What is the solution for y^{''}-6y^'+36y=0 ?

The solution for y^{''}-6y^'+36y=0 is y=e^{3t}(c_{1}cos(3^{1+1/2}t)+c_{2}sin(3^{1+1/2}t))

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y^{''}-6y^'+36y=0

y^{^{''}} - 6 y^{^{'}} + 36 y = 0

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

Solution steps

y^{''} (t) - 6 y^{'} (t) + 36 y = 0

Solve linear ODE:

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

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

\int \frac{1}{w ^{6}} d w

y = (x + \frac{2}{x}) (7 - 2 x^{3})

\int cos^{4} (x s i) n^{4} x

\int \frac{( 1 + x )}{( 1 - x )} d x

\frac{d y}{d t} - 0.06 y = - 24000, y (0) = 250000

What is the solution for y^{''}-6y^'+36y=0 ?
The solution for y^{''}-6y^'+36y=0 is y=e^{3t}(c_{1}cos(3^{1+1/2}t)+c_{2}sin(3^{1+1/2}t))