Problem 35 - Advanced Engineering Mathematics

Question

(a) The substitutions \(y=y_{1}+u\) and $$\frac{d y}{d x}=\frac{d y_{1}}{d x}+\frac{d u}{d x}$$ lead to $$\begin{aligned} \frac{d y_{1}}{d x}+\frac{d u}{d x} &=P+Q\left(y_{1}+u\right)+R\left(y_{1}+u\right)^{2} \\ &=P+Q y_{1}+R y_{1}^{2}+Q u+2 y_{1} R u+R u^{2} \end{aligned}$$ or $$\frac{d u}{d x}-\left(Q+2 y_{1} R\right) u=R u^{2}.$$ This is a Bernoulli equation with \(n=2\) which can be reduced to the linear equation $$\frac{d w}{d x}+\left(Q+2 y_{1} R\right) w=-R$$ by the substitution \(w=u^{-1}\) (b) Identify \(P(x)=-4 / x^{2}, Q(x)=-1 / x,\) and \(R(x)=1 .\) Then \(\frac{d w}{d x}+\left(-\frac{1}{x}+\frac{4}{x}\right) w=-1 .\) An integrating factor is \(x^{3}\) so that \(x^{3} w=-\frac{1}{4} x^{4}+c\) or \(u=\left[-\frac{1}{4} x+c x^{-3}\right]^{-1} .\) Thus, \(y=\frac{2}{x}+u\).

Step-by-Step Solution

Answer

The solution yields \( y = \frac{2}{x} + u \), where \( u = \left[-\frac{1}{4}x + cx^{-3}\right]^{-1} \).

Step 1: Apply Substitution

We begin by substituting \( y = y_{1} + u \) into the original equation. This gives us two substitutions: \( y = y_{1} + u \) and \( \frac{d y}{d x} = \frac{d y_{1}}{d x} + \frac{d u}{d x} \). These substitutions will help simplify the differential equation.

Step 2: Organize Terms

Substitute the expressions into the given equation: \( \frac{d y_{1}}{d x} + \frac{d u}{d x} = P + Q(y_{1}+u) + R(y_{1}+u)^2 \). Expanding and organizing these terms gives \( \frac{d y_{1}}{d x} + \frac{d u}{d x} = P + Qy_{1} + Ry_{1}^2 + Qu + 2y_{1}Ru + Ru^2 \).

Step 3: Isolate du/dx

Rearrange the equation to focus on \( \frac{d u}{d x} \), yielding: \( \frac{d u}{d x} - (Q + 2y_{1}R)u = Ru^2 \). This is in the Bernoulli form with \( n = 2 \).

Step 4: Substitute for Bernoulli Equation

Convert the equation to a linear form by substituting \( w = u^{-1} \). This gives \( \frac{d w}{d x} + (Q + 2y_{1}R)w = -R \).

Step 5: Identify Functions

Identify the functions from the problem statement: \( P(x) = -\frac{4}{x^2} \), \( Q(x) = -\frac{1}{x} \), and \( R(x) = 1 \). Substitute these functions into the linear equation, resulting in \( \frac{d w}{d x} - \left(-\frac{1}{x} + \frac{4}{x}\right) w = -1 \).

Step 6: Integrating Factor

Find the integrating factor, which is \( x^{3} \), to solve the linear differential equation. Multiply throughout by the integrating factor to get: \( x^3 \frac{d w}{d x} + x^2 w = -x^3 \).

Step 7: Solve the Linear Equation

Integrating both sides after finding the integrating factor leads to \( x^3 w = -\frac{1}{4}x^4 + c \), where \( c \) is the constant of integration.

Step 8: Back Substitution

Solve for \( u \) by back-substituting \( w = u^{-1} \), yielding \( u = \left[-\frac{1}{4}x + cx^{-3}\right]^{-1} \). Finally, substitute back to get \( y = \frac{2}{x} + u \).

Key Concepts

Differential Equations

Differential equations are equations that involve an unknown function and its derivatives. They are used to describe various phenomena in engineering, physics, finance, and other fields. An equation is considered a differential equation if it includes derivatives of one or more dependent variables with respect to one or more independent variables.
For instance, in our exercise, we deal with the differential equation involving the derivatives of a function of one variable, typically expressed as \( \frac{dy}{dx} \). This represents how the function \( y \) changes as \( x \) changes, which is central to understanding how different quantities relate over time or space.
In a Bernoulli differential equation, a subset of nonlinear differential equations, the structure generally involves the term \( u^n \), where \( n \) is a constant. The equation can take the shape \( \frac{du}{dx} + a(x)u = b(x)u^n \). Such equations require specific methods to solve, such as substitution or transforming the equation to a different form, leading us into the substitution methods discussed in the next section.

Substitution Methods

Substitution methods are powerful techniques used to simplify differential equations, making them easier to solve. By using substitutions, we can transform a complex problem into a more manageable form.
In this exercise, we employ a substitution by letting \( y = y_1 + u \) and \( \frac{dy}{dx} = \frac{dy_1}{dx} + \frac{du}{dx} \). This helps to separate the problem into components that highlight the nonlinear parts of the solution embeddable within the differential equation.
This substitution allows us to identify the underlying form of the Bernoulli equation \( \frac{du}{dx} - (Q + 2y_1R)u = Ru^2 \). By further substituting \( w = u^{-1} \), the equation transforms into a linear form, which is much simpler to solve. This step is critical for handling nonlinear equations, converting them to linear differential equations, for which there are established solving techniques.

Integrating Factor

The integrating factor is a crucial tool in solving linear differential equations. It transforms a non-separable linear equation into one that can be easily integrated. This is done by multiplying every term in the equation by this factor.
In our exercise, once the Bernoulli equation was transformed using a substitution, it took the form \( \frac{dw}{dx} + (Q + 2y_1R)w = -R \). The function that multiplies the derivative, here in our example is the sum \( (Q - 2y_1R) \), is used to find the integrating factor.
Specifically, for our transformed equation, the integrating factor was identified as \( x^3 \). We multiply the entire equation by \( x^3 \) to enable integration: \( x^3 \frac{dw}{dx} + x^2w = -x^3 \). After this, integrating both sides gives the solution in terms of \( w \), leading us finally to solve for \( u \), and then substituting back to find \( y \).
This process illustrates the power of integrating factors for solving differential equations, as they simplify the integration process and help in finding the general solution.