What is the solution for 2y^{''}-4y^'+5y=0 ?

The solution for 2y^{''}-4y^'+5y=0 is y=e^t(c_{1}cos(sqrt(3/2)t)+c_{2}sin(sqrt(3/2)t))

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2y^{''}-4y^'+5y=0

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

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

Solution steps

2 y^{''} (t) - 4 y^{'} (t) + 5 y = 0

Solve linear ODE:

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

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

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} - 63 y = 0

25 y^{^{''}} + 80 y^{^{'}} + 64 y = 0, y (0) = 8

y^{^{''}} - 12 y^{^{'}} + 36 y = 0, y (0) = 9, y^{^{'}} (0) = 5

7 y^{^{''}} - 8 y^{^{'}} + y = 0, y (0) = 7, y^{^{'}} (0) = \frac{1}{7}

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

What is the solution for 2y^{''}-4y^'+5y=0 ?
The solution for 2y^{''}-4y^'+5y=0 is y=e^t(c_{1}cos(sqrt(3/2)t)+c_{2}sin(sqrt(3/2)t))