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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

x = y^{2} - 5, x = 3 - y^{2}

arcsin (ln (x))

f (x) = 5 + 4 x - x^{3}

n = 0 \sum \infty (- 1)^{n} (n^{3})

\frac{\partial}{\partial x} (x^{8} y^{8} - x^{7} y^{7})

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