What is the solution for y^{''}+4y=1,6ty(0)=3,y^'(0)=-6 ?

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

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y^{''}+4y=1,6ty(0)=3,y^'(0)=-6

y^{^{''}} + 4 y = 1, 6 t y (0) = 3, y^{^{'}} (0) = - 6

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

Solution steps

y^{''} (t) + 4 y = 1

Solve linear ODE:

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

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

y^{^{''}} + 2 y^{^{'}} + 2 y = a cos (b t)

y^{^{''''}} + 2 y^{^{''}} + y = 9 + cos (4 t)

y^{^{''}} + 10 y^{^{'}} + 16 y = 40 cos (4 t)

y^{^{''}} - 2 y^{^{'}} + y = \frac{13 e ^{t}}{1 + t ^{2}}

y^{^{''}} + 36 y = 7 sec (6 x)

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