What is the solution for (y-x4)dx+(x+y4)dy=0 ?

The solution for (y-x4)dx+(x+y4)dy=0 is y=(-x+sqrt(17x^2+c_{1)})/4 ,y=(-x-sqrt(17x^2+c_{1)})/4

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(y-x4)dx+(x+y4)dy=0

(y - x 4) d x + (x + y 4) d y = 0

y = \frac{- x + 17 x ^{2} + c _{1}}{4}, y = \frac{- x - 17 x ^{2} + c _{1}}{4}

Solution steps

(y - x 4) d x + (x + y 4) d y = 0

Solve exact ODE:

y = \frac{- x + 17 x ^{2} + c _{1}}{4}, y = \frac{- x - 17 x ^{2} + c _{1}}{4}

y = \frac{- x + 17 x ^{2} + c _{1}}{4}, y = \frac{- x - 17 x ^{2} + c _{1}}{4}

\frac{( 1 + y )}{( 1 + x )}

x^{2} - 1

\frac{d y}{d x} = 4 x e^{2 x}

\frac{\partial}{\partial x} (x + ln (5 x + y))

\int 5 x + 1 d x

What is the solution for (y-x4)dx+(x+y4)dy=0 ?
The solution for (y-x4)dx+(x+y4)dy=0 is y=(-x+sqrt(17x^2+c_{1)})/4 ,y=(-x-sqrt(17x^2+c_{1)})/4