Differential equations- Pop Growth

[dm59]

dy/dt=(.5+sint)y/5 what is time t when the population has doubled?

 

[sutupidmath]

$$\frac{dy}{dt}=\frac{(.5+sint)y}{5}$$ Well, you first need to solve this one, and come up with a function that predicts population at any point in time t. Do you know how to solve this diff, eq?

$$\frac{dy}{dt}=\frac{(.5+sint)y}{5}=>\frac{dy}{y}=\frac{1}{5}(.5+sint)dt$$

Now integrate on both sides, and solve for y. I guess Y=f(t) or sth like that.

SO there is another info. that you have been provided with, Y(t)=2Yo, where Yo is your initial population. Have you been provided with initial population?or with another info. on this problem?

 

[Dick]

$$\frac{dy}{dt}=\frac{(.5+sint)y}{5}$$ Well, you first need to solve this one, and come up with a function that predicts population at any point in time t. Do you know how to solve this diff, eq?

$$\frac{dy}{dt}=\frac{(.5+sint)y}{5}=>\frac{dy}{y}=\frac{1}{5}(.5+sint)dt$$

Now integrate on both sides, and solve for y. I guess Y=f(t) or sth like that.

SO there is another info. that you have been provided with, Y(t)=2Yo, where Yo is your initial population. Have you been provided with initial population?or with another info. on this problem?

You don't need the initial population. The ode is linear. But you do need to write down the solution, as stupidmath points out. The general solution will have the form C*f(t). So you just need to solve C*f(t)=2*C*f(0) for t. You don't need to know C. It cancels.

 

[dm59]

Got it!

Thank both of you so much for your help.