Problem 7 - Thomas Calculus

Question

Solve the differential equations. \(2 y^{\prime}=e^{x / 2}+y\)

Step-by-Step Solution

Answer

The solution is \( y = \frac{1}{2}e^{x/2} + Ce^{-x/2} \).

Step 1: Rearrange the Differential Equation

Start by isolating the derivative in the given differential equation. We rearrange the equation as follows:\[2y' = e^{x/2} + y \]Divide every term by 2 to isolate the derivative:\[y' = \frac{e^{x/2}}{2} + \frac{y}{2} \]This form is more convenient for further analysis and solution.

Step 2: Recognize the Differential Equation Type

The equation \( y' = \frac{e^{x/2}}{2} + \frac{y}{2} \) is a first-order linear differential equation of the form:\[ y' + P(x)y = Q(x) \]Here, \( P(x) = \frac{1}{2} \) and \( Q(x) = \frac{e^{x/2}}{2} \). This recognition allows us to use an integrating factor to solve the equation.

Step 3: Calculate the Integrating Factor

The integrating factor \( \mu(x) \) is given by:\[ \mu(x) = e^{\int P(x) \, dx} \]Substitute \( P(x) = \frac{1}{2} \):\[ \mu(x) = e^{\int \frac{1}{2} \, dx} = e^{\frac{x}{2}} \]This integrating factor will be used to convert the differential equation into an exact form.

Step 4: Multiply Equation by the Integrating Factor

Multiply the entire differential equation by the integrating factor \( e^{x/2} \):\[e^{x/2}y' + \frac{1}{2}e^{x/2}y = \frac{1}{2}e^{x/2}e^{x/2} \]Simplifying gives:\[e^{x/2}y' + \frac{1}{2}e^{x/2}y = \frac{1}{2}e^{x} \]

Step 5: Recognize the Left Side as a Derivative

Note that the left side of the equation from the multiplying process can be written as the derivative of \( e^{x/2}y \):\[\frac{d}{dx}(e^{x/2}y) = \frac{1}{2}e^{x}\]This is because the derivative of the product \( e^{x/2}y \) is what we have on the left of the equation after multiplying.

Step 6: Integrate Both Sides

Integrate both sides with respect to \( x \):\[\int \frac{d}{dx}(e^{x/2}y) \, dx = \int \frac{1}{2}e^{x} \, dx\]This simplifies to:\[e^{x/2}y = \frac{1}{2}\int e^{x} \, dx\]\[ e^{x/2}y = \frac{1}{2}e^{x} + C\]

Step 7: Solve for y

Divide both sides by \( e^{x/2} \) to solve for \( y \):\[y = \frac{1}{2}e^{x/2} + Ce^{-x/2} \]This is the general solution of the given differential equation, where \( C \) is a constant of integration.

Key Concepts

Integrating Factor

When solving first-order linear differential equations, the integrating factor is a powerful tool that simplifies the process. This factor transforms the differential equation into an exact differential equation, which is much easier to solve. It uses the form

• \( y' + P(x)y = Q(x) \)

where \( P(x) \) and \( Q(x) \) can be any functions of \( x \).
The integrating factor \( \mu(x) \) is calculated using the expression:

• \( \mu(x) = e^{\int P(x) \, dx} \)

For the equation we are considering, \( P(x) = \frac{1}{2} \). Therefore, the integrating factor is found to be

• \( \mu(x) = e^{\frac{x}{2}} \)

By multiplying the entire differential equation by this integrating factor, we help consolidate terms into a simpler expression that resembles a derivative of a product, greatly easing the path to finding the solution.

General Solution

Finding the general solution of a differential equation involves integrating the equation to find a family of solutions. This family includes arbitrary constants which can take any value. For the specific equation we solved, the general solution was derived by multiplying through by the integrating factor. We ended with an expression in the form:

• \( \frac{d}{dx}(e^{x/2}y) = \frac{1}{2}e^{x} \)

By integrating both sides with respect to \( x \), we acquired

• \( e^{x/2}y = \frac{1}{2}e^{x} + C \)

where \( C \) is an arbitrary constant. Finally, to solve for \( y \), we divided by \( e^{x/2} \) leading to the solution:

• \( y = \frac{1}{2}e^{x/2} + Ce^{-x/2} \)

This represents all possible solutions to the differential equation, with the constant \( C \) determining specific solutions based on initial or boundary conditions.

Product Rule

The product rule is a fundamental concept in calculus, particularly in differentiation, used when finding the derivative of two multiplied functions. It states:

• If \( u(x) \) and \( v(x) \) are functions of \( x \), then \( \frac{d}{dx}[u(x)v(x)] = u(x)v'(x) + v(x)u'(x) \).

In the context of solving differential equations using an integrating factor, the product rule comes into play when recognizing that the left side of the equation, after multiplying by the integrating factor, can be expressed as a derivative. For example, in our step-by-step solution, the expression

• \( e^{x/2}y' + \frac{1}{2}e^{x/2}y \)

was rewritten as

• \( \frac{d}{dx}(e^{x/2}y) \)

This recognition simplifies integration because rather than integrating each term individually, you integrate the whole as a product of a derivative, streamlining the process and solving for \( y \) effectively. This method showcases the power of the product rule in simplifying complex expressions.