What is the solution for (x-2y)dy+(y+4x)dx=0 ?

The solution for (x-2y)dy+(y+4x)dx=0 is y=-(-x+sqrt(9x^2-c_{1)})/2 ,y=-(-x-sqrt(9x^2-c_{1)})/2

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(x-2y)dy+(y+4x)dx=0

(x - 2 y) d y + (y + 4 x) d x = 0

y = - \frac{- x + 9 x ^{2} - c _{1}}{2}, y = - \frac{- x - 9 x ^{2} - c _{1}}{2}

Solution steps

(x - 2 y) d y + (y + 4 x) d x = 0

Solve exact ODE:

y = - \frac{- x + 9 x ^{2} - c _{1}}{2}, y = - \frac{- x - 9 x ^{2} - c _{1}}{2}

y = - \frac{- x + 9 x ^{2} - c _{1}}{2}, y = - \frac{- x - 9 x ^{2} - c _{1}}{2}

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What is the solution for (x-2y)dy+(y+4x)dx=0 ?
The solution for (x-2y)dy+(y+4x)dx=0 is y=-(-x+sqrt(9x^2-c_{1)})/2 ,y=-(-x-sqrt(9x^2-c_{1)})/2