What is the solution for e^{-2u+v^2}((1+u)du+2uvdv)=0 ?

The solution for e^{-2u+v^2}((1+u)du+2uvdv)=0 is v=sqrt(-u-ln(u)+c_{1)},v=-sqrt(-u-ln(u)+c_{1)}

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e^{-2u+v^2}((1+u)du+2uvdv)=0

e^{- 2 u + v^{2}} ((1 + u) d u + 2 uv d v) = 0

v = - u - ln (u) + c_{1}, v = - - u - ln (u) + c_{1}

Solution steps

e^{- 2 u + v^{2}} ((1 + u) d u + 2 uv d v) = 0

Solve separable ODE:

v = - u - ln (u) + c_{1}, v = - - u - ln (u) + c_{1}

v = - u - ln (u) + c_{1}, v = - - u - ln (u) + c_{1}

d er i v a t i v e 5 cos (x - \frac{16}{25})

\int x^{3} cos (\frac{x}{2}) d x

- 9 x^{2} y + y^{^{'}} = 9 x^{2}

\int cos (t^{5}) d t

\frac{\partial}{\partial y} (4 y^{3} cos (2 x))

What is the solution for e^{-2u+v^2}((1+u)du+2uvdv)=0 ?
The solution for e^{-2u+v^2}((1+u)du+2uvdv)=0 is v=sqrt(-u-ln(u)+c_{1)},v=-sqrt(-u-ln(u)+c_{1)}