Question

The equation \(y^{\prime}(t)+a y=b y^{p},\) where \(a, b,\) and \(p\) are real numbers, is called a Bernoulli equation. Unless \(p=1,\) the equation is nonlinear and would appear to be difficult to solve-except for a small miracle. By making the change of variables \(v(t)=(y(t))^{1-p},\) the equation can be made linear. Carry out the following steps. a. Letting \(v=y^{1-p},\) show that \(y^{\prime}(t)=\frac{y(t)^{p}}{1-p} v^{\prime}(t)\). b. Substitute this expression for \(y^{\prime}(t)\) into the differential equation and simplify to obtain the new (linear) equation \(v^{\prime}(t)+a(1-p) v=b(1-p),\) which can be solved using the methods of this section. The solution \(y\) of the original equation can then be found from \(v\).

Short answer

#Solution# #tag_title#b. Solve the linear equation in terms of v and find y(t)#tag_content#We have obtained the linear differential equation: \[ v^{\prime}(t) + a(1-p)v(t) = b(1-p) \] To solve this equation, first find the integrating factor e^(a(1-p)t). Next, multiply both sides of the equation by this integrating factor: \[ e^{a(1-p)t}v^{\prime}(t) + a(1-p)e^{a(1-p)t}v(t) = b(1-p)e^{a(1-p)t} \] The left-hand side is now the derivative of a product: (e^(a(1-p)t)v(t))'. Then, integrate both sides with respect to t: \[ \int \frac{d}{dt}(e^{a(1-p)t}v(t)) dt = \int b(1-p)e^{a(1-p)t} dt \] We get: \[ e^{a(1-p)t}v(t) = \frac{b(1-p)}{a(1-p)}e^{a(1-p)t} + C \] Now, solve for v(t): \[ v(t) = \frac{b(1-p)}{a(1-p)} + Ce^{-a(1-p)t} \] Finally, we know that v(t) = (y(t))^(1-p), so we can rewrite the solution in terms of y(t): \[ (y(t))^{1-p} = \frac{b(1-p)}{a(1-p)} + Ce^{-a(1-p)t} \] To solve for y(t), raise both sides to the power of 1/(1-p): \[ y(t) = \left[\frac{b(1-p)}{a(1-p)} + Ce^{-a(1-p)t}\right]^{1/(1-p)} \] This is the solution to our original differential equation in terms of y(t).

Step-by-step solution

a. Find the relationship between \(y^{\prime}(t)\) and \(v^{\prime}(t)\)

We are given the change of variables \(v(t) = (y(t))^{1-p}\). First, we will compute the derivative of \(v(t)\) with respect to \(t\). Using the chain rule: \[ v^{\prime}(t) = \frac{d}{dt}(y(t))^{1-p} = (1-p)(y(t))^{-p}\cdot y^{\prime}(t) \] Now, we need to find the expression for \(y^{\prime}(t)\) in terms of v. To do this, let's solve for y in the change of variables expression first, then substitute into the expression for \(v^{\prime}(t)\) above: \[ y(t) = (v(t))^{1/(1-p)} \] Substituting y(t) into the expression for \(v^{\prime}(t)\): \[ v^{\prime}(t) = (1-p)((v(t))^{1/(1-p)})^{-p}y^{\prime}(t) \] Now, we can rearrange to find \(y^{\prime}(t)\): \[ y^{\prime}(t) = \frac{(y(t))^p}{1-p}v^{\prime}(t) \]

b. Substitute the expression for \(y^{\prime}(t)\) into the original equation and simplify to obtain the linear equation in terms of v and its derivative

Now, let's substitute the expression we found for \(y^{\prime}(t)\) into the original differential equation \(y^{\prime}(t) + ay = by^p\): \[ \frac{(y(t))^p}{1-p}v^{\prime}(t) + a(y(t)) = b(y(t))^p \] Since we want the equation in terms of v, substitute \(y(t) = (v(t))^{1/(1-p)}\) and rearrange the equation to isolate \(v^{\prime}(t)\): \[ \frac{(v(t))^{1/(1-p)p}}{1-p}v^{\prime}(t) + a(v(t))^{1/(1-p)} = b(v(t))^{1/(1-p)p} \] This simplifies to: \[ v^{\prime}(t) + a(1-p)v(t) = b(1-p) \] Now, we have a linear differential equation in terms of v and its derivative. Solve this equation using the standard methods from the section such as integrating factors, and once v(t) is determined, we can use the connection between y(t) and v(t) to find the solution y(t).

Key concepts

Nonlinear Differential Equations

Differential equations can be classified based on their linearity. In a linear differential equation, the dependent variable and all its derivatives appear to the first power, and no products of these appear together. Meanwhile, a nonlinear differential equation like the Bernoulli equation includes terms where the dependent variable or its derivatives are raised to a power other than one. This nonlinearity makes the equation more complex and harder to solve directly. For example, in the Bernoulli equation, the term \(y^p\) indicates nonlinearity if \(p eq 1\). Typically, solving these types of equations requires special techniques, since straightforward methods applicable to linear equations aren't directly usable. Understanding the nature of the equation helps determine the best approach to take, whether that's a change of variables or another method to proceed with finding a solution.

Change of Variables

The change of variables is a key technique used to simplify differential equations. In the context of a Bernoulli equation, the change of variables is used to transform a nonlinear equation into a linear form, which is easier to solve. We start by setting \(v = y^{1-p}\). This transformation allows us to manipulate the original equation such that the ensuing equation in terms of \(v\) and its derivative becomes linear. This is achieved by:

• Computing the derivative \(v'\) in terms of \(y\) and its derivative \(y'\).
• Rewriting the nonlinear term \(y^p\) in a form that incorporates the new variable \(v\).

Once done correctly, this change of variable essentially "straightens out" the curve of complexity, making the problem a matter of applying solutions of linear differential equations.

Linearization Techniques

Linearization techniques help deal with nonlinear differential equations by simplifying them into a form where standard linear methods can be applied. These techniques play a crucial role in solving equations like the Bernoulli equation. Linearization of a nonlinear equation typically involves expressing the nonlinear part in terms of a new variable, as done by letting \(v = y^{1-p}\) in our example. Once the substitution is made, it leads to a linear equation:\[v'(t) + a(1-p)v = b(1-p)\]This is a first-order linear differential equation that can be tackled using known methods, like the integrating factor method, to solve for \(v(t)\). With learning to express nonlinear situations in a linear form, mathematical complexities become far more manageable.

Solution Methods for Differential Equations

Solving differential equations involves various well-established techniques, especially when those equations are transformed into a linear form. After initializing the substitution \(v = y^{1-p}\), the Bernoulli equation becomes linear, enabling us to use known solution techniques.The most common method used to solve first-order linear differential equations is the integrating factor method. It involves:

• Identifying a function, called an integrating factor, that simplifies the equation's integration.
• Multiplying through the linear equation by this factor.
• Solving the modified equation, which typically presents as an easier integral to compute.

Once the equation is solved for \(v(t)\), one can revert to the original variable from \(v\) using \(y = v^{1/(1-p)}\), hence retrieving the solution to the original problem. Mastery of these practices allows students to handle a wide array of differential equations efficiently.