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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d}{d x} (\frac{e ^{x}}{3 + e ^{x}})

\frac{d}{d x} (7 sin (x) - 2 x^{3})

\int \frac{x ^{2} + 36 x + 36}{x ^{3} - 4 x} d x

x \to 1 lim (\frac{x - 1}{ln ( x )})

\frac{d}{d x} ((2 x^{2} + 3) (5 x + 3))

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