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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:

$$\frac{dy}{dt}-\frac{1}{t+1}y(t)=4t^2+4t$$

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.)

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  • $\begingroup$ 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
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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 : $$\frac{1}{t+1}\frac{dy}{dt}-\frac{y}{(t+1)^2}=\frac{4t^2+4t}{t+1}=4t$$ $$\frac{d}{dt}\left(\frac{y}{t+1}\right)=4t$$ $$\frac{y}{t+1}=2t^2+c$$ $$y=2t^2(t+1)+c\:(t+1)$$

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  • $\begingroup$ 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

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