What is the solution for y^{''}+y=3sin(2t),y(0)=12,y^'(0)=1 ?

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

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y^{''}+y=3sin(2t),y(0)=12,y^'(0)=1

y^{^{''}} + y = 3 sin (2 t), y (0) = 12, y^{^{'}} (0) = 1

y = 12 cos (t) + 3 sin (t) - sin (2 t)

Solution steps

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

Solve linear ODE:

y = 12 cos (t) + 3 sin (t) - sin (2 t)

y = 12 cos (t) + 3 sin (t) - sin (2 t)

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

y^{^{''}} - 8 y^{^{'}} + 16 y = 16 e^{4 x}

\frac{d ^{3} y}{d x ^{3}} + y = 3 + e^{- x} + 5 e^{2 x}

\frac{d ^{2} y}{d x ^{2}} - 3 \frac{d y}{d x} + 2 y = 2 e^{3 x} + 4 x

y^{^{''}} - 3 y^{^{'}} = e^{3 x} + 2 x - 1

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