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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{\partial}{\partial x} (x^{2} + y^{2})

\int_{- 1}^{1} sec^{3} (x) d x

f (x) = \frac{( 2 x + 1 )}{x + 2}, at x = 1

y = \frac{3 x - 4}{2 x + 3}

\frac{\partial}{\partial x} (x^{6} y^{6} + 9 x^{4} y)

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