What is the solution for (y^2+3xy)dx-3x^2dy=0 ?

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

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(y^2+3xy)dx-3x^2dy=0

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

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

Solution steps

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

Solve Bernoulli ODE:

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

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

1 + s

f (x) = 23 sin (x) + 2 cos (x)

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

y = ln (ln (x^{32}))

\frac{d}{d t} (cos (t) (1 - sin (t)))

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