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

The solution for x^2y^{''}+3xy^'-4y=0 is y=c_{1}x^{-1+sqrt(5)}+c_{2}x^{-1-sqrt(5)}

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

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

y = c_{1} x^{- 1 + 5} + c_{2} x^{- 1 - 5}

Solution steps

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

Solve linear ODE:

y = c_{1} x^{- 1 + 5} + c_{2} x^{- 1 - 5}

y = c_{1} x^{- 1 + 5} + c_{2} x^{- 1 - 5}

\int \frac{( x ^{2} + x - 6 )}{( x ^{2} - 4 x + 3 )} d x

f (x) = (\frac{x + 1}{x - 1})^{3}, at x = 2

\int 6 x^{2} e^{x^{3}} d x

\int \frac{1}{3 - 2 x - x ^{2}} d x

\frac{\partial}{\partial x} (x y e^{yz})

What is the solution for x^2y^{''}+3xy^'-4y=0 ?
The solution for x^2y^{''}+3xy^'-4y=0 is y=c_{1}x^{-1+sqrt(5)}+c_{2}x^{-1-sqrt(5)}