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 D^{4} - 3 D^{3} - 2 D^{2}) y = 0

\frac{1}{2} y^{^{''}} + 3 y^{^{'}} + 4 y = 0, y (0) = 0, y^{^{'}} (0) = - \frac{5}{3}

y^{^{''}} - 4 y^{^{'}} - 5 y = 0, y (0) = 4, y^{^{'}} (0) = 14

y^{^{''}} + y = 0, y (0) = 11, y^{^{'}} (0) = 33

\frac{1}{20} y^{^{''}} + 3 y^{^{'}} + 80 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))