Chapter 2 - First Order Differential Equations - 2.1 Linear Equations; Method of Integrating Factors - Problems - Page 40: 29

Finding the integrating factor: $\mu(t)=\exp(\int 1/4\ dt)$ $\mu(t)=e^{t/4}$ Finding the y-function: $y=\frac{1}{\mu(t)}[\int \mu(s)(3+2\cos 2s)ds+c]$ $y=e^{-t/4}[4/65\ e^{t/4}(16\sin 2t+2\cos 2t+195)+c]$ $y=4/65\ (16\sin 2t+2\cos 2t+195)+ce^{-t/4}$ $y(0)=788/65+c=0\rightarrow c=-788/65$ a) $y=4/65\ (16\sin 2t+2\cos 2t+195-197e^{-t/4})$ At $t\rightarrow+\infty$, the exponential term goes to zero, so the value will be between a minimum when $\sin(2t)=-1, \cos(2t)=0$ and amaximum when $\sin(2t)=1, \cos(2t)=0$, i.e., $y(t\rightarrow +\infty) \rightarrow [716/65,\ 844/65]$ b) $12=4/65\ (16\sin 2t+2\cos 2t+195-197e^{-t/4})$ $16\sin 2t+2\cos 2t-197e^{-t/4}=0$ The first solution is $t=10.066$

Finding the integrating factor: $\mu(t)=\exp(\int 1/4\ dt)$ $\mu(t)=e^{t/4}$ Finding the y-function: $y=\frac{1}{\mu(t)}[\int \mu(s)(3+2\cos 2s)ds+c]$ $y=e^{-t/4}[4/65\ e^{t/4}(16\sin 2t+2\cos 2t+195)+c]$ $y=4/65\ (16\sin 2t+2\cos 2t+195)+ce^{-t/4}$ $y(0)=788/65+c=0\rightarrow c=-788/65$ a) $y=4/65\ (16\sin 2t+2\cos 2t+195-197e^{-t/4})$ At $t\rightarrow+\infty$, the exponential term goes to zero, so the value will be between a minimum when $\sin(2t)=-1, \cos(2t)=0$ and amaximum when $\sin(2t)=1, \cos(2t)=0$, i.e., $y(t\rightarrow +\infty) \rightarrow [716/65,\ 844/65]$ b) $12=4/65\ (16\sin 2t+2\cos 2t+195-197e^{-t/4})$ $16\sin 2t+2\cos 2t-197e^{-t/4}=0$ The first solution is $t=10.066$