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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y = x^{4} + 3 e^{x}, (0, 3)

\frac{x ^{3} - 3}{x ^{2} + 2}

\int_{1}^{5} x d x

\int x^{2} (ln (x))^{2} d x

\frac{\partial}{\partial x} (e^{- (x - 3)^{2}})

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