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

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

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

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

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

Solution steps

y^{''} (t) - y^{'} (t) + 2 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 x} (e^{y} + x + x^{3} + 5)

\int_{- 1}^{1} (t^{3} - 9 t) d t

\frac{2}{x - 6}

y^{^{'}} - 2 x y = x y^{2}

\int 12 sin (x) d x

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