for linear y^{''}+4y=3cos(t),y(0)=0,y^'(0)=0 is y=cos(t)-cos(2t)

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

linear

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

y = cos (t) - cos (2 t)

Solution steps

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

Solve linear ODE:

y = cos (t) - cos (2 t)

y = cos (t) - cos (2 t)

y^{^{''}} - y = 6 t + 8

y^{^{''}} + 4 y = sin (t) (t - 2 π)

\frac{d ^{2} y}{d t ^{2}} + y = e^{- 2 t} sin (t)

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

4 y^{^{''}} + 27 y - 26 = 0

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