What is the solution for 2y^{''}-6y^'+5y=0 ?

The solution for 2y^{''}-6y^'+5y=0 is y=e^{(3t)/2}(c_{1}cos(t/2)+c_{2}sin(t/2))

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2y^{''}-6y^'+5y=0

2 y^{^{''}} - 6 y^{^{'}} + 5 y = 0

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

Solution steps

2 y^{''} (t) - 6 y^{'} (t) + 5 y = 0

Solve linear ODE:

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

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

\int cot (15 x) d x

\frac{\partial}{\partial s} (s^{2} + t^{2})

n = 0 \sum \infty n sin (n \cdot π)

\frac{s + 1}{( s ( s ^{2} + s + 1 ) )}

\frac{\partial}{\partial x} (e^{c o s (x)})

What is the solution for 2y^{''}-6y^'+5y=0 ?
The solution for 2y^{''}-6y^'+5y=0 is y=e^{(3t)/2}(c_{1}cos(t/2)+c_{2}sin(t/2))