# Dealing with absolute value in the process of solving differential equations

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.)

• umm, should it be $dy/dt$ not $dy/dy$? Feb 13 '18 at 4:22
• @user254433 it is, not used to using the notation for equations so in the process I wrote $dy$ twice, already fixed it, thank you! 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 : $$\frac{1}{t+1}\frac{dy}{dt}-\frac{y}{(t+1)^2}=\frac{4t^2+4t}{t+1}=4t$$ $$\frac{d}{dt}\left(\frac{y}{t+1}\right)=4t$$ $$\frac{y}{t+1}=2t^2+c$$ $$y=2t^2(t+1)+c\:(t+1)$$
• 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. Feb 13 '18 at 12:29