What is the solution for 2y^{''}-2y^'+25y=0 ?

The solution for 2y^{''}-2y^'+25y=0 is y=e^{t/2}(c_{1}cos((7t)/2)+c_{2}sin((7t)/2))

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2y^{''}-2y^'+25y=0

2 y^{^{''}} - 2 y^{^{'}} + 25 y = 0

y = e^{\frac{t}{2}} (c_{1} cos (\frac{7 t}{2}) + c_{2} sin (\frac{7 t}{2}))

Solution steps

2 y^{''} (t) - 2 y^{'} (t) + 25 y = 0

Solve linear ODE:

y = e^{\frac{t}{2}} (c_{1} cos (\frac{7 t}{2}) + c_{2} sin (\frac{7 t}{2}))

y = e^{\frac{t}{2}} (c_{1} cos (\frac{7 t}{2}) + c_{2} sin (\frac{7 t}{2}))

\frac{\partial}{\partial y} (x^{2} + 2 x y + y^{2})

\int_{0}^{5} sin (π x) d x

\frac{d}{d x} (2 x^{2} y cos (x))

\frac{s}{s ^{2} + s + 25}

y^{^{''}} - 12 y^{^{'}} + 40 y = 0

What is the solution for 2y^{''}-2y^'+25y=0 ?
The solution for 2y^{''}-2y^'+25y=0 is y=e^{t/2}(c_{1}cos((7t)/2)+c_{2}sin((7t)/2))