What is the solution for ((2y)/x+2x)+(2ln(x)-3)y^'=0 ?

The solution for ((2y)/x+2x)+(2ln(x)-3)y^'=0 is y=-(x^2)/(2ln(x)-3)+(c_{1})/(2ln(x)-3)

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((2y)/x+2x)+(2ln(x)-3)y^'=0

(\frac{2 y}{x} + 2 x) + (2 ln (x) - 3) y^{^{'}} = 0

y = - \frac{x ^{2}}{2 ln ( x ) - 3} + \frac{c _{1}}{2 ln ( x ) - 3}

Solution steps

(\frac{2 y}{x} + 2 x) + (2 ln (x) - 3) y^{'} (x) = 0

Solve linear ODE:

y = - \frac{x ^{2}}{2 ln ( x ) - 3} + \frac{c _{1}}{2 ln ( x ) - 3}

y = - \frac{x ^{2}}{2 ln ( x ) - 3} + \frac{c _{1}}{2 ln ( x ) - 3}

\frac{\partial}{\partial x} (\frac{26 x ^{2}}{( x - y ) ^{2}})

1 + 4 x^{2} - 2 x^{3}

\frac{d}{d x} (sin (x - 1) (x^{3}))

x - 4

\int x cos (9 x) d x

What is the solution for ((2y)/x+2x)+(2ln(x)-3)y^'=0 ?
The solution for ((2y)/x+2x)+(2ln(x)-3)y^'=0 is y=-(x^2)/(2ln(x)-3)+(c_{1})/(2ln(x)-3)