What is the solution for y^{''}+4y=16t,y^0=3,y^'0=-6 ?

The solution for y^{''}+4y=16t,y^0=3,y^'0=-6 is y=c_{1}cos(2t)+c_{2}sin(2t)+4t

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y^{''}+4y=16t,y^0=3,y^'0=-6

y^{^{''}} + 4 y = 16 t, y^{0} (0) = 3, y^{^{'}} 0 = - 6

y = c_{1} cos (2 t) + c_{2} sin (2 t) + 4 t

Solution steps

y^{''} (t) + 4 y = 16 t

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t) + 4 t

y = c_{1} cos (2 t) + c_{2} sin (2 t) + 4 t

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

y^{^{''}} - 7 y = t^{2}

(y^{^{''}} + 2 y^{^{'}} + 5) = e^{x}

y^{^{'''}} + y^{^{'}} = x + sin (x) + cos (x)

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

What is the solution for y^{''}+4y=16t,y^0=3,y^'0=-6 ?
The solution for y^{''}+4y=16t,y^0=3,y^'0=-6 is y=c_{1}cos(2t)+c_{2}sin(2t)+4t