What is the solution for x^2y^{''}+5xy^'+4y=0 ?

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

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x^2y^{''}+5xy^'+4y=0

x^{2} y^{^{''}} + 5 x y^{^{'}} + 4 y = 0

y = \frac{c _{1}}{x ^{2}} + \frac{c _{2} ln ( x )}{x ^{2}}

Solution steps

x^{2} y^{''} (x) + 5 x y^{'} (x) + 4 y = 0

Solve linear ODE:

y = \frac{c _{1}}{x ^{2}} + \frac{c _{2} ln ( x )}{x ^{2}}

y = \frac{c _{1}}{x ^{2}} + \frac{c _{2} ln ( x )}{x ^{2}}

\frac{d y}{d x} = \frac{1}{x ^{2}} - \frac{2 y}{x}

\frac{x ^{3}}{( x + 1 ) ^{2}}

\frac{d}{d θ} (2 cos (θ) + 2 sin (θ))

3 (1 + x^{2}) \frac{d y}{d x} = 2 x y (y^{3} - 1)

e^{- 2 s} \cdot \frac{1}{s - 3}

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