4
$\begingroup$

I am having trouble finding the integrating factor for turning the below differential equation into an exact one (Tenenbaum and Pollard, exercise 10, problem 6). Any hints and suggestions would extremely helpful and lead me to the solution.

Solve the differential equation :

$$ (x^2-y^2-y)dx - (x^2-y^2-x)dy=0$$

My attempt:

The coefficients of $dx$ and $dy$ are not homogenous functions. Further,

$ \begin{align} P(x,y) &= x^2 - y^2 - y \\ \frac{\partial P(x,y)}{\partial y}&=-2y-1 \\ Q(x,y) &= -(x^2-y^2-x)\\ \frac{\partial Q(x,y)}{\partial x}&=-(2x-1) \\ \therefore \frac{\partial P}{\partial y} \neq \frac{\partial Q}{\partial x} \end{align} $

The given differential equation is not exact. We have :

$\begin{align} & \frac{\partial P}{\partial y} - \frac{\partial Q}{\partial x} \\ =& -2y-1+(2x-1)\\ =&(2x-2y-2) \end{align}$.

Moreover,

$\begin{align} & yQ-xP\\ = & -y(x^2-y^2-x)-x(x^2-y^2-y)\\ = & -x^2 y + y^3 + xy - x^3 + xy^2 + xy\\ = & y^3 - x^3 + xy - xy(x - y - 1) \end{align}$

It doesn't look like $(\partial P / \partial y - \partial Q / \partial x)/(yQ-xP)$ will be a function of $u=xy$ alone.

Also,

$\begin{align} & yQ+xP\\ = & -y(x^2-y^2-x)+x(x^2-y^2-y)\\ = & -x^2 y + y^3 + xy + x^3 - xy^2 - xy\\ = & y^3 + x^3 - x^2 y - x y^2 \\ = & (y + x)(y^2 + x^2 - xy) - xy(y + x) \\ = & (y + x)(y^2 + x^2 - 2xy)\\ = & (y + x)(y - x)^2 \end{align}$

It doesn't look like $y^2(\partial P / \partial y - \partial Q / \partial x)/(yQ+xP)$ will be a function of $u=x/y$ or $x^2(\partial P / \partial y - \partial Q / \partial x)/(yQ+xP)$ will be a function of $u=y/x$ alone.

$\endgroup$
  • 1
    $\begingroup$ The implicit solution is given by $$-\frac{e^{2x-2y(x)}(y(x)+x)}{2(x-y(x))}=C$$ $\endgroup$ – Dr. Sonnhard Graubner Jun 1 at 16:32
  • $\begingroup$ @Dr.SonnhardGraubner, thanks! How do I find the integrating factor for this differential equation? $\endgroup$ – Quasar Jun 1 at 16:37
5
$\begingroup$

The given differential equation is not exact and I think you can't find the integrating factor by known way and an easier way, rather you can solve it as follows

${}$

$(x^2-y^2-y)dx - (x^2-y^2-x)dy=0$

$\implies (x^2-y^2)(dx-dy)+xdy-ydx=0$

$\implies (1-\frac{y^2}{x^2})(dx-dy)+\frac{xdy-ydx}{x^2}=0$

$\implies (1-\frac{y^2}{x^2})(dx-dy)+d(\frac{y}{x})=0$

$\implies (dx-dy)+ \frac{d(\frac{y}{x})}{1-\frac{y^2}{x^2}}=0$

Integrating,

$$x-y+\tanh^{-1} (\frac{y}{x}) =c \quad \text{when} \quad |y|\lt|x|$$, or in other form $$x-y+\log|\frac{1+\frac{y}{x}}{1-\frac{y}{x}}|=c$$ where $c$ is integrating constant.

$\endgroup$
3
$\begingroup$

Start with $$ (x^2-y^2-y)\,\mathrm{d}x - (x^2-y^2-x)\,\mathrm{d}y=0$$ We note that this is $$ (x^2-y^2)\,\mathrm{d}(x-y)+(x\,\mathrm{d}y-y\,\mathrm{d}x)=0 $$ So $(u,v)=(x+y,x-y)$ is a reasonable choice to try. Simplifying the resulting equation gives $$ (2u+1)v\,\mathrm{d}u=u\,\mathrm{d}v $$ which is a separable equation.

$\endgroup$
1
$\begingroup$

The part of $\dfrac{\partial P}{\partial y} - \dfrac{\partial Q}{\partial x} = 2(x-y-1)=2(z-1)$ shows that you may assume a new variable $z=x-y$ which which shows the denominator of integrating factor might be of the form $$\dfrac{\partial I}{\partial x}Q - \dfrac{\partial I}{\partial y}P = P+Q=z$$ so $$p(z)=\dfrac{2(z-1)}{z}$$

$\endgroup$
1
$\begingroup$

$(x^2-y^2-y)~dx-(x^2-y^2-x)~dy=0$

$(-x^2+y^2+x)~dy=(-x^2+y^2+y)~dx$

$(y^2-x^2+x)\dfrac{dy}{dx}=y^2-x^2+y$

You can consider as the ODE of the type http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20EquationsSecond%20Edition%5Cc2972_fm.pdf#page=169

The general solution can express as $\begin{cases}x=C|t|e^{4t}-t\\y=C|t|e^{4t}+t\end{cases}$

$\endgroup$
  • 1
    $\begingroup$ Your given link does not work. $\endgroup$ – nmasanta Jun 3 at 5:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.