Solve $ye^ydx+(1+xe^y)dy=0$ Solve $ye^ydx+(1+xe^y)dy=0$
I try to solve it by finding the integration factor $\mu$.
Denote $P=ye^y, Q=1+xe^y$.
The ode isn't exact since $P_y\neq Q_x$ .
$\frac{Q_x-P_y}{-P}=\frac{-ye^y}{-ye^y}=1$ , if $\mu=1$ I got the same ode.
How can I solve this?
Help, please.
Thanks !
 A: Hint: Rewrite it as an ODE using $\dfrac{dx}{dy}$. That means you treat $x = f(y)$ and then solve for $y$ in terms of $x$ by taking inverse.
A: Multiply both sides by $e^{-y}$ to get:
$$ydx+(e^{-y}+x)dy=0\tag 1$$
Note that $(1)$ is an exact ODE and that infact LHS of $(1)$ can be written as $d(xy-e^{-y})$.
So solution to the ode is: $xy-e^{-y}=c$, where $c$ is any constant.
Edit: The given equation is not an exact ODE. Let's try to make it an ODE by multiplying by a function $\mu$. If such a function exists then the following (equation $(2)$) is an exact ODE:
$$\mu ye^ydx+\mu(1+xe^y)dy=0\tag 2$$
Since $(2)$ is an exact ODE, it follows that $(\mu ye^y)_y=(\mu (1+xe^y))_x \tag 3$
Simplifying $(3)$ gives: $\mu_y (ye^y)+(e^y+ye^y)\mu=\mu_x(1+xe^y)+e^y\mu\tag 4$
Assuming $\mu$ to be free from $x$, it follows that $\mu_x=0$ , which alongwith $(4)$ gives: $$\mu_y(ye^y)+\mu(ye^y)=0$$, which is satisfied by $\mu=e^{-y}$.
A: The given equation can be written as $$ye^y\frac{dx}{dy}+1+xe^y=0$$Or,$$\frac{dx}{dy}+\frac1yx=-\frac{e^{-y}}y$$This is of the form$$\frac{dx}{dy}+Px=Q$$Thus, $$\text{Integrating factor}=e^{\int\frac1ydy}=y$$Therefore, the solution is$$xy=-\int e^{-y}dy+c\\\implies xy=e^{-y}+c$$Verification: Differentiating the obtained solution w.r.t. $y$ $$x+y\frac{dx}{dy}=-e^{-y}$$ Multiplying by $e^y$ $$xe^y+ye^y\frac{dx}{dy}=-1\\\implies (xe^y+1)dy+ye^ydx=0$$
A: $$ye^ydx+(1+xe^y)dy=0$$
$$e^y(ydx+xdy)+dy=0$$
$$e^yd(xy)+dy=0$$
This is separable.
$$d(xy)=-e^{-y}dy$$
Integrate.
