Problem 6 - Calculus Early Transcendentals

Question

Solve the differential equation by separation of variables. Where reasonable, express the family of solutions as explicit functions of x. $$ y^{\prime}=-x y $$

Step-by-Step Solution

Answer

The solution is \( y = C_1 e^{-\frac{1}{2}x^2} \).

Step 1: Write Down the Given Differential Equation

The differential equation given is \( y' = -xy \). We interpret this equation as indicating that the derivative of \( y \) with respect to \( x \) is equal to \( -xy \).

Step 2: Separate the Variables

Separate the variables by rewriting the equation in the form \( \frac{dy}{y} = -x \, dx \). This allows us to integrate each side with respect to its independent variable.

Step 3: Integrate Both Sides

Integrate the left side with respect to \( y \) and the right side with respect to \( x \):\[ \int \frac{1}{y} \, dy = \int -x \, dx \]This gives us:\[ \ln |y| = -\frac{1}{2}x^2 + C \] where \( C \) is the constant of integration.

Step 4: Solve for y

To express \( y \) as an explicit function of \( x \), exponentiate both sides of the equation:\[ |y| = e^{-\frac{1}{2}x^2 + C} \]Therefore, \( y = \pm e^C e^{-\frac{1}{2}x^2} \). Let \( C_1 = \pm e^C \).Thus, the solution can be rewritten as:\[ y = C_1 e^{-\frac{1}{2}x^2} \]

Key Concepts

Separation of Variables

Separation of variables is a technique used to solve differential equations. It involves rearranging the equation so that each variable and its differential are on separate sides. This allows for simpler integration. In our exercise, we start with the differential equation \( y' = -xy \). To use separation of variables, we'll rewrite it as \( \frac{dy}{y} = -x \, dx \).
This step is crucial because it isolates each variable alongside its corresponding differential component. By doing this:

• We enable independent integration of the variables, which simplifies solving the equation.
• We prep the problem for the integration process, paving the way for finding explicit solutions to the differential equation.

Understanding separation of variables is essential because it simplifies complex differential equations and often leads directly to an explicit solution.

Integration

Integration involves finding the antiderivative of a function. Once we have separated the variables, we can integrate both sides of our equation. The equation \( \frac{dy}{y} = -x \, dx \) allows us to apply integration:
\[ \int \frac{1}{y} \, dy = \int -x \, dx \]
On the left, we integrate \( \frac{1}{y} \), which results in \( \ln |y| \). On the right, integrating \( -x \) yields \( -\frac{1}{2}x^2 + C \), where \( C \) is the constant of integration.

• Indefinite integration requires the inclusion of a constant of integration, "C," since the antiderivative is non-unique.
• Integration reverses the process of differentiation, allowing us to uncover original functions from their rates of change.

This process transforms the separated variables into a solvable equation, which is key in moving toward expressing \( y \) explicitly.

Exponential Functions

Exponential functions are mathematical expressions where variables appear as exponents. After integrating and obtaining the equation \( \ln |y| = -\frac{1}{2}x^2 + C \), we'll use exponentiation to solve for \( y \):
Exponentiating both sides, we have:\[ |y| = e^{-\frac{1}{2}x^2 + C} \]
Exponentiating eliminates the natural logarithm, simplifying our expression to an exponential form. Since \( e^{a+b} = e^a \cdot e^b \), this can be further simplified:\[ y = \pm e^Ce^{-\frac{1}{2}x^2} \], where \( C_1 = \pm e^C \).
Exponential functions are powerful:

• They grow or decay at rates proportional to their current value, which is essential in modeling real-world phenomena.
• They help simplify the transition from logarithmic to explicit form.

Understanding exponential functions illuminates how solutions behave, particularly in growth or decay scenarios in differential equations.