What is the solution for t^2y^{''}+ty^'+4y=0 ?

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

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t^2y^{''}+ty^'+4y=0

t^{2} y^{^{''}} + t y^{^{'}} + 4 y = 0

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

Solution steps

t^{2} y^{''} (t) + t y^{'} (t) + 4 y = 0

Solve linear ODE:

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

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

\frac{d}{d x} (x^{3} + x^{2} y + 4 y^{2}) = 6

\int 3 x^{3} cos (5 x) d x

x \to 5 lim (x^{2} + 11)

x \to + (- 3) - lim (2 x^{2} - 6)

\int \frac{100 - x ^{2}}{x} d x

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