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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d}{d x} (cos (x) + cot (x) + csc (x) + 3 e^{x})

n = 0 \sum \infty (\frac{2}{3})^{- n}

f (x) = \frac{7 ^{x}}{x ^{3}}

\frac{\partial}{\partial y} (x^{3} y^{3})

\int_{1}^{3} \frac{1}{1 + x ^{2}} d x

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