What is the solution for 4y^{''}-12y^'+73y=0 ?

The solution for 4y^{''}-12y^'+73y=0 is y=e^{(3t)/2}(c_{1}cos(4t)+c_{2}sin(4t))

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4y^{''}-12y^'+73y=0

4 y^{^{''}} - 12 y^{^{'}} + 73 y = 0

y = e^{\frac{3 t}{2}} (c_{1} cos (4 t) + c_{2} sin (4 t))

Solution steps

4 y^{''} (t) - 12 y^{'} (t) + 73 y = 0

Solve linear ODE:

y = e^{\frac{3 t}{2}} (c_{1} cos (4 t) + c_{2} sin (4 t))

y = e^{\frac{3 t}{2}} (c_{1} cos (4 t) + c_{2} sin (4 t))

81 y^{^{''}} - 4 y = 0

y^{^{'''}} + 3 y^{^{''}} - 7 y^{^{'}} + 3 y = 0

y^{^{''}} + (e - 1) y^{^{'}} - ey = 0

y^{^{''}} - k^{2} y = 0

5 y^{^{'''}} - 4 y^{^{''}} - 61 y^{^{'}} - 12 y = 0

What is the solution for 4y^{''}-12y^'+73y=0 ?
The solution for 4y^{''}-12y^'+73y=0 is y=e^{(3t)/2}(c_{1}cos(4t)+c_{2}sin(4t))