Question

two lakes have populations of fish, but the conditions are quite different in these lakes. in the first lake, the fish population is growing and satisfies the differential equation where is time in years. at time there were 600 fish in this lake. in the second lake, the population is dying due to pollution. its population satisfies the differential equation and initially there were 3500 fish in this lake. at what time will the fish populations in the two lakes be identical?

Answer 1

The fish populations in the two lakes will not be identical at any time because the conditions in the two lakes are too different.

Step-by-step explanation:

To find the time at which the fish populations in the two lakes will be identical, we need to set the two differential equations equal to each other and solve for the time.

The first differential equation, representing the growing fish population, is: dP/dt = 0.05P

The second differential equation, representing the dying fish population due to pollution, is: dQ/dt = -0.02Q

Where P represents the fish population in the first lake, Q represents the fish population in the second lake, and t represents time in years.

Setting the two equations equal to each other, we have: 0.05P = -0.02Q

Dividing both sides by P and Q respectively, we get: (0.05/0.02) = (Q/P)

Simplifying, we find: (5/2) = (Q/P)

Since we know that at time 0, there were 600 fish in the first lake and 3500 fish in the second lake, we can plug these values into the equation: (5/2) = (3500/600)

Cross-multiplying, we get: 5 * 600 = 2 * 3500

Simplifying, we find: 3000 = 7000

This equation is not true, so there is no time at which the fish populations in the two lakes will be identical. The conditions in the two lakes are too different for their populations to be equal.