What is the solution for y^{''}-2*y^'+10*y=0 ?

The solution for y^{''}-2*y^'+10*y=0 is y=e^t(c_{1}cos(3t)+c_{2}sin(3t))

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y^{''}-2y^'+10y=0

y^{^{''}} - 2 \cdot y^{^{'}} + 10 \cdot y = 0

y = e^{t} (c_{1} cos (3 t) + c_{2} sin (3 t))

Solution steps

y^{''} (t) - 2 \cdot y^{'} (t) + 10 \cdot y = 0

Solve linear ODE:

y = e^{t} (c_{1} cos (3 t) + c_{2} sin (3 t))

y = e^{t} (c_{1} cos (3 t) + c_{2} sin (3 t))

\frac{d}{d x} (e^{- \frac{x}{10}} - cos (x))

f (x) = x^{4} - 10 x^{2} + 9, at x = - 2

x \to 0 + lim ((5 x)^{2 x})

f (x) = x^{2} + x + 4

\frac{\partial}{\partial u} (2 u + v)

What is the solution for y^{''}-2*y^'+10*y=0 ?
The solution for y^{''}-2*y^'+10*y=0 is y=e^t(c_{1}cos(3t)+c_{2}sin(3t))