A trick to find a solution for : $ydx + (x+x^2y^4)dy = 0$

Question

In the first part of my question I already proved that if $P(x,y)dx + Q(x,y)dy = 0$ is an exact equation and its solution is $F(x,y)=c$, then for each differentiable function $\mu(t)$ we get that the following equation is also exact:

$$\mu(F(x,y))P(x,y)dx+\mu(F(x,y))Q(x,y)dy = 0 $$

Then, second part says: Let $P(x,y)dx+Q(x,y)dy=0$ be an exact equation, and let $F(x,y)=c$ be its solution, which applies $\nabla F = (P,Q)$. now lets take a look at: $$Pdx + (Q+Q_1)dy=0$$ which is not exact, and we suggest to find an integration factor of the form $\mu = \mu(F(x,y))$.

Use this way to find a solution for: $$ydx + (x+x^2y^4)dy = 0$$

so, I thought that what we really need to deal with is: $$\mu(F(x,y))P(x,y)dx+\mu(F(x,y))(Q_1 + Q)dy = 0 $$

And I marked $\widetilde{Q}(x,y)= \mu(F(x,y))(Q_1 + Q) $

And also marked: $\widetilde{P}(x,y)= \mu(F(x,y))P(x,y) $

and then I wanted to prove that the follow partial derivatives are equal: $$\widetilde{P_y}^\prime(x,y)= \widetilde{Q_x}^\prime(x,y)$$

then, here comes the uncertain part of my solution, I derived the upper equation in this way:

$\frac{du}{dt}\frac{dF}{dy}P(x,y) + P_y^\prime$$\mu(F(x,y))$ = $\frac{dy}{dt}\frac{dF}{dx}(Q_1+Q) + (Q_1+Q)_x ^\prime\mu(F(x,y)) $

and I all could get from here by some calculations is:

• $P_y^\prime$ = $ (Q_1)_{x} ^\prime + Q_x^\prime$

• $\widetilde{F}_y ^\prime$$P(x,y)$= $\widetilde{F}_x ^\prime$$(Q_1 + Q)$

And what I mean by $\widetilde{F}_y ^\prime$ and $\widetilde{F}_x ^\prime$ is the solution of the $new$ equation.

for somehow, I got stuck here, how do I continue ? and if there are mistakes what are they? Any kind of help would be appreciated.

Answer 1

$$ydx + (x+x^2y^4)dy = 0$$ $$ydx + xdy + x^2y^4dy = 0$$ $ydx+xdy=d(xy)$ $$d(xy)+x^2y^4dy=0$$ let $z=xy$ $$dz+z^2y^2dy=0$$ $$\frac{dz}{z^2}=-y^2dy$$ $$-\frac{1}{z}=-\frac{1}{3}y^3+constant$$ $$z=\frac{3}{y^3+C}$$ $$x=\frac{3}{y(y^3+C)}$$ Solving $y^4+Cy-\frac{3}{x}=0$ for $y$ would lead to $y(x)$

Answer 2

$$y dx+(x+x^2 y^4)dy =0$$ Method of integrating factor :

$p(x,y)dx+q(x,y)dy=0$ where $p=y$ and $q=x+x^2 y^4$

$$\frac{1}{xp-yq}\left( \frac{\partial q}{\partial x}-\frac{\partial p}{\partial y} \right) = \frac{1+2xy^4-1}{xy-y(x+x^2y^4)}=-\frac{2}{xy}=f(xy)$$ With reference to http://mathworld.wolfram.com/OrdinaryDifferentialEquation.html , eq.(12), the integrating factor is $xy$-dependant, i.e. : $\mu(xy)$ , to be not confused with $\mu(x,y)$.

The exact ODE is $dF=P(x,y)dx+Q(x,y)dy=0$ where $P=y\, \mu(xy)$ and $Q=(x+x^2y^4)\mu(xy)$ $$ \frac{\partial ^2 F}{\partial x \partial y}=\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x} $$ $$ \mu+xy\mu'=(1+2xy^4)\mu+(x+x^2y^4)t\mu' $$ $$xy\mu'+2\mu(xy)=0$$ $$\mu=\frac{1}{(xy)^2}$$ The exact ODE is : $$\frac{1}{(xy)^2}\left(y\:dx+(x+x^2y^4)dy\right)=0$$ $$d\left( -\frac{1}{xy}+\frac{1}{3}y^3\right)=0$$ $$-\frac{1}{xy}+\frac{1}{3}y^3=constant$$