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

The solution for y^{''}-y^'+4y=0 is y=e^{t/2}(c_{1}cos((sqrt(15)t)/2)+c_{2}sin((sqrt(15)t)/2))

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

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

y = e^{\frac{t}{2}} (c_{1} cos (\frac{15 t}{2}) + c_{2} sin (\frac{15 t}{2}))

Solution steps

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

Solve linear ODE:

y = e^{\frac{t}{2}} (c_{1} cos (\frac{15 t}{2}) + c_{2} sin (\frac{15 t}{2}))

y = e^{\frac{t}{2}} (c_{1} cos (\frac{15 t}{2}) + c_{2} sin (\frac{15 t}{2}))

y^{^{'''}} - 13 y^{^{''}} + 50 y^{^{'}} - 56 y = 0

\frac{d ^{2} f}{d x ^{2}} = f (x), f^{^{'}} (π) = 0, f^{^{'}} (0) = 0

y^{^{''}} - 12 y^{^{'}} + 32 y = 0, y (0) = - 1, y^{^{'}} (0) = - 2

y^{^{''}} + 25 y = 0, y^{^{'}} (π) = 5, y (0) = 4

y^{^{''}} + 64 y = 0, y (0) = 4, y^{^{'}} (0) = 1

What is the solution for y^{''}-y^'+4y=0 ?
The solution for y^{''}-y^'+4y=0 is y=e^{t/2}(c_{1}cos((sqrt(15)t)/2)+c_{2}sin((sqrt(15)t)/2))