Problem 12 - Calculus

Question

In Exercises \(2-28,\) use separation of variables to find the solutions to the differential equations subject to the given initial conditions. $$\frac{d z}{d y}=z y, \quad z=1 \text { when } y=0$$

Step-by-Step Solution

Answer

The solution is \( z(y) = e^{\frac{y^2}{2}} \).

Step 1: Write the Differential Equation

We are given the differential equation \( \frac{dz}{dy} = zy \) with the initial condition \( z = 1 \) when \( y = 0 \). The goal is to find the solution \( z(y) \).

Step 2: Separate the Variables

To separate the variables, rewrite the equation such that all \( z \) terms are on one side and all \( y \) terms are on the other side. Divide both sides by \( z \) and multiply by \( dy \): \( \frac{1}{z} dz = y \, dy \).

Step 3: Integrate Both Sides

Integrate both sides of the equation with respect to their own variable: \( \int \frac{1}{z} \, dz = \int y \, dy \). This results in \( \ln|z| = \frac{y^2}{2} + C \), where \( C \) is the constant of integration.

Step 4: Solve for z

To solve for \( z \), exponentiate both sides of the equation to remove the natural logarithm: \( z = e^{\frac{y^2}{2} + C} \). This simplifies to \( z = Ce^{\frac{y^2}{2}} \) since \( e^C \) is a constant.

Step 5: Apply the Initial Condition

Use the initial condition \( z = 1 \) when \( y = 0 \) to find \( C \). Substitute into the equation: \( 1 = Ce^{0} \). Therefore, \( C = 1 \).

Step 6: Write the General Solution

From the previous step, we have \( z = 1 \times e^{\frac{y^2}{2}} \). So, the solution to the differential equation is \( z(y) = e^{\frac{y^2}{2}} \).

Key Concepts

Differential Equations

A differential equation is an equation that involves an unknown function and its derivatives. These equations help us understand how different quantities change with respect to one another. In the exercise provided, the differential equation is \( \frac{dz}{dy} = zy \). Here, \( z \) is the unknown function of \( y \), and the equation relates how \( z \) changes as \( y \) changes.

Differential equations are prevalent in modeling real-world phenomena, such as population growth, motion, and heat conduction. Learning how to solve them is crucial for understanding many processes in engineering and the sciences. To solve them, one common method we use is separation of variables, which is effective for certain kinds of differential equations. By transforming the equation to isolate terms involving each variable, we can separate the equation into two integrals, each involving only one of the variables. This allows us to simplify and solve the equation step-by-step, as illustrated in the exercise.

Integration

Integration is a fundamental concept in calculus used to find areas under curves or the accumulation of quantities. In solving differential equations using the separation of variables method, integration is a critical step. Once we separate the differential equation, we integrate each side with respect to its corresponding variable.

In the solution provided, after separating variables, we have the equation \( \int \frac{1}{z} \, dz = \int y \, dy \). The integration of \( \frac{1}{z} \) with respect to \( z \) gives us \( \ln|z| \), while the integration of \( y \) with respect to \( y \) results in \( \frac{y^2}{2} \). Obtaining these integrals allows us to express the solution to the differential equation in a more explicit form that we can manipulate further.

Becoming proficient in integration helps solve a wide variety of problems, not only in mathematics but also in physics, engineering, and other sciences, where calculating quantities accumulated over time or space is essential.

Initial Conditions

Initial conditions are values given for the unknown function and its derivatives at specific points, and they are crucial in solving differential equations. They allow us to find the particular solution that fits the specific situation described by the equation.

In the provided exercise, the initial condition is \( z = 1 \) when \( y = 0 \). This initial condition is used to determine the constant of integration in the general solution. After solving the separated and integrated equation, we have \( z = Ce^{\frac{y^2}{2}} \). By substituting the initial condition into this equation, \( 1 = Ce^{0} \), we find that \( C = 1 \), giving us the specific solution \( z(y) = e^{\frac{y^2}{2}} \).

Understanding and applying initial conditions correctly ensures that solutions to differential equations are not just general, but specific to the problem context, providing precise and usable information for analysis.