Problem 77 - Calculus for Biology and Medicine

Question

Solve by rewriting the differential equation as an equation for \(\frac{d x}{d y}\) : $$ \frac{d y}{d x}=\frac{y}{y^{2}+1}, \text { for } x \geq 0 \text { with } y(0)=1 $$

Step-by-Step Solution

Answer

The differential equation \( \frac{dy}{dx} = \frac{y}{y^2+1} \) becomes \( \frac{dx}{dy} = y + \frac{1}{y} \) with the particular solution \( x = \frac{y^2}{2} + \ln|y| - \frac{1}{2} \).

Step 1: Inverse Variables in the Differential

To rewrite the differential equation as an equation for \( \frac{dx}{dy} \), we start by noting the original equation \( \frac{dy}{dx} = \frac{y}{y^2 + 1} \). Since we need \( \frac{dx}{dy} \), take the reciprocal of \( \frac{dy}{dx} \), resulting in \( \frac{dx}{dy} = \frac{y^2 + 1}{y} \).

Step 2: Simplify the Inverse Equation

Simplify \( \frac{dx}{dy} = \frac{y^2 + 1}{y} \) by dividing every term in the numerator by \( y \). This gives \( \frac{dx}{dy} = y + \frac{1}{y} \).

Step 3: Define the General Solution for the New Equation

Integrate both sides with respect to \( y \) to find \( x \) in terms of \( y \). From \( \frac{dx}{dy} = y + \frac{1}{y} \), integrate: \( x = \int (y + \frac{1}{y}) \, dy = \int y \, dy + \int \frac{1}{y} \, dy \).

Step 4: Integrate Each Term Separately

Calculate the integrals: \( \int y \, dy = \frac{y^2}{2} \) and \( \int \frac{1}{y} \, dy = \ln|y| \). Thus, we have \( x(y) = \frac{y^2}{2} + \ln|y| + C \), where \( C \) is the integration constant.

Step 5: Apply the Initial Condition

Use the initial condition \( y(0) = 1 \) to find \( C \). Substitute \( y = 1 \) and \( x = 0 \) into the equation \( x(y) = \frac{y^2}{2} + \ln|y| + C \), giving \( 0 = \frac{1^2}{2} + \ln(1) + C \). Solving, we find \( C = -\frac{1}{2} \).

Step 6: Write the Particular Solution

Substitute \( C = -\frac{1}{2} \) back into the general solution: \( x = \frac{y^2}{2} + \ln|y| - \frac{1}{2} \). This is the particular solution using the given initial conditions.

Key Concepts

Reciprocal Functions in Differential Equations

When dealing with differential equations, it's often necessary to switch from \( \frac{dy}{dx} \) to \( \frac{dx}{dy} \), especially when performing integration is more straightforward in terms of the alternative variable. A common method is using the reciprocal function.
Imagine the scenario where you have to write an expression for \( \frac{dx}{dy} \). You start with the equation \( \frac{dy}{dx} = \frac{y}{y^2 + 1} \) and recast it in terms of \( dx \) by using the reciprocal function, turning it into \( \frac{dx}{dy} = \frac{y^2 + 1}{y} \).

• Reciprocal functions play a crucial role because they allow re-expression of differential equations by flipping the numerator and the denominator.
• This manipulation is not just a mathematical trick; it's a strategic approach to simplify integration or other algebraic manipulations.

Using reciprocal functions effectively can help manage complex differential equations by changing the perspective of integration.

Understanding Integration in Differential Equations

Integration is a powerful tool in solving differential equations. In the exercise, after rewriting the equation using the reciprocal, integration helps transition from differentiating to finding a meaningful equation linking variables.
Integrating \( \frac{dx}{dy} = y + \frac{1}{y} \) involves adding up small changes in \( x \) as \( y \) changes. Integration decomposes into two parts:

• The integral of \( y \) yields \( \int y \, dy = \frac{y^2}{2} \), representing accumulated area under \( y \) over the interval.
• The integral of \( \frac{1}{y} \), \( \int \frac{1}{y} \, dy = \ln|y| \), provides a logarithmic growth expression.

Together, these integrals help construct the equation \( x(y) = \frac{y^2}{2} + \ln|y| + C \). Thus, integration links derivative-based expressions with more familiar functions, making them easier to analyze.

Applying Initial Conditions

Initial conditions are crucial for finding a particular solution to differential equations. In the problem, the initial condition given is \( y(0) = 1 \). This tells us exactly where one particular solution should begin on the curve described by the general solution.
When applying the initial condition to the integrated equation \( x(y) = \frac{y^2}{2} + \ln|y| + C \), you substitute \( y = 1 \) and \( x = 0 \) to solve for \( C \).

• This process fixes the constant, tailoring the general function to match real-world constraints or specific problem requirements.
• The calculated \( C = -\frac{1}{2} \) is then substituted back into the equation, resulting in the particular solution: \( x = \frac{y^2}{2} + \ln|y| - \frac{1}{2} \).

Using initial conditions sharpens the solution from broad possibilities to a precise fit, making it applicable and realistic.