What is the solution for y^{''}+9y=6cos(3x) ?

The solution for y^{''}+9y=6cos(3x) is y=c_{1}cos(3x)+c_{2}sin(3x)+xsin(3x)

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y^{''}+9y=6cos(3x)

y^{^{''}} + 9 y = 6 cos (3 x)

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

Solution steps

y^{''} (x) + 9 y = 6 cos (3 x)

Solve linear ODE:

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

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

(D^{2} - 3 D + 2) y = e^{3 x}

\frac{5}{6} q^{^{''}} + 10 q^{^{'}} + 60 q = 600

\frac{d ^{2} y}{d t ^{2}} - 4 y = t cos (2 t), t = 0

x^{^{''}} + \frac{1.0889}{19.6} x = cos (2 t)

0.001 y^{^{''}} + 10 y = 10 cos (100 x)

What is the solution for y^{''}+9y=6cos(3x) ?
The solution for y^{''}+9y=6cos(3x) is y=c_{1}cos(3x)+c_{2}sin(3x)+xsin(3x)