What is the solution for y^{''}+y=-16*cos(3x) ?

The solution for y^{''}+y=-16*cos(3x) is y=c_{1}cos(x)+c_{2}sin(x)+2cos(3x)

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y^{''}+y=-16*cos(3x)

y^{^{''}} + y = - 16 \cdot cos (3 x)

y = c_{1} cos (x) + c_{2} sin (x) + 2 cos (3 x)

Solution steps

y^{''} (x) + y = - 16 \cdot cos (3 x)

Solve linear ODE:

y = c_{1} cos (x) + c_{2} sin (x) + 2 cos (3 x)

y = c_{1} cos (x) + c_{2} sin (x) + 2 cos (3 x)

y^{^{''}} - y = \frac{e ^{x}}{x ^{2}}

y^{^{''}} - 9 y = 1, y^{^{'}} (0) = 10

y^{^{''}} + y = 9 e^{- 4 x}, y (0) = 0, y^{^{'}} (0) = 0

x^{^{''}} + 2 x^{^{'}} + 5 x = 10 cos (3 t)

y^{^{''}} - y^{^{'}} = 1, y (0) = 19, y^{^{'}} (0) = 1

What is the solution for y^{''}+y=-16*cos(3x) ?
The solution for y^{''}+y=-16*cos(3x) is y=c_{1}cos(x)+c_{2}sin(x)+2cos(3x)