Problem 44 - Calculus

Question

Solve the differential equations in Problems \(44-47 .\) Assume \(x, y, t > 0\). $$\frac{d y}{d t}=y(2-y), \quad y(0)=1$$

Step-by-Step Solution

Answer

The solution is \(y = \frac{2e^{2t}}{1 + e^{2t}}\).

Step 1: Recognize the Type of Differential Equation

The given equation \(\frac{dy}{dt}=y(2-y)\) is a first-order separable differential equation. We can separate variables and solve it by integration.

Step 2: Separate Variables

Rearrange the equation to isolate \(y\) terms on one side and \(t\) terms on the other side: \(\frac{dy}{y(2-y)} = dt\).

Step 3: Perform the Partial Fraction Decomposition

Write \(\frac{1}{y(2-y)}\) as the sum of two partial fractions: \(\frac{1}{y(2-y)} = \frac{A}{y} + \frac{B}{2-y}\). Solve for \(A\) and \(B\) to get \(A = \frac{1}{2}, B = \frac{1}{2}\).

Step 4: Integrate Both Sides

Integrate both sides with respect to their respective variables:\(\int \left( \frac{1}{2y} + \frac{1}{2(2-y)} \right) dy = \int dt\).This gives \(\frac{1}{2} \ln |y| - \frac{1}{2} \ln |2-y| = t + C\).

Step 5: Simplify and Solve for the Constant

Combine the logs and simplify: \(\frac{1}{2}\ln\left(\frac{y}{2-y}\right)=t+C\). Set initial condition \(y(0)=1\). Substitute into the equation to solve for \(C\):\(\frac{1}{2} \ln\left(\frac{1}{1}\right) = 0 + C\), so \(C=0\).

Step 6: Solve for \(y\) in Terms of \(t\)

Exponentiate to simplify: \[\ln\left(\frac{y}{2-y}\right) = 2t\]\[\frac{y}{2-y} = e^{2t}\].Rearrange to solve for \(y\): \[y = \frac{2e^{2t}}{1 + e^{2t}}\].

Step 7: Verify the Solution with Initial Condition

Verify that the solution satisfies the initial condition. Substitute \(t=0\) to confirm \(y(0)=1\): \[y(0) = \frac{2e^{0}}{1+e^{0}} = \frac{2}{2} = 1\]. The solution is correct.

Key Concepts

Separable Differential Equation

A separable differential equation is one where the variables can be separated on either side of the equation, allowing for straightforward integration. In this particular problem, we started with the equation \(\frac{dy}{dt}=y(2-y)\). This equation is defined as a first-order separable differential equation. It means both the dependent variable \(y\) and independent variable \(t\) can be separated and manipulated independently to solve the equation.
To do this, we rearrange the equation such that all the \(y\) terms are on one side and all the \(t\) terms on the other: \(\frac{dy}{y(2-y)} = dt\). By doing so, we're setting ourselves up to perform integration on both sides.

• Step-by-step: Identify terms that depend on \(y\) and \(t\).
• Reorganize via multiplication or division to achieve separation.
• Prepare to integrate each side with respect to its variable.

Understanding this process is crucial when tackling separable differential equations, as it forms the cornerstone of solving them through integration.

Initial Value Problem

An initial value problem is a differential equation accompanied by an initial condition. This initial condition is a specific solution to the problem, at a particular point, which allows you to find an exact, particular solution, rather than a general one.
In our exercise, we have the differential equation \(\frac{dy}{dt} = y(2-y)\) with an initial condition \(y(0)=1\). This initial condition tells us that when \(t=0\), the value of \(y\) should be \(1\). This added piece of information allows us to solve for the constant of integration \(C\) after integrating both sides of the equation.

• Why it's important: Initial conditions provide specific solutions instead of general solutions.
• They tie the solution to a real-world condition or scenario.
• Helps verify the solution by substituting back to check against the original condition.

By setting \(t=0\) in our integrated equation, we could easily find \(C=0\), confirming our solution satisfies the initial requirement \(y(0)=1\).

Partial Fraction Decomposition

Partial fraction decomposition is a technique used when integrating rational functions, which are fractions where both the numerator and the denominator are polynomials. This method simplifies the integration process by splitting a complex fraction into simpler, more manageable ones.
In our problem, we encounter the fraction \(\frac{1}{y(2-y)}\). To integrate this, we decompose it into partial fractions: \(\frac{1}{y(2-y)} = \frac{A}{y} + \frac{B}{2-y}\). Solving for \(A\) and \(B\) gives us \(A = \frac{1}{2}\) and \(B = \frac{1}{2}\). This decomposition allows the subsequent integration to be straightforward.

• Steps involved: Express the fraction as the sum of simpler fractions with unknown coefficients.
• Solve for these coefficients, often by substituting convenient values or by comparing coefficients.
• Integrate the simpler fractions individually.

This technique is especially helpful as it breaks down the problem, making it feasible to integrate more complex polynomials by turning them into a sum of more basic terms.