What is the solution for t^2y^{''}+ty^'+36y=0 ?

The solution for t^2y^{''}+ty^'+36y=0 is y=c_{1}cos(6ln(t))+c_{2}sin(6ln(t))

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t^2y^{''}+ty^'+36y=0

t^{2} y^{^{''}} + t y^{^{'}} + 36 y = 0

y = c_{1} cos (6 ln (t)) + c_{2} sin (6 ln (t))

Solution steps

t^{2} y^{''} (t) + t y^{'} (t) + 36 y = 0

Solve linear ODE:

y = c_{1} cos (6 ln (t)) + c_{2} sin (6 ln (t))

y = c_{1} cos (6 ln (t)) + c_{2} sin (6 ln (t))

2 x + \frac{1}{x}

\int \frac{6}{x ^{2}} d x

x \cdot f (x)

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

\frac{1}{( s ^{2} ( s ^{2} + 4 ) )}

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