Problem 60 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

In Exercises 59 through 62, solve the given initial value problem. \(\frac{d y}{d x}=x(x-1)\), where \(y=1\) when \(x=1\)

Step-by-Step Solution

Answer

The solution is \(y = \frac{x^3}{3} - \frac{x^2}{2} + \frac{7}{6}\).

Step 1 - Write Down the Given Differential Equation

The differential equation given is \(\frac{d y}{d x}=x(x-1)\). We need to solve this initial value problem.

Step 2 - Integrate Both Sides

To solve the differential equation, integrate both sides with respect to \(x\): \(\frac{d y}{d x} = x(x-1) \Rightarrow \int \frac{d y}{d x} dx = \int x(x-1) dx\).

Step 3 - Simplify the Integral

Simplify the integral on the right-hand side: \(\frac{dy}{dx} = x(x-1) = x^2 - x\) so we need to find \(\int (x^2 - x) dx\).

Step 4 - Calculate the Primitive Function

Integrate term by term: \(\begin{align*} y & = \int (x^2 - x) dx \ \ & = \int x^2 dx - \int x dx \ \ & = \frac{x^3}{3} - \frac{x^2}{2} + C \end{align*}\)

Step 5 - Apply Initial Conditions

Plug in the initial condition \(y=1\) when \(x=1\) to solve for \(C\): \(\begin{align*} 1 & = \frac{1^3}{3} - \frac{1^2}{2} + C \ \ & = \frac{1}{3} - \frac{1}{2} + C \ \ & = -\frac{1}{6} + C \Rightarrow C = \frac{7}{6} \end{align*}\)

Step 6 - Write the Final Solution

Combine the constants back into the general solution: \(\begin{align*} y & = \frac{x^3}{3} - \frac{x^2}{2} + \frac{7}{6} \end{align*}\)

Key Concepts

Understanding Differential Equations

A differential equation is an equation that involves an unknown function and its derivatives. In simpler terms, it shows a relationship between a function and how it changes. For example, if we have \(\frac{dy}{dx}=f(x)\), it means that the rate of change of the function \(y\) with respect to variable \(x\) is given by \(f(x)\). Differential equations are used in various fields like physics, engineering, and economics to model complex systems. They can be solved using techniques like separation of variables, integrating factors, or by applying initial conditions, depending on the problem.

Basics of Integration

Integration is a fundamental concept in calculus that can be thought of as the reverse process of differentiation. While differentiation is used to find the rate of change of a function, integration helps us find the original function from its rate of change. For example, if given the derivative \(f'(x)\), integration helps us find \(f(x)\). There are standard rules to integrate functions, like \int x^n dx=\frac{x^{n+1}}{n+1}\ for any real number \(n\) not equal to \(−1\). In the exercise, we integrate \( x^2 - x \) term by term, yielding \ (\frac{x^3}{3} - \frac{x^2}{2} + C) \ as the integral.

Applying Initial Conditions

Initial conditions are values given for the function and its derivatives at a specific point. They are crucial in solving differential equations because they help us find the constants of integration, making our solution unique. For example, in the exercise, we are given \( y = 1 \) when \( x = 1 \). We use this information to solve for the constant \( C \) in our integrated function. By substituting these values into the general solution, we can find the exact value of \( C \), ensuring our solution fits the given condition perfectly.