What is the solution for y^{''}+9y=cos(3t),y(0)=2,y^'(0)=5 ?

The solution for y^{''}+9y=cos(3t),y(0)=2,y^'(0)=5 is y=2cos(3t)+5/3 sin(3t)+(tsin(3t))/6

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y^{''}+9y=cos(3t),y(0)=2,y^'(0)=5

y^{^{''}} + 9 y = cos (3 t), y (0) = 2, y^{^{'}} (0) = 5

y = 2 cos (3 t) + \frac{5}{3} sin (3 t) + \frac{t sin ( 3 t )}{6}

Solution steps

y^{''} (t) + 9 y = cos (3 t)

Solve linear ODE:

y = 2 cos (3 t) + \frac{5}{3} sin (3 t) + \frac{t sin ( 3 t )}{6}

y = 2 cos (3 t) + \frac{5}{3} sin (3 t) + \frac{t sin ( 3 t )}{6}

y^{^{''}} - y^{^{'}} - 6 y = (x^{2} + 3 x) e^{2 x} + cos (2 x)

\frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} + 12 y = 20 cos (3 x)

y^{^{''}} + 6 y^{^{'}} + 13 y = e^{- 3 x} \cdot cos (2 x)

y^{^{''}} - y^{^{'}} + y = 6 sin (t) - 3 e^{2 t}

y^{^{''}} - 6 y^{^{'}} + 9 y = 18 x - 3

What is the solution for y^{''}+9y=cos(3t),y(0)=2,y^'(0)=5 ?
The solution for y^{''}+9y=cos(3t),y(0)=2,y^'(0)=5 is y=2cos(3t)+5/3 sin(3t)+(tsin(3t))/6