What is the solution for y^{''}+9y=18cos(3x) ?

The solution for y^{''}+9y=18cos(3x) is y=c_{1}cos(3x)+c_{2}sin(3x)+3xsin(3x)

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y^{''}+9y=18cos(3x)

y^{^{''}} + 9 y = 18 cos (3 x)

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

Solution steps

y^{''} (x) + 9 y = 18 cos (3 x)

Solve linear ODE:

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

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

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

D^{2} y = 81 y

y^{^{''}} - y^{^{'}} - 6 y = 50 sin (x) - 5 e^{- 2 x}

\frac{d ^{2} x}{d t ^{2}} + 2 \frac{d x}{d t} + 5 x = 1

y^{^{''}} + y^{^{'}} = c

What is the solution for y^{''}+9y=18cos(3x) ?
The solution for y^{''}+9y=18cos(3x) is y=c_{1}cos(3x)+c_{2}sin(3x)+3xsin(3x)