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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

20 y^{^{''}} + 80 y^{^{'}} + 100 y = 0

2 y^{^{''}} + 3 y - 2 y = 0

\frac{d ^{2} y}{d x ^{2}} - 10 \frac{d y}{d x} - 3 y = 0

y^{^{''}} + 15 y = 0, y (0) = sin (k t)

y^{^{''}} - \frac{1}{c} y = 0

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