What is the solution for y^{''}-4y^'+9y=0 ?

The solution for y^{''}-4y^'+9y=0 is y=e^{2t}(c_{1}cos(sqrt(5)t)+c_{2}sin(sqrt(5)t))

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y^{''}-4y^'+9y=0

y^{^{''}} - 4 y^{^{'}} + 9 y = 0

y = e^{2 t} (c_{1} cos (5 t) + c_{2} sin (5 t))

Solution steps

y^{''} (t) - 4 y^{'} (t) + 9 y = 0

Solve linear ODE:

y = e^{2 t} (c_{1} cos (5 t) + c_{2} sin (5 t))

y = e^{2 t} (c_{1} cos (5 t) + c_{2} sin (5 t))

\int_{0}^{2} x^{2} x^{3} + 1 d x

t an g e n t f (x) = \frac{7}{4 - x}, at x = - 1

\frac{\partial}{\partial x} (y^{4} sin (2 x))

\int 24 y (1 - x - y) d y

\frac{d}{d x} (\frac{( 3 x + 1 )}{x ^{\frac{2}{3}}})

What is the solution for y^{''}-4y^'+9y=0 ?
The solution for y^{''}-4y^'+9y=0 is y=e^{2t}(c_{1}cos(sqrt(5)t)+c_{2}sin(sqrt(5)t))