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[tex]\frac{dy}{dt}=r(t)y - k, \qquad y(0)=y_0[/tex]

where the growth rate [tex]r(t)[/tex] is given by

[tex]r(t)=\frac{1+\sin t}{5}[/tex]

and [tex]k[/tex] represents the rate of predation.

(a) Suppose that [tex]k=\frac{1}{5}[/tex]. Plot [tex]y[/tex] versus [tex]t[/tex] for several values of [tex]y_0[/tex] between 1/2 and 1.

(b) Estimate the critical initial population [tex]y_c[/tex] below which the population will become extinct.

(c) Choose other values of [tex]k[/tex] and find the corresponding [tex]y_c[/tex] for each one.

(d) Use that data you have found in parts (a) and (b) to plot [tex]y_c[/tex] versus [tex]k[/tex]."

I'm trying to apply the Method of Integrating Factors, but I'm stuck. Here's what I have:

[tex]\frac{dy}{dt}- r(t)y = - k[/tex]

[tex]\mu = \exp \left[ -\frac{1}{5} \int \left( 1 + \sin t \right) \: dt \right] = \exp \left( \frac{\cos t}{5} - \frac{t}{5} \right)[/tex]

[tex]y(t) = \exp \left( \frac{t}{5} - \frac{\cos t}{5} \right) \int -k \exp \left( \frac{\cos t}{5} - \frac{t}{5} \right) \: dt[/tex]

[tex]y(t) = -k \exp \left( \frac{t}{5} - \frac{\cos t}{5} \right) \underbrace{\int \exp \left( \frac{\cos t}{5} - \frac{t}{5} \right) \: dt} _{\mbox{?}}[/tex]

Any help is highly appreciated.