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(2x^4+y^4)dy-x^3ydx=0

Solution

(2 x^{4} + y^{4}) d y - x^{3} y d x = 0

Solution

y = \frac{- 1 + 1 - 16 c _{1} x ^{4}}{4 2 ^{3} 4 c _{1}}, y = - \frac{- 1 + 1 - 16 c _{1} x ^{4}}{4 2 ^{3} 4 c _{1}}, y = - \frac{- 1 + 1 - 16 c _{1} x ^{4}}{2 2 c _{1}}, y = - - \frac{- 1 + 1 - 16 c _{1} x ^{4}}{2 2 c _{1}}, y = \frac{- 1 - 1 - 16 c _{1} x ^{4}}{4 2 ^{3} 4 c _{1}}, y = - \frac{- 1 - 1 - 16 c _{1} x ^{4}}{4 2 ^{3} 4 c _{1}}, y = - \frac{- 1 - 1 - 16 c _{1} x ^{4}}{2 2 c _{1}}, y = - - \frac{- 1 - 1 - 16 c _{1} x ^{4}}{2 2 c _{1}}

Solution steps

(2 x^{4} + y^{4}) d y - x^{3} y d x = 0

Solve exact ODE:

y = \frac{- 1 + 1 - 16 c _{1} x ^{4}}{4 2 ^{3} 4 c _{1}}, y = - \frac{- 1 + 1 - 16 c _{1} x ^{4}}{4 2 ^{3} 4 c _{1}}, y = - \frac{- 1 + 1 - 16 c _{1} x ^{4}}{2 2 c _{1}}, y = - - \frac{- 1 + 1 - 16 c _{1} x ^{4}}{2 2 c _{1}}, y = \frac{- 1 - 1 - 16 c _{1} x ^{4}}{4 2 ^{3} 4 c _{1}}, y = - \frac{- 1 - 1 - 16 c _{1} x ^{4}}{4 2 ^{3} 4 c _{1}}, y = - \frac{- 1 - 1 - 16 c _{1} x ^{4}}{2 2 c _{1}}, y = - - \frac{- 1 - 1 - 16 c _{1} x ^{4}}{2 2 c _{1}}

y = \frac{- 1 + 1 - 16 c _{1} x ^{4}}{4 2 ^{3} 4 c _{1}}, y = - \frac{- 1 + 1 - 16 c _{1} x ^{4}}{4 2 ^{3} 4 c _{1}}, y = - \frac{- 1 + 1 - 16 c _{1} x ^{4}}{2 2 c _{1}}, y = - - \frac{- 1 + 1 - 16 c _{1} x ^{4}}{2 2 c _{1}}, y = \frac{- 1 - 1 - 16 c _{1} x ^{4}}{4 2 ^{3} 4 c _{1}}, y = - \frac{- 1 - 1 - 16 c _{1} x ^{4}}{4 2 ^{3} 4 c _{1}}, y = - \frac{- 1 - 1 - 16 c _{1} x ^{4}}{2 2 c _{1}}, y = - - \frac{- 1 - 1 - 16 c _{1} x ^{4}}{2 2 c _{1}}

Graph

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Frequently Asked Questions (FAQ)

What is the solution for (2x^4+y^4)dy-x^3ydx=0 ?
The solution for (2x^4+y^4)dy-x^3ydx=0 is