What is the solution for y^{''}+9y=cos(3t),y(0)=5,y^'(0)=4 ?

The solution for y^{''}+9y=cos(3t),y(0)=5,y^'(0)=4 is y=5cos(3t)+4/3 sin(3t)+(tsin(3t))/6

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y^{''}+9y=cos(3t),y(0)=5,y^'(0)=4

y^{^{''}} + 9 y = cos (3 t), y (0) = 5, y^{^{'}} (0) = 4

y = 5 cos (3 t) + \frac{4}{3} sin (3 t) + \frac{t sin ( 3 t )}{6}

Solution steps

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

Solve linear ODE:

y = 5 cos (3 t) + \frac{4}{3} sin (3 t) + \frac{t sin ( 3 t )}{6}

y = 5 cos (3 t) + \frac{4}{3} sin (3 t) + \frac{t sin ( 3 t )}{6}

y^{^{''}} - 4 y^{^{'}} + 4 y = 2 t^{2} + 4 + e^{2 t} + t sin (2 t)

y^{^{''}} + 9 y = 4 (\frac{\frac{u ^{5}}{3} π - u ^{13}}{3 π})

y^{^{''}} + y^{^{'}} = sinh (t)

y^{^{''}} - 2 \cdot y^{^{'}} + 2 \cdot y = cos (t), y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} + 6 y = t, y (0) = - 3, y^{^{'}} (0) = 0

What is the solution for y^{''}+9y=cos(3t),y(0)=5,y^'(0)=4 ?
The solution for y^{''}+9y=cos(3t),y(0)=5,y^'(0)=4 is y=5cos(3t)+4/3 sin(3t)+(tsin(3t))/6