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

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

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

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

y = c_{1} x + \frac{c _{2}}{x ^{4}}

Solution steps

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

Solve linear ODE:

y = c_{1} x + \frac{c _{2}}{x ^{4}}

y = c_{1} x + \frac{c _{2}}{x ^{4}}

\int (2 cos (4 y) - 1)^{\frac{1}{2}} (sin (4 y)) d y

f (x) = x + \frac{2}{x}

\int_{0}^{\frac{π}{2}} cos^{4} (x) d x

\int \frac{5 x + 3}{x ^{2} + 4 x + 4} d x

\frac{d}{d x} (2^{3^{x^{2}}})

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