Solving a Differential Equation with Boundary Conditions and Limits

[prolong199]

Im studying for my final coming up and i can't figure out the differential equation in the practice exam, can someone please help me?

y'=(y-2)^2(y-4) if y(0)=3, the limit of y(t) as t goes to infinity is?

a) infinity
b) 4
c) 2
d) 0
e) -infinity

thanks.

 

[prolong199]

The solutions say the answer is c) 2 but i have no idea. I tried expanding it out with the y terms and taking over to the left with dy, and the constant on the right with dt, then integrate is this correct?

 

[AKG]

It is 2. Draw a graph of y' as a function of y, and start with your finger on the y-axis at y=3. Move your finger along the axis at a speed dictated by the value of y' for whatever y you're at. At y = 3, for example, y' is negative, so start moving left. What would happen if you reached 2, where the speed is 0? You'd stop. And you'd stay there. So either you keep tending towards 2, or you reach 2 and stay there. Either way, the limit is 2. There's some way to make all this rigorous, but you're the one studying this stuff so you figure it out.

 

[teclo]

Im studying for my final coming up and i can't figure out the differential equation in the practice exam, can someone please help me?

y'=(y-2)^2(y-4) if y(0)=3, the limit of y(t) as t goes to infinity is?

a) infinity
b) 4
c) 2
d) 0
e) -infinity

thanks.

the first thing you'd want to do is get y as a function of t.

whoops, i missed your second post. yes, you want to solve for the function y{t} and take the limit as it goes to infinity. you have the right approach, the integration just gets a little messy (but not bad really, assuming that your functions are really multiplied by each other and not one to the power of the other)