What is the solution for 4y^{''}-24y^'+37y=0,y(0)=-9 ?

The solution for 4y^{''}-24y^'+37y=0,y(0)=-9 is y=e^{3t}(-9cos(t/2)+c_{2}sin(t/2))

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4y^{''}-24y^'+37y=0,y(0)=-9

4 y^{^{''}} - 24 y^{^{'}} + 37 y = 0, y (0) = - 9

y = e^{3 t} (- 9 cos (\frac{t}{2}) + c_{2} sin (\frac{t}{2}))

Solution steps

4 y^{''} (t) - 24 y^{'} (t) + 37 y = 0

Solve linear ODE:

y = e^{3 t} (- 9 cos (\frac{t}{2}) + c_{2} sin (\frac{t}{2}))

y = e^{3 t} (- 9 cos (\frac{t}{2}) + c_{2} sin (\frac{t}{2}))

y^{^{''}} = - 4 y, y (0) = 1, y^{^{'}} (0) = 0

8 y^{^{''}} + 16 y^{^{'}} + 128 y = 0, y (0) = 5, y^{^{'}} (0) = 8

(d^{2} + d + 1) y = 0

4 y^{^{''}} - 4 y^{^{'}} - 3 y = 0, y (- 2) = e, y^{^{'}} (- 2) = - \frac{e}{2}

y^{^{'''}} - 3 y^{^{''}} + 2 y^{^{'}} + 6 y = 0

What is the solution for 4y^{''}-24y^'+37y=0,y(0)=-9 ?
The solution for 4y^{''}-24y^'+37y=0,y(0)=-9 is y=e^{3t}(-9cos(t/2)+c_{2}sin(t/2))