What is the solution for y^{''}+(2y^')/x-(6y)/(x^2)=0 ?

The solution for y^{''}+(2y^')/x-(6y)/(x^2)=0 is y=c_{1}x^2+(c_{2})/(x^3)

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y^{''}+(2y^')/x-(6y)/(x^2)=0

y^{^{''}} + \frac{2 y ^{^{'}}}{x} - \frac{6 y}{x ^{2}} = 0

y = c_{1} x^{2} + \frac{c _{2}}{x ^{3}}

Solution steps

y^{''} (x) + \frac{2 y ^{'} ( x )}{x} - \frac{6 y}{x ^{2}} = 0

Solve linear ODE:

y = c_{1} x^{2} + \frac{c _{2}}{x ^{3}}

y = c_{1} x^{2} + \frac{c _{2}}{x ^{3}}

x \to 1 lim (1 - cos (π) x)

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\int_{0}^{\frac{3 π}{2}} ∣ sin (x) ∣ d x

\int \frac{( ln ( z ) ) ^{2}}{z} d z

What is the solution for y^{''}+(2y^')/x-(6y)/(x^2)=0 ?
The solution for y^{''}+(2y^')/x-(6y)/(x^2)=0 is y=c_{1}x^2+(c_{2})/(x^3)