What is the solution for y^{''}+3/x y^'-(15)/(x^2)y=0 ?

The solution for y^{''}+3/x y^'-(15)/(x^2)y=0 is y=c_{1}x^3+(c_{2})/(x^5)

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y^{''}+3/x y^'-(15)/(x^2)y=0

y^{^{''}} + \frac{3}{x} y^{^{'}} - \frac{15}{x ^{2}} y = 0

y = c_{1} x^{3} + \frac{c _{2}}{x ^{5}}

Solution steps

y^{''} (x) + \frac{3}{x} y^{'} (x) - \frac{15}{x ^{2}} y = 0

Solve linear ODE:

y = c_{1} x^{3} + \frac{c _{2}}{x ^{5}}

y = c_{1} x^{3} + \frac{c _{2}}{x ^{5}}

\int \frac{x + 1}{x - 3} d x

\int tan^{5} (6 x) sec^{4} (6 x) d x

\frac{d}{d x} (in (x sin (x)))

f (x) = \frac{x}{x - 5}

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

What is the solution for y^{''}+3/x y^'-(15)/(x^2)y=0 ?
The solution for y^{''}+3/x y^'-(15)/(x^2)y=0 is y=c_{1}x^3+(c_{2})/(x^5)