What is the solution for t^2y^{''}+ty^'+16y=0 ?

The solution for t^2y^{''}+ty^'+16y=0 is y=c_{1}cos(4ln(t))+c_{2}sin(4ln(t))

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t^2y^{''}+ty^'+16y=0

t^{2} y^{^{''}} + t y^{^{'}} + 16 y = 0

y = c_{1} cos (4 ln (t)) + c_{2} sin (4 ln (t))

Solution steps

t^{2} y^{''} (t) + t y^{'} (t) + 16 y = 0

Solve linear ODE:

y = c_{1} cos (4 ln (t)) + c_{2} sin (4 ln (t))

y = c_{1} cos (4 ln (t)) + c_{2} sin (4 ln (t))

\int \frac{( 2 x - 1 ) ( x + 3 )}{6} d x

(0, 0), (2, 1)

\int \frac{\frac{1}{x ^{2}}}{( a - \frac{1}{x} ) ^{k}} d x

\frac{\partial}{\partial t} (t e^{- 2 x} + cos (x))

f (x) = x^{2}, at x = 1.9

What is the solution for t^2y^{''}+ty^'+16y=0 ?
The solution for t^2y^{''}+ty^'+16y=0 is y=c_{1}cos(4ln(t))+c_{2}sin(4ln(t))