What is the solution for y^{''}-5y^'+7y=0 ?

The solution for y^{''}-5y^'+7y=0 is y=e^{(5t)/2}(c_{1}cos((sqrt(3)t)/2)+c_{2}sin((sqrt(3)t)/2))

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y^{''}-5y^'+7y=0

y^{^{''}} - 5 y^{^{'}} + 7 y = 0

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

Solution steps

y^{''} (t) - 5 y^{'} (t) + 7 y = 0

Solve linear ODE:

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

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

2 y^{^{''}} + 2 y^{^{'}} + 5 y = 0, y (0) = 1, y^{^{'}} (0) = - 5

y^{^{'''}} + 3 y^{^{''}} - 54 y = 0

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

y^{^{'''}} - 6 y^{^{''}} - 36 y^{^{'}} + 216 y = 0

4 y^{^{''}} - 8 y^{^{'}} + 5 y = 0

What is the solution for y^{''}-5y^'+7y=0 ?
The solution for y^{''}-5y^'+7y=0 is y=e^{(5t)/2}(c_{1}cos((sqrt(3)t)/2)+c_{2}sin((sqrt(3)t)/2))