What is the solution for 4y^{''}+9y=15 ?

The solution for 4y^{''}+9y=15 is y=c_{1}cos((3t)/2)+c_{2}sin((3t)/2)+5/3

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4y^{''}+9y=15

4 y^{^{''}} + 9 y = 15

y = c_{1} cos (\frac{3 t}{2}) + c_{2} sin (\frac{3 t}{2}) + \frac{5}{3}

Solution steps

4 y^{''} (t) + 9 y = 15

Solve linear ODE:

y = c_{1} cos (\frac{3 t}{2}) + c_{2} sin (\frac{3 t}{2}) + \frac{5}{3}

y = c_{1} cos (\frac{3 t}{2}) + c_{2} sin (\frac{3 t}{2}) + \frac{5}{3}

y^{^{''}} - 3 y^{^{'}} + 2 y = sin (x), y (0) = 0, y^{^{'}} (0) = 2

y^{^{''}} + y = sec^{3} (x)

y^{^{''}} + 4 y^{^{'}} + 3 y = 9 e^{- 3 x}

y^{^{''}} + 2 y^{^{'}} + y = 3 e^{- x}

y^{^{''}} - 4 y^{^{'}} + 4 y = t^{3} e^{2 t}

What is the solution for 4y^{''}+9y=15 ?
The solution for 4y^{''}+9y=15 is y=c_{1}cos((3t)/2)+c_{2}sin((3t)/2)+5/3