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). $$ t^{2} y^{\prime}+2 t y-y^{3}=0, \quad t>0 $$

Short answer

Question: Solve the Bernoulli equation \(t^2y' + 2ty - y^3 = 0\) for \(t>0\). Answer: The solution for the given Bernoulli equation is \(y(t) = \frac{1}{\sqrt{C}}t^{-2}\), where \(C\) is a constant.

Step-by-step solution

Identify p(t), q(t), and n

For the given equation, we notice that it can be written in the Bernoulli form as follows: $$ t^2y' + 2ty = y^3 $$ Comparing with \(y' + p(t)y = q(t)y^n\), we have: 1. \(p(t) = \frac{2}{t}\). 2. \(q(t) = \frac{1}{t^2}\). 3. \(n = 3\).

Perform the substitution

According to Problem 27(b), we make the substitution: $$ z = y^{1-n} = y^{-2} $$ Which implies: $$ y = z^{-\frac{1}{2}} $$ Now, differentiate \(y\) with respect to \(t\): $$ \frac{dy}{dt} = \frac{d}{dt}\left(z^{-\frac{1}{2}}\right) = -\frac{1}{2}z^{-\frac{3}{2}}\frac{dz}{dt} $$ Now, substitute this result and \(y = z^{-\frac{1}{2}}\) back into the original equation: $$ t^2 \left(-\frac{1}{2}z^{-\frac{3}{2}}\frac{dz}{dt}\right) + 2t\left(z^{-\frac{1}{2}}\right) - \left(z^{-\frac{1}{2}}\right)^3 = 0 $$ This simplifies to: $$ -t^2 z^{-\frac{3}{2}}\frac{dz}{dt} + 2tz^{-\frac{1}{2}} - z^{\frac{3}{2}} = 0 $$ Multiplying both sides by \(-z^{\frac{3}{2}}\), we obtain: $$ t^2\frac{dz}{dt} - 2tz + z^3 = 0 $$

Solve the transformed linear equation

The transformed equation is: $$ t^2\frac{dz}{dt} - 2tz + z^3 = 0 $$ We can rewrite it as: $$ \frac{dz}{dt} - \frac{2}{t}z + z^3 = 0 $$ Now, perform following substitution: 1. \(v = z^2\) 2. \(\frac{dv}{dt} = 2z\frac{dz}{dt}\) So, the equation becomes: $$ t^2\frac{dv}{dt} - 4tz + 2z^3 = 0 $$ Now the equation is: $$ \frac{dv}{dt} - \frac{4}{t}v = 0 $$ This is a first-order linear differential equation with form: \(\frac{dv}{dt} + k(t)v = 0\). We can solve this using the integrating factor method. The integrating factor is: \(I(t) = e^{\int{k(t)dt}} = e^{\int{-\frac{4}{t}dt}} = e^{-4\ln{t}} = t^{-4}\) Multiply both sides by the integrating factor: $$ t^{-4}\frac{dv}{dt} - 4t^{-5}v = 0 $$ The left side is the derivative of the product \((t^{-4}v)\): $$ \frac{d}{dt}(t^{-4}v) = 0 $$ Integrate both sides with respect to \(t\): $$ t^{-4}v = C $$ Where \(C\) is a constant. Now, solve for \(v\): $$ v(t) = Ct^4 $$

Reverse the substitution and find the solution for y

Now, let's reverse the substitution for \(z\) and \(y\). We have \(v = z^2\), so: $$ z(t) = \sqrt{Ct^4} $$ Now reverse the substitution for \(y\): \(y = z^{-\frac{1}{2}}\): $$ y(t) = \left(\sqrt{Ct^4}\right)^{-\frac{1}{2}} = \frac{1}{\sqrt{C}}t^{-2} $$ This is the solution for the given Bernoulli equation.

Key concepts

Differential Equation

A differential equation is a mathematical equation that relates a function to its derivatives. These equations are fundamental in expressing physical laws and phenomena, where the rate of change of a quantity is important. The Bernoulli equation, presented in our exercise, is one such differential equation of particular interest.

Differential equations can be broadly classified as ordinary differential equations (ODE), which involve functions of a single variable and their derivatives, and partial differential equations (PDE), involving functions of multiple variables. The complexity of solving a differential equation depends on its type and order. The one provided in the exercise is an ordinary differential equation of the first order.

Solving a differential equation often involves finding an unknown function that satisfies the given relationship between the function and its derivatives. Methods vary according to the type of equation, ranging from separation of variables to more advanced techniques like the Laplace transform. The critical understanding is that each method hinges on reducing the complexity of the original equation to a more manageable form that can then be integrated to find the solution.

Integrating Factor Method

The integrating factor method is a technique used to solve linear first-order differential equations of the form \(\frac{dy}{dx} + p(x)y = q(x)\). The purpose of the method is to create a left-hand side in the differential equation that represents the derivative of a product of functions. This is achieved by multiplying through by an integrating factor that is typically an exponential function of an integral involving the coefficient of \(y\) in the original equation.

The integrating factor, which is crucial for this method, is given by \( e^{\int{p(x)dx}} \) where \(p(x)\) is a function of \(x\). Calculating this factor correctly and applying it to each side of the equation allows the simplification of the equation such that the left side becomes the derivative of a product of two functions. This technique was used in Step 3 of our exercise to simplify the otherwise intractable differential equation. Once the sides represent a product's derivative, integrating both sides yields a simpler equation that is typically solvable for the unknown function.

Change of Variables

The change of variables is a powerful method for simplifying differential equations, especially for non-linear ones like the Bernoulli equation in our exercise. This method involves substituting a new variable that reduces the equation to a more solvable form, often transforming a non-linear equation into a linear one.

In the context of the Bernoulli equation, this involves substituting a new variable \(z\) for the original dependent variable \(y\) such that \(z = y^{1 - n}\), where \(n\) is a number that characterizes the particular Bernoulli equation. This transformation is incredibly useful because it takes the non-linear term \(y^n\) and turns it into a term linear in \(z\), as seen in Step 2 of our exercise.

Once this transformation is in place, the new variable often leads to a linear differential equation that falls into a category of problems that have well-understood solution methods, such as the integrating factor method mentioned earlier. Ultimately, once the linear equation is solved for the new variable, one more change of variables is needed to express the solution back in terms of the original variable, completing the solving process as demonstrated in Step 4 of the given exercise.