Currently learning how to solve linear differential equations using the idea of the Product Rule of differentiation and finding the integrating factor. I keep encountering the same problem of not being sure how to deal with absolute values that appear in the process of reaching to a solution. For example:


Integrating factor: $$g(t)=-\frac{1}{t+1}$$ $$M(t)=e^{\int g(t)dt}$$ $$M(t)=e^{\int-\frac{1}{t+1}dt}$$ $$M(t)=e^{-\ln |t+1|}$$ $$M(t)=\frac{1}{|t+1|}$$

Solving the equation: $$\frac{dy}{dt}-\frac{1}{t+1}y(t)=4t^2+4t$$ $$M(t)\frac{dy}{dt}-M(t)\frac{1}{t+1}y(t)=M(t)(4t^2+4t)$$ $$\left(\frac{1}{|t+1|}\right)\frac{dy}{dt}-\left(\frac{1}{|t+1|}\right)\left(\frac{1}{t+1}\right)y(t)=\left(\frac{1}{|t+1|}\right)(4t^2+4t)$$

How do I finish solving this equation?

(I've never completely understood the concept of absolute value, so when dealing with problems like this I don't know how to get rid of it.)

  • $\begingroup$ umm, should it be $dy/dt$ not $dy/dy$? $\endgroup$
    – user254433
    Feb 13 '18 at 4:22
  • $\begingroup$ @user254433 it is, not used to using the notation for equations so in the process I wrote $dy$ twice, already fixed it, thank you! $\endgroup$ Feb 13 '18 at 12:19

Obviously, your equation isn't $\quad \frac{dy}{dy}-\frac{1}{t+1}=4t^2+4t\quad$ but is : $$\frac{dy}{dt}-\frac{1}{t+1}y(t)=4t^2+4t$$ The integrating factor is $\quad \frac{1}{t+1}\quad$ so that : $$\frac{1}{t+1}\frac{dy}{dt}-\frac{y}{(t+1)^2}=\frac{4t^2+4t}{t+1}=4t$$ $$\frac{d}{dt}\left(\frac{y}{t+1}\right)=4t$$ $$\frac{y}{t+1}=2t^2+c$$ $$y=2t^2(t+1)+c\:(t+1)$$

  • $\begingroup$ How did you find the integrating factor? If you calculated it like I did it should’ve been inside an absolute value. How did you remove it? (That is the core question of my post.) $\endgroup$ Feb 13 '18 at 12:18
  • $\begingroup$ When you have an integration factor $\mu$, you can take $C\mu$ with any constant $C\neq 0$ as other integrating factor. This changes nothing in the further calculus. Thus, $\mu$ or $-\mu$ or $|\mu|$ are equivalent integrating factors. $\endgroup$
    – JJacquelin
    Feb 13 '18 at 12:29

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.