Problem 20 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

\(\frac{d y}{d x}+3 x^{2} y=x^{2} ; y=2\) when \(x=0\)

Step-by-Step Solution

Answer

The solution is \( y = \frac{1}{3} + \frac{5}{3} e^{-x^3} \).

Step 1: Identify the Type of Differential Equation

This is a first-order linear differential equation of the form \( \frac{d y}{d x} + P(x)y = Q(x) \). In this case, \(P(x) = 3x^2 \) and \( Q(x) = x^2 \).

Step 2: Find the Integrating Factor

The integrating factor \( \mu(x) \) is given by \( e^{\int P(x) \, dx} \). Here, \( P(x) = 3x^2 \), so calculate \( \mu(x) = e^{\int 3x^2 \, dx} = e^{x^3} \).

Step 3: Multiply Both Sides by the Integrating Factor

Multiply the entire differential equation by the integrating factor: \( e^{x^3} \frac{d y}{d x} + 3x^2 e^{x^3} y = x^2 e^{x^3} \).

Step 4: Simplify to Recognize a Product Derivative

Observe that \( e^{x^3} \frac{d y}{d x} + 3x^2 e^{x^3} y = \frac{d}{dx} (y e^{x^3}) \), so the equation simplifies to \( \frac{d}{dx} (y e^{x^3}) = x^2 e^{x^3} \).

Step 5: Integrate Both Sides

Integrate both sides with respect to \( x \): \(\int \frac{d}{dx} (y e^{x^3}) \, dx = \int x^2 e^{x^3} \, dx \). This gives \( y e^{x^3} = \int x^2 e^{x^3} \, dx \). Let \( u = x^3 \), then \( du = 3x^2 \, dx \), so \( \int x^2 e^{x^3} \, dx = \frac{1}{3} \int e^u \, du = \frac{1}{3} e^{x^3} + C \).

Step 6: Solve for \( y \)

Solve for \( y \) by dividing both sides by \( e^{x^3} \): \( y e^{x^3} = \frac{1}{3} e^{x^3} + C \) becomes \( y = \frac{1}{3} + Ce^{-x^3} \).

Step 7: Apply the Initial Condition

Use the initial condition \( y = 2 \) when \( x = 0 \) to solve for \( C \): \( 2 = \frac{1}{3} + C \), so \( C = \frac{5}{3} \).

Step 8: Write the Final Solution

Substitute \( C = \frac{5}{3} \) into the general solution to get: \( y = \frac{1}{3} + \frac{5}{3} e^{-x^3} \).

Key Concepts

Integrating Factor

When solving first-order linear differential equations, the integrating factor is a useful tool. It converts the equation into a form where we can easily integrate both sides.
Consider a differential equation of the form: \ \ \frac{d y}{d x} + P(x)y = Q(x).
Here, the integrating factor, denoted as \( \ \mu(x)\), is given by: \ \ \ \mu(x) = e^{\ \int P(x) \, dx }.
The purpose of this factor is to multiply through the entire differential equation, which helps to simplify it into a product rule form. This makes it easier to integrate both sides of the equation.
For our specific problem, we identify \( P(x) = 3x^2 \), which gives us the integrating factor as: \ \ \ \mu(x) = e^{\int 3x^2 \, dx } = e^{ x^3 }.
After finding the integrating factor, we multiply it through the entire differential equation, simplifying our problem significantly.

Initial Conditions

Initial conditions are additional pieces of information that allow us to find a unique solution to a differential equation.
They usually come in the form of \( y( x_0 ) = y_0 \). This tells us that when \( x = x_0 \), the value of \( y \) is \( y_0 \).
In our example, the initial condition given is \( y(0) = 2 \).
This means when \( x = 0 \), \( y \) is equal to 2.
We use this initial condition to solve for the constant \( C \) in our general solution.
For our differential equation, after finding the general solution: \ y = \frac{1}{3} + Ce^{-x^3}.
We substitute \( x = 0 \) and \( y = 2 \) to find \( C \):
\begin{aligned} y( 0 ) &= \frac{1}{3} + C e^{0} \ \ 2 &= \frac{1}{3} + C \ \ C &= \frac{5}{3}. \end{aligned}
Thus, our final solution becomes: \ y = \frac{1}{3} + \frac{5}{3}e^{-x^3}. \

Differentiation and Integration

Differentiation and integration are core processes in calculus used to solve differential equations.
Differentiation involves finding the derivative of a function, which represents the rate of change.
In our differential equation: \frac{d y}{d x} + 3x^2 y = x^2, \Typical practice involves simplifying it using integrating factors. Integration, on the other hand, involves finding the antiderivative.
This process helps us determine the original function given its derivative.
After transforming our differential equation using the integrating factor, we identified: \ \begin{aligned} \frac{d}{dx} (y e^{x^3}) &= x^2 e^{x^3}, \ \text{We then integrated both sides:} \ \ \ \begin{aligned} \ \text{Left side:} \ \ \int \frac{d}{dx} (y e^{x^3}) \, dx \&= y e^{x^3}, \ \ \ \ \ \text{Right side:} \ \ \ \int x^2 e^{x^3} \, dx .
Changing variables by setting \( u = x^3 \) and \( du = 3x^2 \, dx \), the integral becomes: \ \frac{1}{3} \, \int e^u \, du = \frac{1}{3} e^{x^3} + C. \Solving this gives our general solution where we can apply the initial condition to find the specific solution to our problem.