0
$\begingroup$

I'm struggling to understand how this step works when solving a differential equation by integrating factors.

The question is, solve: $$x\frac{dy}{dx} - 3y = x^4 \cos(x), $$for $ x > 0$.

I don't understand how to get from step (a) to step (b).

$$(a): \frac{1}{x^3}y' - \frac{3}{x^4}y = \cos(x),\\ (b): \frac{d}{dx}\frac{1}{x^3}y = \cos(x).$$

In the notes it mentions that the 2nd term is a derivative of the first term but it doesn't really help me understand what's going on.

Thanks.

$\endgroup$
  • $\begingroup$ All integrating factor questions are the same. A simpler example might be to see that $y' + xy = \frac{d}{dx}(xy)$ $\endgroup$ – Kaynex Dec 28 '16 at 22:47
2
$\begingroup$

We have a differential equation $$a(x)y'+b(x)y=c(x)$$ Divide through by $a(x)$ to obtain $$y'+p(x)y=q(x)$$ Now imagine that we have a convenient term $k(x)$ such that $$k(x)y'+p(x)k(x)y=(k(x)y)'=k'(x)y+y'k(x)$$ This yields a differential equation for $k(x)$, i.e. $$k'(x)=k(x)p(x)$$ A separable equation with solution $k(x)=Ce^{\int p(x)}$. Now, because of the way that we have constructed $k(x)$, when we multiply $y'+p(x)y=q(x)$ through by $k(x)$, we obtain $$k(x)y'+k(x)p(x)y=(yk(x))'=q(x)$$ Which can be solved for $y$ by integrating both sides and dividing by $k(x)$. In your specific case $p(x)=-\frac 3x$ and $q(x)=x^3\cos x$.

$\endgroup$
  • $\begingroup$ Well, explained, but I think $p(x)$ is missing a minus sign. $\endgroup$ – Mike Dec 29 '16 at 0:29
  • $\begingroup$ @Mike good catch, had not noticed. $\endgroup$ – Guacho Perez Dec 29 '16 at 1:10
1
$\begingroup$

Use the product rule on $\frac{1}{x^3}y$ and you'll get exactly the terms from part (a).

$\endgroup$
1
$\begingroup$

You can just check that by performing the differentiation in $(b)$ (the notation is kind of misleading): $$ \frac{\mathrm{d}}{\mathrm{d} x} \left( \frac{1}{x^3} y(x) \right) = \frac{-3}{x^4} y(x) + \frac{1}{x^3} y'(x). $$

$\endgroup$

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.