Problem 50 - Calculus for Biology and Medicine

Question

In Problems, solve each differential equation with the given initial condition. $$ \frac{d y}{d x}=\frac{y^{2}}{x}, \text { with } y(1)=1 . $$

Step-by-Step Solution

Answer

The solution is \( y = -\frac{1}{\ln|x| - 1} \).

Step 1: Separate Variables

To solve the differential equation \( \frac{d y}{d x} = \frac{y^2}{x} \), we first separate the variables to isolate \( y \) terms on one side and \( x \) terms on the other. This gives us: \( \frac{d y}{y^2} = \frac{d x}{x} \).

Step 2: Integrate Both Sides

Now, we integrate both sides of the equation. The left side becomes \( \int \frac{1}{y^2} \, dy = -\frac{1}{y} + C_1 \) and the right side becomes \( \int \frac{1}{x} \, dx = \ln|x| + C_2 \). Thus, \( -\frac{1}{y} = \ln|x| + C \) where \( C = C_2 - C_1 \).

Step 3: Solve for y

Rearrange the equation to solve for \( y \): \( -\frac{1}{y} = \ln|x| + C \) implies \( y = -\frac{1}{\ln|x| + C} \).

Step 4: Apply Initial Condition

Use the given initial condition \( y(1) = 1 \) to find the particular solution. Substitute \( x = 1 \) and \( y = 1 \) in the equation: \( 1 = -\frac{1}{\ln(1) + C} \). Since \( \ln(1) = 0 \), we have \( 1 = -\frac{1}{C} \) which gives \( C = -1 \).

Step 5: Write Final Solution

Substitute \( C = -1 \) back into the solution for \( y \): \( y = -\frac{1}{\ln|x| - 1} \). This is the particular solution to the differential equation with the given initial condition.

Key Concepts

Understanding Initial Conditions

In differential equations, an initial condition is a given value that helps to determine a specific solution out of a family of possible solutions. You can think of a differential equation as a set of instructions that describes how something changes. But without an initial condition, we don't know the starting point. In our example, the differential equation \( \frac{dy}{dx} = \frac{y^2}{x} \) could have many solutions. However, by specifying that \( y(1) = 1 \), we pinpoint exactly which path the solution should take from the general family.

• The initial condition \( y(1) = 1 \) means when \( x \) is 1, \( y \) should also be 1.
• This allows us to solve for the constant \( C \) after integration.

Once this initial condition is applied, it tailors the equation to reflect the specific behavior starting at that point. This is crucial for finding out how systems behave at any other point.

Exploring Separation of Variables

Separation of variables is a handy technique for solving differential equations. It involves rearranging the equation to separate all the \( y \) terms and \( x \) terms onto opposite sides of the equation. In our example, we start with the equation \( \frac{dy}{dx} = \frac{y^2}{x} \).
The goal is to separate the variables to have all \( y \) terms on one side and all \( x \) terms on the other side. This results in:

• \( \frac{dy}{y^2} = \frac{dx}{x} \)

This technique is quite powerful as it turns a more complex equation into a simpler one that can be integrated easily. By doing this, it opens the door for us to use integration to solve for \( y \) in terms of \( x \). This method works excellently when variables can be neatly separated, as seen here.

Journey Through Integration

Integration is a fundamental operation used to solve differential equations once the variables are separated. In our differential equation \( \frac{dy}{dx} = \frac{y^2}{x} \), after separating variables, we are left with:

• \( \int \frac{1}{y^2} \, dy = \int \frac{1}{x} \, dx \)

The process of integrating these expressions helps us find relationships between \( x \) and \( y \). The integration of \( \frac{1}{y^2} \, dy \) yields \( -\frac{1}{y} \, + \, C_1 \) and \( \frac{1}{x} \, dx \) results in \( \ln|x| + C_2 \).
Integration helps unravel the complexity by turning the problem of change into a problem of accumulation - here, accumulating the rate of change, in a sense.

• After integration, the constants \( C_1 \) and \( C_2 \) might seem arbitrary.
• The difference \( C = C_2 - C_1 \) is solved using the initial condition.

Integrating both sides allows us to form an equation that relates \( x \) and \( y \), which we then solve with the information given by the initial condition.