What is the solution for e^{2x-y}dx+e^{y-2x}dy=0 ?

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

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e^{2x-y}dx+e^{y-2x}dy=0

e^{2 x - y} d x + e^{y - 2 x} d y = 0

y = \frac{ln ( - \frac{1}{2} e ^{4 x} + c _{1} )}{2}

Solution steps

e^{2 x - y} d x + e^{y - 2 x} d y = 0

Solve separable ODE:

y = \frac{ln ( - \frac{1}{2} e ^{4 x} + c _{1} )}{2}

y = \frac{ln ( - \frac{1}{2} e ^{4 x} + c _{1} )}{2}

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

(y + y cos (x) y) d x + (x + x cos (x) y) d y = 0

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

x (1 + y^{2}) d x - y (1 + x^{2}) d y = 0

\frac{d x}{d t} = 125 - \frac{x}{100}

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