What is the solution for ydu+(2y-u)dy=0 ?

The solution for ydu+(2y-u)dy=0 is y(u)=-u/(2\W(-\frac{u){c_{1}})}

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ydu+(2y-u)dy=0

y d u + (2 y - u) d y = 0

y (u) = - \frac{u}{2 W ( - \frac{u}{c _{1}} )}

Solution steps

y (u) d u + (2 y (u) - u) d y (u) = 0

Solve using substitution:

y (u) = - \frac{u}{2 W ( - \frac{u}{c _{1}} )}

y (u) = - \frac{u}{2 W ( - \frac{u}{c _{1}} )}

\int \frac{1}{4 x + 2} d x

k = 0 \sum \infty (2 k + 1) x^{2 k}

y^{^{'}} - y^{2} = 1

n = 1 \sum \infty (- \frac{1}{4})^{n}

\frac{d y}{d x} - \frac{y}{x} = 8 x e^{x}

What is the solution for ydu+(2y-u)dy=0 ?
The solution for ydu+(2y-u)dy=0 is y(u)=-u/(2\W(-\frac{u){c_{1}})}