Problem 34 - Thomas Calculus

Question

This last equation is linear in the (unknown) dependent variable u. Solve the differential equations. \(y^{\prime}-y=x y^{2}\)

Step-by-Step Solution

Answer

The solution is \(y = \frac{1}{1-x+Ce^{-x}}\).

Step 1: Identify the Type of Differential Equation

The given differential equation is \(y' - y = xy^2\). This is a first-order non-linear differential equation due to the \(y^2\) term on the right-hand side.

Step 2: Rearrange the Equation

Rewrite the equation in the form \(y' = y + xy^2\), so that we have it expressed as \(y' = f(x, y)\), facilitating further analysis.

Step 3: Check for Recognizable Form

Recognize that this equation is a Bernoulli equation, which has the general form \(y' + p(x)y = q(x)y^n\). Here, \(n = 2\), \(p(x) = -1\), and \(q(x) = x\).

Step 4: Use the Bernoulli Equation Method

For a Bernoulli equation of the form \(y' + p(x)y = q(x)y^n\), perform the substitution \(v = y^{1-n} = y^{-1}\). This leads to \(v' = -y^{-2}y'\).

Step 5: Transform the Equation

Substitute \(v = y^{-1}\), leading to \(y = v^{-1}\) and \(y' = -v^{-2}v'\). Substitute these into the equation and simplify to obtain \(-v^{-2}v' - v^{-1} = xv^{-2}\).

Step 6: Simplify and Rearrange

Simplifying gives \(v' + v = -x\). This is a linear first-order differential equation in \(v\).

Step 7: Solve the Linear Differential Equation

The equation \(v' + v = -x\) is a linear first-order differential equation. Solve it using the integrating factor method, where the integrating factor \(\mu(x) = e^{\int 1 \, dx} = e^x\).

Step 8: Apply the Integrating Factor

Multiply through by the integrating factor: \(e^x v' + e^x v = -xe^x\), which simplifies to \((e^x v)' = -xe^x\).

Step 9: Integrate Both Sides

Integrate both sides with respect to \(x\): \(e^x v = \int -xe^x \, dx + C\). Use integration by parts for \(\int -xe^x \, dx\).

Step 10: Calculate the Integral

Applying integration by parts, let \(u = -x\) and \(dv = e^x dx\), then \(du = -dx\) and \(v = e^x\). Thus, \(\int -xe^x \, dx = -xe^x - \int -e^x dx = -xe^x + e^x = e^x(1-x)\).

Step 11: Solve for v

Substitute the integral solution into the integrated equation: \(e^x v = e^x (1-x) + C\). Divide through by \(e^x\) to solve for \(v\): \(v = 1 - x + Ce^{-x}\).

Step 12: Back-Substitute to Find y

Recall the substitution \(v = y^{-1}\), hence \(y = v^{-1}\). Substitute for \(v\) to obtain \(y = \frac{1}{1-x+Ce^{-x}}\).

Key Concepts

First-Order Differential Equations

In mathematics, a **first-order differential equation** is an equation involving a function and its derivative. The function is dependent on a single variable. These types of equations are called "first-order" because the highest derivative of the function is the first derivative.

• The general form of a first-order differential equation is given by: \[ y' = f(x,y) \]
• It seeks to determine the function \(y\) that satisfies this relationship.

First-order differential equations appear in many real-world applications such as modeling population growth, analyzing circuits, and modeling motion dynamics.
To solve first-order differential equations, various methods can be employed depending on the form of the equation. For instance, in our problem, although the equation is initially non-linear, it can be transformed into a linear form. Once transformed, it can be tackled using an appropriate solving technique, like the integrating factor method.

Integrating Factor Method

The **Integrating Factor Method** is a powerful technique used for solving linear first-order differential equations of the standard form:

• \( y' + P(x)y = Q(x) \)

An integrating factor, denoted \( \mu(x) \), is a function that simplifies the process of solving these types of equations.

• It is generally defined as:\[ \mu(x) = e^{\int P(x) \, dx} \]
• By multiplying both sides of the differential equation by the integrating factor, the left-hand side becomes the derivative of the product \( \mu(x)y \).
• Thus, the equation becomes easier to integrate directly.

This method is particularly useful when the equation cannot be solved by simpler methods such as separation of variables. In the original exercise, the integrating factor transforms the equation into a directly integrable form, leading to a straightforward solution.

Non-Linear Differential Equations

A **non-linear differential equation** involves a function and its derivatives that do not possess linear properties. In this context, a non-linear relationship can involve terms like \( y^2 \), \( xy \), or any products of the function and its derivatives.

• Non-linear differential equations are notoriously more complex to solve analytically compared to their linear counterparts.
• Such equations often require a creative approach, advanced methods, or numerical solutions.

In the provided exercise, the initial non-linear equation \( y' - y = xy^2 \) posed a challenge because of the term \( xy^2 \). Recognizing it as a form of the Bernoulli equation, a method of transforming such non-linear equations into a solvable linear form was employed. This transformation relies on substituting variables to simplify the equation, allowing it to be addressed using techniques suitable for linear equations.