What is the solution for y^{''}-2y^'+9y=0 ?

The solution for y^{''}-2y^'+9y=0 is y=e^t(c_{1}cos(2^{1+1/2}t)+c_{2}sin(2^{1+1/2}t))

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y^{''}-2y^'+9y=0

y^{^{''}} - 2 y^{^{'}} + 9 y = 0

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

Solution steps

y^{''} (t) - 2 y^{'} (t) + 9 y = 0

Solve linear ODE:

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

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

\frac{\partial}{\partial x} (e^{x y} + \frac{y}{x})

g (x) = \frac{9}{x ^{3}}

\frac{\partial}{\partial x} (x^{2} - 3 x y^{2})

\int \frac{25}{1 + 25 x ^{2}} d x

\frac{\partial}{\partial y} (x^{2} cos (y))

What is the solution for y^{''}-2y^'+9y=0 ?
The solution for y^{''}-2y^'+9y=0 is y=e^t(c_{1}cos(2^{1+1/2}t)+c_{2}sin(2^{1+1/2}t))