What is the solution for t^2y^{''}+ty^'+25y=0 ?

The solution for t^2y^{''}+ty^'+25y=0 is y=c_{1}cos(5ln(t))+c_{2}sin(5ln(t))

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t^2y^{''}+ty^'+25y=0

t^{2} y^{^{''}} + t y^{^{'}} + 25 y = 0

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

Solution steps

t^{2} y^{''} (t) + t y^{'} (t) + 25 y = 0

Solve linear ODE:

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

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

\frac{d y}{d x} = 3 x

x \to - 1 lim (- x^{2} + 1)

y^{^{''}} - 12 y^{^{'}} + 36 y = 0, y (0) = - 2, y^{^{'}} (0) = 3

\int_{0}^{1} \frac{2}{t ^{2} + 1} d t

\frac{e ^{x}}{4 x ^{2} + 3}

What is the solution for t^2y^{''}+ty^'+25y=0 ?
The solution for t^2y^{''}+ty^'+25y=0 is y=c_{1}cos(5ln(t))+c_{2}sin(5ln(t))