What is the solution for y^{''}-5/x y^'+8/(x^2)y=0 ?

The solution for y^{''}-5/x y^'+8/(x^2)y=0 is y=c_{1}x^4+c_{2}x^2

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y^{''}-5/x y^'+8/(x^2)y=0

y^{^{''}} - \frac{5}{x} y^{^{'}} + \frac{8}{x ^{2}} y = 0

y = c_{1} x^{4} + c_{2} x^{2}

Solution steps

y^{''} (x) - \frac{5}{x} y^{'} (x) + \frac{8}{x ^{2}} y = 0

Solve linear ODE:

y = c_{1} x^{4} + c_{2} x^{2}

y = c_{1} x^{4} + c_{2} x^{2}

7 t \frac{d y}{d t} - 3 y = t

y = 4 - x^{2}, (3, - 5)

\frac{d}{d x} (\frac{3}{( 4 x ^{2} - 1 ) ^{4}})

\frac{d}{d x} (x^{3} + \frac{6}{7} x - 10)

\int 7 tan^{3} (2 x) sec^{5} (2 x) d x

What is the solution for y^{''}-5/x y^'+8/(x^2)y=0 ?
The solution for y^{''}-5/x y^'+8/(x^2)y=0 is y=c_{1}x^4+c_{2}x^2