Solving first order non linear diferrential equation

Question

I am trying to solve: $$y dx+(xy+2x-ye^y) dy=0$$

I tried to use the substitution $u =yx$ but it didn't work. Also can you tell me what methods to use classically in this case please.

Answer 1

Hint :

Let $R(x,y) = y$ and $S(x,y) = 2x - e^y + xy$. The given equation is not exact, because : $R_y \neq S_x$.

This means that you need to find an integrating factor $μ(y)$ :

$$\frac{\partial}{\partial y}\big[μ(y)R(x,y)\big] =\frac{\partial}{\partial x}\big[μ(y)S(x,y)\big]$$

$$\Rightarrow$$

$$y\frac{dμ(y)}{dy} + μ(y) = (y+2)μ(y)$$

Then, after you've found such $μ$, you'll multiply both sides of your ODE and it will be an exact differential equation, which you can proceed to solving with the standard method for exact equations.

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Elaboration of the hint :

Working on the last named equation, you get :

$$\frac{\frac{\partial μ(y)}{\partial y}}{μ(y)} = \frac{1}{y} + 1 \Rightarrow \ln(μ(y)) = y + \ln(y) \Leftrightarrow μ(y) = ye^y$$

Now, if you multiply both sides by this $μ(y)$, your equation will be exact (you need to double check it, so repeat the elaboration in the begining of the hint, but now with new functions).

Define $f(x,y)$ such that : $\frac{\partial}{\partial x} P(x,y)$ and $\frac{\partial}{\partial y} Q(x,y)$ where $P,Q$ are the corresponding functions used to show the exactness of the equation as above.

The solution then will be given by $f(x,y) = c_1$, where $c_1$ is an arbitrary constant.

Integrate the following expression to find $f(x,y)$ with respect to $x$ :

$$\frac{\partial}{\partial x}f(x,y) = P(x,y)$$

and then, the $f(x,y)$ that you will find will have a "constant" function $g(y)$ with respect to $y$. Differentiation $f(x,y)$ with respect to $y$ and substitution in the expression :

$$\frac{\partial}{\partial y}f(x,y) = Q(x,y)$$

will allow you then to integrate the function $g(y)$ and find the complete expression of $f(x,y)$, which you have to solve with respect to $y$ in order to find the exact function that is the solution to the differential equation we started from.

Answer 2

$$y +(xy+2x-ye^y) \frac{dy}{dx}=0$$ $$y \frac{dx}{dy}+(xy+2x-ye^y) =0$$ $$y \frac{dx}{dy}+(y+2)x =ye^y$$ This a simple first order linear ODE considering the function $x(y)$

To make it more obvious, let $x=Y$ and $y=X$ : $$X \frac{dY}{dX}+(X+2)Y =Xe^X$$ I suppose that you know how to solve it, leading to : $$Y=\frac{c}{e^X X^2}+\bigg(\frac12-\frac{1}{2X}+\frac{1}{4X^2}\bigg)e^X$$ Coming back to the original notation, the solution is expressed on implicit form, that is the function $x(y)$ : $$x=\frac{c}{e^y y^2}+\bigg(\frac12-\frac{1}{2y}+\frac{1}{4y^2}\bigg)e^y$$ As far as I know, there is no closed form (made of a finite number of standard functions) for the inverse function $y(x)$.

Answer 3

Consider the equation to be $$y x'+(xy+2x-ye^y) =0$$ To get rid of the unpleasnt term in brackets, let $$xy+2x-ye^y=z \implies x=\frac{z+ye^y }{y+2}\implies x'=\frac{z'+ye^y +e^y}{y+2}-\frac{z+ye^y }{(y+2)^2}$$ Replace in the initial equation $$y (y+2) z'+(y^2+3y+4) z=- y (y^2+2y+2)e^y$$ The homogeneous part is quite simple; then continue with the variation of parameters.

Answer 4

$$y dx+(xy+2x-ye^y) dy=0$$ divide by $y\ne0$ $$ dx+(x+2\frac xy-e^y) dy=0$$ x is now the function $$ x'=-(x+2\frac xy-e^y) $$ $$ x'+x(1+\frac 2y)=e^y $$ Which is simple to solve