What is the solution for y^{''}+y=t-3sin(2t),y(0)=1,y^'(0)=-3 ?

The solution for y^{''}+y=t-3sin(2t),y(0)=1,y^'(0)=-3 is y=cos(t)-6sin(t)+t+sin(2t)

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y^{''}+y=t-3sin(2t),y(0)=1,y^'(0)=-3

y^{^{''}} + y = t - 3 sin (2 t), y (0) = 1, y^{^{'}} (0) = - 3

y = cos (t) - 6 sin (t) + t + sin (2 t)

Solution steps

y^{''} (t) + y = t - 3 sin (2 t)

Solve linear ODE:

y = cos (t) - 6 sin (t) + t + sin (2 t)

y = cos (t) - 6 sin (t) + t + sin (2 t)

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

y^{^{''}} + 81 y = 8 sin (9 t)

x^{^{''}} + 2 x^{^{'}} + 5 x = 5 cos (t)

y^{^{''}} + 4 y = 4 csc^{2} (2 x)

y^{^{''}} + 2 y^{^{'}} - 3 y = sin (2 t)

What is the solution for y^{''}+y=t-3sin(2t),y(0)=1,y^'(0)=-3 ?
The solution for y^{''}+y=t-3sin(2t),y(0)=1,y^'(0)=-3 is y=cos(t)-6sin(t)+t+sin(2t)