What is the solution for (2x-y)+(2y-x)y^'=0,y(1)=3 ?

The solution for (2x-y)+(2y-x)y^'=0,y(1)=3 is y=(x+sqrt(-3x^2+28))/2

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(2x-y)+(2y-x)y^'=0,y(1)=3

(2 x - y) + (2 y - x) y^{^{'}} = 0, y (1) = 3

y = \frac{x + - 3 x ^{2} + 28}{2}

Solution steps

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

Solve exact ODE:

y = \frac{x + - 3 x ^{2} + 28}{2}

y = \frac{x + - 3 x ^{2} + 28}{2}

\frac{\partial}{\partial y} (e^{x y} sin (z))

x \to 1 lim (x)

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

\int (\frac{1}{x} + e^{x}) d x

y^{^{''}} + 64 y = 7 tan (8 x)

What is the solution for (2x-y)+(2y-x)y^'=0,y(1)=3 ?
The solution for (2x-y)+(2y-x)y^'=0,y(1)=3 is y=(x+sqrt(-3x^2+28))/2