What is the solution for y^{''}+4y=3cos(2) ?

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

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y^{''}+4y=3cos(2)

y^{^{''}} + 4 y = 3 cos (2)

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{3 cos ( 2 )}{4}

Solution steps

y^{''} (t) + 4 y = 3 cos (2)

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{3 cos ( 2 )}{4}

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{3 cos ( 2 )}{4}

y^{^{''}} + 2 y^{^{'}} - 3 y = 5 e^{x} sin (2 x)

y^{^{''}} - 4 y^{^{'}} + 5 y = e^{2} x, y (0) = 1, y^{^{'}} (0) = 3

8 y^{^{''}} - y^{^{'}} + y + x^{2} = 0

y^{^{'''}} - 9 y^{^{''}} + 27 y^{^{'}} - 27 y = 8 e^{3 x}

\frac{d ^{2} x}{d t ^{2}} + 81 x = 8 sin (9 t)

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