Question

Sometimes it is possible to solve a nonlinear equation by making a change of the dependent variable that converts it into a linear equation. The most important such equation has the form $$ y^{\prime}+p(t) y=q(t) y^{n} $$ and is called a Bernoulli equation after Jakob Bernoulli. the given equation is a Bernoulli equation. In each case solve it by using the substitution mentioned in Problem 27(b). \(y^{\prime}=r y-k y^{2}, r>0\) and \(k>0 .\) This equation is important in population dynamics an is discussed in detail in Section 2.5 .

Short answer

Question: Solve the Bernoulli equation \(y^{\prime} = ry - ky^2\) using the substitution mentioned in the corresponding Problem 27(b). Answer: The solution to the given Bernoulli equation is \(y = \frac{1}{\frac{-k}{r} + Ce^{-rt}}\).

Step-by-step solution

Rewrite the equation in the standard Bernoulli form

We are given the equation as $$y^{\prime} = ry - ky^2$$ To rewrite it in the standard Bernoulli equation form, we need to rewrite it as: $$y^{\prime} - ry = - ky^2$$ So, we have \(p(t) = -r\), \(q(t) = -k\), and \(n = 2\).

Apply the substitution

According to the hint of using the substitution mentioned in Problem 27(b), we need to apply the substitution \(z = y^{1-n} = y^{1-2} = y^{-1}\). This means that \(y = z^{-1}\). Now, let's find the derivative of the substitution: $$z^{\prime} = \frac{d}{dt}y^{-1} = -y^{-2} y^{\prime}$$. Now we will replace y and its derivative in the original equation with the transformed variable z: $$-y^{-2} y^{\prime} - (-r)y = - ky^2$$

Simplify the transformed equation

Replacing \(y^{\prime}\) with \(-y^2 z^{\prime}\) and \(y\) with \(z^{-1}\), we have: $$z^{-1} z^{\prime} + rz^{-1} = -kz^2$$ Multiplying the equation by \(z^2\), we get: $$z^{\prime} + r z = -k$$ Now, we have a transformed linear equation that we can solve using an integrating factor.

Solve the linear equation

Let's find an integrating factor for the linear equation with z. The integrating factor is given by: $$\mu(t) = e^{\int rt dt} = e^{r \int t dt} = e^{rt}$$ Now multiply the transformed linear equation by the integrating factor: $$e^{rt} z^{\prime} + r e^{rt} z = - e^{rt} k$$ Now, we can observe that the left side of the equation is the derivative of \((e^{rt} z)\) with respect to t. So, we can write it as: $$\frac{d}{dt}(e^{rt} z) = - e^{rt} k$$ Now, integrate both sides of the equation with respect to t: $$\int \frac{d}{dt}(e^{rt} z) dt = \int - e^{rt} k dt$$ $$e^{rt} z = \frac{- k}{r} e^{rt} + C$$ Now, solve for z: $$z = \frac{- k}{r} + Ce^{-rt}$$

Revert the substitution and obtain the solution for y

Now, we need to revert the substitution. Recall that \(y = z^{-1}\). So, the solution for y is given by: $$y = \frac{1}{z} = \frac{1}{\frac{- k}{r} + Ce^{-rt}}$$ And this is the solution to the given Bernoulli equation.