4
$\begingroup$

An ODE of the form $M(x,y)dx+N(x,y)dy=0$ is called "good" if $\frac{\partial (M(x,y))}{\partial y}=\frac{\partial (N(x,y))}{\partial x}$

We are given the differential equation $(3x^2y+2xy+y^3)dx+(x^2+y^2)dy=0$. This ODE is not "good". We are asked to find $\mu (x,y)$ such that:

$$\mu (x,y)(3x^2y+2xy+y^3)dx+\mu (x,y)(x^2+y^2)dy=0, (*)$$ is "good".

What I did:

if the equation $(*)$ is good then $\mu_y (x,y) M(x,y)+\mu (x,y)M_y (x,y)=\mu_x (x,y)N(x,y)+\mu (x,y)N_x (x,y)$

so we get

$\mu_y(x,y)(3x^2y+2xy+y^3)+\mu(x,y)(3x^2+2x+3y^2)=\mu_x(x,y) (x^2+y^2)+2x \mu(x,y)$

And now I'm stuck.

Even if we were to guess $\mu_x(x,y)=0$ or $\mu_y(x,y)=0$ we will never get something like $\frac{\mu_y}{\mu}=\phi(y)$ or $\frac{\mu_x}{\mu}=\psi(x)$. $\mu$ seems to depend on both variables and unless the above restrictions apply (which they don't here) I don't know how to find $\mu$. Please help.

$\endgroup$
  • 3
    $\begingroup$ The standard terminology for 'good' is 'exact'. $\endgroup$ – Git Gud Nov 23 '14 at 18:16
  • 2
    $\begingroup$ I'm pretty sure it is called either "exact" or "total". Never heard of "good". $\endgroup$ – UserX Nov 23 '14 at 18:17
  • $\begingroup$ Well, as long as the question was understandable... $\endgroup$ – Oria Gruber Nov 23 '14 at 18:18
  • 2
    $\begingroup$ And $\mu$ is called the 'integrating factor' $\endgroup$ – RHP Nov 23 '14 at 18:18
  • 2
    $\begingroup$ Try $\mu(x,y)=e^{3x}$. $\endgroup$ – Git Gud Nov 23 '14 at 18:21
3
$\begingroup$

Notice that: $$ \frac{M_y - N_x}{N} = \frac{(3x^2 + 2x + 3y^2) - (2x)}{x^2 + y^2} = 3 $$ which is independent of $y$ (and also $x$, by conincidence). This suggests that we guess that $\mu_y = 0$ so that $\mu$ is a function of $x$ only. Thus, we obtain: $$ \mu M_y = \mu_x N + \mu N_x \iff \frac{d\mu}{dx} = \frac{M_y - N_x}{N} \cdot \mu = 3\mu $$ This ODE is separable/linear and can be easily solved to obtain $\mu(x) = e^{3x}$.

$\endgroup$
1
$\begingroup$

Solution 1

Hint

Make the ansatz $\mu(x)$(the reason is explained in Adriano's answer)

$$(\mu (x)M(x,y))_y=(\mu (x)N(x,y))_x\iff (3x^2+2x+3y^2)\mu (x)=\mu (x)_x (x^2+y^2)+2x\mu (x)\iff \frac{\mu (x)_x}{\mu (x)}=3 \iff \mu (x)=e^{3x}$$

Solution 2

Hint

$$3x^2y+(x^2+y^2)y'+y^3+2xy=0\stackrel{y\colon =xv}{\iff} (x^2+x^2v^2)(xv'+v)+x^3v^3+3x^3v++2x^2v\;\stackrel{\text{simplify}}{\iff}\; v'=\frac{-v^3-xv^3-3v-3xv}{x(v^2+1)}\iff v'=-\frac{v(x+1)(v^2+3)}{x(v^2+1)}\iff \frac{v'(v^2+1)}{(v^2+3)v}=-\frac{x+1}{x}$$

Notation in case 1;

$X(a,b)_a\colon =\frac{\partial X(a,b)}{\partial a}$

Notation in case 2;

$(•)'\colon=\frac{\mathrm{d}•}{\mathrm{d}x}$

$\endgroup$

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.