Chapter 2 - First Order Differential Equations - 2.1 Linear Equations; Method of Integrating Factors - Problems - Page 41: 33

Finding the integrating factor: $\mu(t)=\exp(\int a\ dt)$ $\mu(t)=e^{at}$ Finding the y function: $y(t)=1/\mu(t) [\int \mu(s)(be^{-\lambda s})ds+c]$ $y(t)=e^{-at}[\frac{b}{a-\lambda}(e^{(a-\lambda)t})+c]$ $y(t)=\frac{b}{a-\lambda}e^{-\lambda t}+ce^{-at}$ $y(0)=y_0=\frac{b}{a-\lambda}+c\rightarrow c=y_0-\frac{b}{a-\lambda}$ $y(t)=\frac{b}{a-\lambda}(e^{-\lambda t}-e^{-at})+y_0e^{-at}$ For $a=\lambda$ Applying the L'Hospital rule on the first term: $\lim_{a\rightarrow \lambda/t\rightarrow +\infty} \dfrac{e^{-\lambda t}-e^{-at}}{a-\lambda}=\lim_{a\rightarrow \lambda/t\rightarrow +\infty}\dfrac{te^{-at}}{1}=te^{-\lambda t}\rightarrow0 $ For the second term, the exponential goes to 0, so $lim_{t\rightarrow+\infty} y(t)=0$ For $a\ne \lambda$: Since both a and $\lambda$ are positive constants, all the exponential terms go to zero as t approaches infinity, so $lim_{t\rightarrow+\infty} y(t)=0$

Finding the integrating factor: $\mu(t)=\exp(\int a\ dt)$ $\mu(t)=e^{at}$ Finding the y function: $y(t)=1/\mu(t) [\int \mu(s)(be^{-\lambda s})ds+c]$ $y(t)=e^{-at}[\frac{b}{a-\lambda}(e^{(a-\lambda)t})+c]$ $y(t)=\frac{b}{a-\lambda}e^{-\lambda t}+ce^{-at}$ $y(0)=y_0=\frac{b}{a-\lambda}+c\rightarrow c=y_0-\frac{b}{a-\lambda}$ $y(t)=\frac{b}{a-\lambda}(e^{-\lambda t}-e^{-at})+y_0e^{-at}$ For $a=\lambda$ Applying the L'Hospital rule on the first term: $\lim_{a\rightarrow \lambda/t\rightarrow +\infty} \dfrac{e^{-\lambda t}-e^{-at}}{a-\lambda}=\lim_{a\rightarrow \lambda/t\rightarrow +\infty}\dfrac{te^{-at}}{1}=te^{-\lambda t}\rightarrow0 $ For the second term, the exponential goes to 0, so $lim_{t\rightarrow+\infty} y(t)=0$ For $a\ne \lambda$: Since both a and $\lambda$ are positive constants, all the exponential terms go to zero as t approaches infinity, so $lim_{t\rightarrow+\infty} y(t)=0$