Problem 33 - Calculus Single Variable

Question

Find the solution of the given initial value problem. $$ y^{\prime}(x)-x y^{2}=x y(4 x-y) \quad y(0)=1 $$

Step-by-Step Solution

Answer

The solution to the initial value problem is \(y = e^{\frac{4}{3}x^3}\).

Step 1: Identify the equation type

The given equation is a first-order differential equation of the form \(y'(x) - xy^2 = xy(4x - y)\). We notice that it is a type of separable differential equation as it can be rearranged to separate variables on each side.

Step 2: Rewrite the equation

Rearrange the given equation to isolate terms involving \(y\) and \(x\) on different sides.\[ y'(x) = xy^2 + xy(4x - y) = xy^2 + 4x^2y - xy^2 = 4x^2y\]Thus, the simplified differential equation is:\[y'(x) = 4x^2y\].

Step 3: Separate variables

Separate variables, bringing all \(y\)-terms to one side and \(x\)-terms to another:\[ \frac{dy}{y} = 4x^2 dx\]

Step 4: Integrate both sides

Integrate both sides of the equation:\[ \int \frac{dy}{y} = \int 4x^2 \, dx\]This gives:\[ \ln |y| = \frac{4}{3}x^3 + C\]

Step 5: Solve for y

Exponentiate both sides to solve for \(y\):\[ y = e^{\left(\frac{4}{3}x^3 + C\right)} = e^{\left(\frac{4}{3}x^3\right)} \cdot e^C\]Let \(e^C = C_1\), where \(C_1\) is a constant, so\(y = C_1 e^{\frac{4}{3}x^3}\).

Step 6: Apply the initial condition

Use the initial condition \(y(0)=1\) to solve for the constant \(C_1\):\[ 1 = C_1 e^{\frac{4}{3}(0)^3} = C_1\]Thus, \(C_1 = 1\). So the solution is:\[y = e^{\frac{4}{3}x^3}\].

Key Concepts

Differential Equation

Differential equations are equations involving derivatives of a function. They indicate how a function changes, providing insight into dynamic systems. In this problem, the equation is a first-order differential equation, as it contains the first derivative of the function, noted as \(y'(x)\). These types of equations help us to model real-world phenomena by showing how a particular quantity evolves over time or space according to an established relationship.

• A common form of a differential equation is \(y'(x) = f(x, y)\), where the rate of change of \(y\) depends on both \(x\) and \(y\) itself.
• Differential equations appear in a multitude of scientific disciplines, including physics, biology, and economics.

In this specific exercise, the goal is to find the function \(y(x)\) that satisfies the equation and any given conditions, providing a complete description of the behavior of the system represented by the equation.

Separable Differential Equations

Separable differential equations are a special class of differential equations in which variables can be separated onto different sides of the equation. This property makes them easier to solve since you can integrate each side with respect to its variable independently.In the original problem, the equation is rearranged to easily separate variables:

• The terms involving \(y\) are moved to one side of the equation: \(\frac{dy}{y}\), where the derivative \(\frac{dy}{y}\) just involves \(y\).
• The terms involving \(x\) are isolated on the other side: \(4x^2 \, dx\), where the differential \(dx\) only involves \(x\).

Separation of variables is a powerful technique because, once separated, the problem reduces to integrating simpler functions on either side of the equation. Solutions to separable differential equations often represent exponential growth or decay processes, common in natural and economic systems.

Integration

Integration is an essential operation in calculus, often used to find the original function from its derivative. Given the separated equation \(\frac{dy}{y} = 4x^2 \, dx\), both sides can now be integrated to solve for \(y\).On the left-hand side, when you integrate \(\frac{dy}{y}\), you obtain \(\ln|y|\), since the integral of \(\frac{1}{y}\) with respect to \(y\) is the natural logarithm.On the right-hand side, the integral of \(4x^2\) with respect to \(x\) yields \(\frac{4}{3}x^3\), as obtained from applying basic polynomial integration rules:

• The power rule for integration, \(\int x^n \, dx = \frac{x^{n+1}}{n+1}\), simplifies the process significantly.

After integrating, the equation \(\ln|y| = \frac{4}{3}x^3 + C\) is obtained, where \(C\) is the constant of integration. This constant reflects the various vertical shifts of potential solution curves for \(y\).

Initial Conditions

Initial conditions provide specific numerical values that help us determine the particular solution out of many possible solutions of a differential equation. In many real-world applications, initial conditions are known, making it easier to identify the exact form of the solution that fits the initial scenario described.In our problem, the initial value given is \(y(0) = 1\). Applying this initial condition allows you to find the constant \(C_1\) in the solution \(y = C_1 e^{\frac{4}{3}x^3}\):

• Substituting the initial condition into the solution, you get \(1 = C_1 e^{\frac{4}{3}(0)^3} = C_1\), hence \(C_1 = 1\).
• This specific condition ensures the solution aligns perfectly with the given situation at \(x = 0\), resulting in the particular solution \(y = e^{\frac{4}{3}x^3}\).

Applying initial conditions is crucial for tailoring generic solutions of differential equations to match specified requirements or observations, often represented in initial value problems.