Solve each of the following equations.<br>(4b + 11)(3b

g Suppose that the populationy(t)of a certain kind of fish is given by the logistic modely′= 3y−2y2(a) Solve the Bernoulli equat

Ksju [112]

Answer:

y = $\frac{3}{2 + 3Ce^{-3t} }$

Step-by-step explanation:

As given , y' = 3y - 2y²

⇒y' - 3y = -2y²

Divide by y² in the above equation

⇒ $\frac{y'}{y^{2} } - \frac{3y}{y^{2} } = -\frac{2y^{2} }{y^{2} }$

⇒ $\frac{y'}{y^{2} } - \frac{3}{y } = -2$ ........(1)

Now , let $\frac{1}{y}$ = u

⇒- $\frac{1}{y^{2} } \frac{dy}{dt} = \frac{du}{dt}$

⇒- $\frac{1}{y^{2} }y' = \frac{du}{dt}$

∴ equation (1) becomes

- $\frac{du}{dt}$ - 3u = -2

⇒ $\frac{du}{dt}$ + 3u = 2

It is a linear differential equation

Now,

Integrating factor = I.F = $e^{\int\limits {3} \, dt }$ = $e^{3t}$

∴ The solution becomes

u.(I.F) = $\int\limits {2.(I.F)} \, dt$ + C

⇒u.( $e^{3t}$ ) = $\int\limits {2.(e^{3t} )} \, dt$ + C

⇒u.( $e^{3t}$ ) = $\frac{2e^{3t} }{3}$ + C

⇒u = $\frac{2}{3} + Ce^{-3t}$

As $\frac{1}{y}$ = u

⇒ $\frac{1}{y}$ = $\frac{2}{3} + Ce^{-3t}$

⇒ y = $\frac{1}{\frac{2}{3} + Ce^{-3t} } = \frac{3}{2 + 3Ce^{-3t} }$

⇒ y = $\frac{3}{2 + 3Ce^{-3t} }$

6 0

3 years ago

Why don't we all just work on this together don't we all want to get the answer for this question come on guy's we have to be sm

IgorLugansk [536]

I Agree with you completely, but to be honest with you I Myself is more like the person who provides answers, rather than questions, but for the people who do just ask questions, I do feel you should take what this person said into consideration. ^.^

3 0

4 years ago

Read 2 more answers

Given: AE ≅ CE ; DE ≅ BE Prove: ABCD is a parallelogram. Parallelogram A B C D is shown. Diagonals are drawn from point A to poi

nikklg [1K]

Answer:

The proof is below

Step-by-step explanation:

Given a parallelogram ABCD. Diagonals AC and BD intersect at E. We have to prove that AE is congruent to CE and BE is congruent to DE i.e diagonals of a parallelogram bisect each other.

In ΔACD and ΔBEC

AD=BC (∵Opposite sides of a parallelogram are equal)

∠DAC=∠BCE (∵Alternate angles)

∠ADC=∠CBE (∵Alternate angles)

By ASA rule, ΔACD≅ΔBEC

By CPCT(Corresponding Parts of Congruent triangles)

AE=EC and DE=EB

Hence, AE is conruent to CE and BE is congruent to DE

6 0

3 years ago

The Ransin Sports Company has noted that the size of individual customer orders is normally distributed with a mean of $112 and

Mashutka [201]