What is the solution for y^{''}-3/x y^'+4/(x^2)y=0 ?

The solution for y^{''}-3/x y^'+4/(x^2)y=0 is y=c_{1}x^2+c_{2}ln(x)x^2

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y^{''}-3/x y^'+4/(x^2)y=0

y^{^{''}} - \frac{3}{x} y^{^{'}} + \frac{4}{x ^{2}} y = 0

y = c_{1} x^{2} + c_{2} ln (x) x^{2}

Solution steps

y^{''} (x) - \frac{3}{x} y^{'} (x) + \frac{4}{x ^{2}} y = 0

Solve linear ODE:

y = c_{1} x^{2} + c_{2} ln (x) x^{2}

y = c_{1} x^{2} + c_{2} ln (x) x^{2}

x \to 2 lim (\frac{( ∣ x - 2 ∣ )}{2})

\frac{d}{d x} (8 x - \frac{5}{3 x ^{2}} + e)

\frac{d}{d x} (tan (cos (2 x)))

(1 + ln (x) + \frac{y}{x}) d x = (8 - ln (x)) d y

\int \frac{e ^{x}}{1 + 9 e ^{2 x}} d x

What is the solution for y^{''}-3/x y^'+4/(x^2)y=0 ?
The solution for y^{''}-3/x y^'+4/(x^2)y=0 is y=c_{1}x^2+c_{2}ln(x)x^2