Problem 52 - Calculus for Biology and Medicine

Question

Solve each differential equation with the given initial condition. \(\frac{d x}{d y}=\frac{1}{2} \frac{x}{y}\), with \(x_{0}=2\) if \(y_{0}=3\)

Step-by-Step Solution

Answer

The solution is \( x = \frac{2}{\sqrt{3}} \sqrt{y} \).

Step 1: Identify the form and separate variables

The given differential equation is \( \frac{d x}{d y} = \frac{1}{2} \frac{x}{y} \). This equation can be arranged to separate variables. Start by multiplying both sides by \( dy \) and dividing by \( x \) to get \( \frac{d x}{x} = \frac{1}{2} \frac{dy}{y} \).

Step 2: Integrate both sides

Integrate both sides of the equation. The left side becomes \( \int \frac{d x}{x} = \ln |x| \) and the right side becomes \( \int \frac{1}{2y} \, dy = \frac{1}{2} \ln |y| \). Thus, we have \( \ln |x| = \frac{1}{2} \ln |y| + C \).

Step 3: Solve for x

To solve for \( x \), exponentiate both sides: \( e^{\ln |x|} = e^{\frac{1}{2} \ln |y| + C} \). Simplifying gives \( |x| = Ce^{\frac{1}{2} \ln |y|} = C \cdot y^{1/2} \). Thus, \( x = C y^{1/2} \).

Step 4: Apply initial conditions

Use the initial condition \( x_0 = 2 \) when \( y_0 = 3 \) to find \( C \). Substitute these values into \( x = C y^{1/2} \) to get \( 2 = C (3^{1/2}) \). Therefore, \( C = \frac{2}{\sqrt{3}} \).

Step 5: Write the explicit solution

Substitute \( C \) back into the general solution: \( x = \frac{2}{\sqrt{3}} y^{1/2} \). Therefore, the solution to the differential equation with the given initial condition is \( x = \frac{2}{\sqrt{3}} \sqrt{y} \).

Key Concepts

Separation of Variables

Differential equations can often seem challenging, but the separation of variables technique simplifies them. This method is useful for equations like \( \frac{d x}{d y} = \frac{1}{2} \frac{x}{y} \) where variables can be neatly divided and solved separately. The main idea is to manipulate the equation so that all terms involving one variable (in this case, \( x \)) and its differential are on one side of the equation, and all terms involving the other variable (\( y \)) are on the other side.To begin, you rewrite the equation so it becomes easier to integrate. You multiply both sides by \( dy \) and divide by \( x \) to get:

• \( \frac{d x}{x} = \frac{1}{2} \frac{dy}{y} \)

This step is crucial because it allows integration with respect to each variable. After separation, you have a battle of integrations waiting for you on both sides. It's like an adventure through calculus land!Once you isolate the variables, the path forward involves integrating each side to find a solution connecting \( x \) and \( y \). If ever in doubt, remember that whatever you divide or multiply by, keep balance in your equation.

Integrating Factors

In some cases, differential equations do not directly organize into separable variables. For those occasions, the integrating factor technique comes to the rescue. However, for our specific equation, separation was achievable, so an integrating factor wasn't necessary.If the situation arises where variables refuse to separate, applying an integrating factor can help rewrite the equation in a solvable form. The integrating factor is often a function that simplifies the differential equation, making it easier to integrate. You might find this in linear differential equations where both the dependent variable and its derivative appear unsynchronizable.For example, in a different equation of form \( \frac{d y}{d x} + P(x)y = Q(x) \), you would multiply through by an integrating factor \( \mu(x) \) which is usually \( e^{\int P(x) \, dx} \). This turns the left-hand side into the derivative of a product, making it integrable easily. Don't let the complexity confuse you, embrace the power of integrating factors for those stubborn equations.

Initial Conditions

Initial conditions complete the journey of solving differential equations. They allow us to find specific solutions from a general family of possible results. Once you've separated variables and integrated to find \( \ln |x| = \frac{1}{2} \ln |y| + C \), you have a general solution for that differential equation.The magic happens when you apply the initial conditions. In our case, the initial conditions were \( x_0 = 2 \) for \( y_0 = 3 \). Substituting these values helps determine the constant \( C \). Let's see how it works:

• Start with \( x = C y^{1/2} \).
• Substitute the known \( y_0 = 3 \) into the equation yielding \( 2 = C (3^{1/2}) \).
• This simplifies to find \( C = \frac{2}{\sqrt{3}} \).

Now, by replacing \( C \) back into your solution, you tailor the equation perfectly to your initial conditions:

• \( x = \frac{2}{\sqrt{3}} \sqrt{y} \)

Initial conditions give you the power to distinguish one unique solution among the limitless possibilities afforded by the differential equation framework. They are essential in narrowing down the mathematical narrative to reflect real-world scenarios.