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

The solution for y^'=((x+3y))/((y-3x)) is y=(6x+sqrt(40x^2+c_{1)})/2 ,y=(6x-sqrt(40x^2+c_{1)})/2

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y^'=((x+3y))/((y-3x))

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

y = \frac{6 x + 40 x ^{2} + c _{1}}{2}, y = \frac{6 x - 40 x ^{2} + c _{1}}{2}

Solution steps

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

Solve exact ODE:

y = \frac{6 x + 40 x ^{2} + c _{1}}{2}, y = \frac{6 x - 40 x ^{2} + c _{1}}{2}

y = \frac{6 x + 40 x ^{2} + c _{1}}{2}, y = \frac{6 x - 40 x ^{2} + c _{1}}{2}

sec (5 x)

f (x) = 5 x^{2} + 5 x + 7

\frac{\partial}{\partial x} (arctan (\frac{y}{( x - 1 )}))

\frac{\partial}{\partial x} (62 - 2 x^{2} - 2 y^{2})

\frac{d}{d x} (ln (\frac{1}{1 + x}))

What is the solution for y^'=((x+3y))/((y-3x)) ?
The solution for y^'=((x+3y))/((y-3x)) is y=(6x+sqrt(40x^2+c_{1)})/2 ,y=(6x-sqrt(40x^2+c_{1)})/2