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). \(y^{\prime}=\epsilon y-\sigma y^{3}, \epsilon>0\) and \(\sigma>0 .\) This equation occurs in the study of the stability of fluid flow.

Short answer

Question: Solve the following Bernoulli equation: \(y^{\prime}=\epsilon y-\sigma y^{3}\). Answer: The general solution to the given Bernoulli equation is \(y^{3}(t) = \frac{2}{3}(2\epsilon\frac{1}{y} + \frac{2}{3}\sigma y^3 + C)\).

Step-by-step solution

Perform the given substitution

Let's perform the given substitution to the problem. We have to substitute \(v=y^{1-n}\), and in our case, \(n=3\), thus resulting in: $$ v=y^{1-3}=y^{-2} $$ Now, we need to find \(v^{\prime}\) to substitute it in the original equation. To do this, we will first take the derivative of \(y\) with respect to \(t\) and then take the derivative of \(v\) with respect to \(t\) as well. Therefore, we need to use the chain rule as follows: $$ \frac{dv}{dt}=\frac{d(y^{-2})}{dy} \cdot \frac{dy}{dt}=-2y^{-3}\cdot y^{\prime} $$

Substitute the values in the given equation

We need to substitute the calculated derivatives in the given equation and solve for \(v^{\prime}\). The given equation is: $$ y^{\prime}=\epsilon y-\sigma y^{3} $$ Substitute \(y^{\prime}\) as: $$ y^{\prime}=-\frac{v^{\prime}}{2y^{3}} $$ The equation becomes: $$ -\frac{v^{\prime}}{2y^{3}}=\epsilon y-\sigma y^{3} $$

Solve for \(v^{\prime}\)

Let's solve the above equation for \(v^{\prime}\): $$ v^{\prime}=-2(\epsilon y-\sigma y^{3})y^{3} $$ Now, the equation we have is a linear ordinary differential equation in \(v\): $$ v^{\prime} = -2\epsilon y^4 + 2\sigma y^6 $$

Integrate to find \(v(t)\)

Integrate both sides of the equation with respect to \(t\): $$ \int v^{\prime} dt = \int -2\epsilon y^4 dt + \int 2\sigma y^6 dt $$ To perform integration, we substitute \(v=y^{-2}\) back: $$ \int v^{\prime} dt = -2\epsilon \int y^{-2} dt + 2\sigma \int y^{2} dt $$

Solve the integral equation

Now, solve the integral equation: $$ v(t)=-2\epsilon \int y^{-2} dt + 2\sigma \int y^{2} dt + C $$ By finding the integral, we get: $$ v(t)=2\epsilon\frac{1}{y} + \frac{2}{3}\sigma y^3 + C $$ And solving for \(y(t)\): $$ y^{3}(t) = \frac{2}{3}(2\epsilon\frac{1}{y} + \frac{2}{3}\sigma y^3 + C) $$ Hence, this is the general solution to the given equation.