Problem 27 - Calculus Early Transcendentals: Pearson New International Edition

Question

For the differential equation \(\frac{d y}{d x}-\frac{y}{x}=x^{2}, x>0\), the integrating factor is \(e^{\int(-1 / x) d x}\). The general antiderivative \(\int\left(-\frac{1}{x}\right) d x\) is equal to \(-\ln x+C\). (a) Multiply both sides of the differential equation by \(\exp \left(\int\left(-\frac{1}{x}\right) d x\right)=\exp (-\ln x+C)\), and show that \(\exp (-\ln x+C)\) is an integrating factor for every value of \(C\). (b) Solve the resulting equation for \(y\), and show that the solution agrees with the solution obtained when we assumed that \(C=0\) in the integrating factor.

Step-by-Step Solution

Answer

Solution matches for any integrating factor when C=0; y = \(\frac{x^2}{2} + D\).

Step 1: Find the integrating factor

The given integrating factor is \(e^{\int(-1 / x) d x}\). Let's compute it.\[\int \left(-\frac{1}{x}\right) dx = -\ln |x| + C\]So the integrating factor becomes: \[e^{\int \left(-\frac{1}{x}\right) dx} = e^{-\ln x + C} = e^{C} \cdot \frac{1}{x}\ as \ e^{-\ln x} = \frac{1}{x}\]

Step 2: Verify the integrating factor

Multiplying through the differential equation \(\frac{d y}{d x}-\frac{y}{x}=x^{2}\) by the integrating factor:\[x^{C} \cdot \frac{1}{x} \cdot \left(\frac{d y}{d x} - \frac{y}{x}\right) = x^{C} \cdot \frac{1}{x} \cdot x^{2}\]Simplifies to:\[\frac{d}{dx}(y \cdot x^C) = x^{C+1}\]

Step 3: Integrate the simplified equation

Integrate both sides with respect to \(x\):\[\int \frac{d}{dx}(y \cdot x^C) \, dx = \int x^{C+1} \, dx\]The left becomes:\[ y \cdot x^C \] The right integrates to:\[ \frac{x^{C+2}}{C+2} + D \] where \(D\) is an integration constant.

Step 4: Solve for \(y\)

Solving for \(y\), we have:\[y = \frac{x^{C+2}}{(C+2)x^C} + \frac{D}{x^C}\]This simplifies to:\[y = \frac{x^{2}}{C+2} + Dx^{-C}\]

Step 5: Check with \(C=0\)

Set \(C = 0\) in the derivative equation:\[\frac{d}{dx}(y \cdot x^0) = x\to \int x \, dx = \frac{x^2}{2} + D\to y \cdot 1 = \frac{x^2}{2} + D\]Thus: \[y = \frac{x^2}{2} + D\] This matches the case of \(C = 0\) from our simplified general solution in Step 4.

Key Concepts

Integrating Factor

An integrating factor is a function that, when multiplied to a differential equation, transforms it into a more easily solvable form. For linear first-order differential equations of the form \(\frac{d y}{d x} + P(x)y = Q(x)\), an integrating factor \(\mu(x) = e^{\int P(x) \, dx}\) can be used.
To obtain the integrating factor for the given differential equation \(\frac{d y}{d x} - \frac{y}{x} = x^2\), we start by identifying \(P(x) = -\frac{1}{x}\). The integral \(\int(-\frac{1}{x}) \, dx\) evaluates to \(-\ln |x| + C\), leading to the integrating factor calculated as \(e^{- ext{ln} x + C} = \frac{e^C}{x}\).
The integrating factor simplifies the original differential equation, allowing us to multiply both sides and make solving for \(y\) more straightforward.

• Integrate the obtained expression to find \(\mu(x)\).
• Multiply the differential equation by \(\mu(x)\) to make it exact.

General Antiderivative

A general antiderivative of a function is essentially the function known as an indefinite integral, represented together with a constant of integration. For the expression \(\int \left(-\frac{1}{x}\right) \, dx\), the integral becomes \(-\ln |x| + C\).
This antiderivative includes a constant \(C\) that allows for flexibility, accounting for all possible antiderivatives of the function. This constant is crucial when determining the integrating factor, as seen in our example, where it contributes to simplifying the function to \(\frac{e^C}{x}\).
In the context of solving differential equations, finding the general antiderivative helps to form a family of solutions.

• Remember that the constant \(C\) represents all possible vertical shifts of the antiderivative.
• Use the general antiderivative when solving integrals that result as part of the solution process.

Solution Verification

Verifying the solution to a differential equation is essential to ensure its accuracy. By substituting the solution back into the original equation, we prove whether it satisfies the given problem.
In our example, after calculating the general solution for \(y\), which is \(y = \frac{x^2}{C + 2} + Dx^{-C}\), substituting it back into the differential equation ensures it holds true.
This step checks that each part fits, confirming consistency. When \(C = 0\), we easily see this aligns with \(y = \frac{x^2}{2} + D\), verifying our solution.

• Substitute \(y\) back into your original differential equation.
• Simplify to check if both sides of the equation match.

Variable Separation

Variable separation is a method used for solving differential equations where one can separate variables conveniently on opposite sides of the equation. However, it's primarily functional for equations that can be expressed in a manner allowing rearrangement of all \(x\) terms on one side and \(y\) terms on the other.
While the given exercise doesn't explicitly implement variable separation, understanding it enhances how we manipulate equations. We see this technique's value in simplifying components before applying integrating factors.
Essentially, variables are rearrangeable when the equation takes the form \(g(y)dy = h(x)dx\) and subsequently integrate both parts separately.

• Check if the equation allows for variable separation.
• Apply integration on both sides to find a solution.