What is the solution for 25x^2y^{''}+25xy^'+y=0 ?

The solution for 25x^2y^{''}+25xy^'+y=0 is y=c_{1}cos(1/5 ln(x))+c_{2}sin(1/5 ln(x))

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25x^2y^{''}+25xy^'+y=0

25 x^{2} y^{^{''}} + 25 x y^{^{'}} + y = 0

y = c_{1} cos (\frac{1}{5} ln (x)) + c_{2} sin (\frac{1}{5} ln (x))

Solution steps

25 x^{2} y^{''} (x) + 25 x y^{'} (x) + y = 0

Solve linear ODE:

y = c_{1} cos (\frac{1}{5} ln (x)) + c_{2} sin (\frac{1}{5} ln (x))

y = c_{1} cos (\frac{1}{5} ln (x)) + c_{2} sin (\frac{1}{5} ln (x))

x \to \infty lim (x^{x})

d er i v a t i v e \frac{- 8 x - 2}{9 x - 4}

\frac{\partial}{\partial x} (4 x^{3} + y)

d er i v a t i v e \frac{10 x + 1}{x + 100}

d er i v a t i v e 7^{x^{7}}

What is the solution for 25x^2y^{''}+25xy^'+y=0 ?
The solution for 25x^2y^{''}+25xy^'+y=0 is y=c_{1}cos(1/5 ln(x))+c_{2}sin(1/5 ln(x))