Problem 70 - Thomas Calculus

Question

Solve the initial value problems in Exercises \(67-86\). $$ \frac{d y}{d x}=9 x^{2}-4 x+5, \quad y(-1)=0 $$

Step-by-Step Solution

Answer

The particular solution is \( y = 3x^3 - 2x^2 + 5x + 10 \).

Step 1: Identify the Differential Equation

The given problem involves solving a first-order ordinary differential equation (ODE): \( \frac{dy}{dx} = 9x^2 - 4x + 5 \). We are also given an initial condition \( y(-1) = 0 \).

Step 2: Integrate the Differential Equation

To find the general solution, integrate the right-hand side: \[ y = \int (9x^2 - 4x + 5) \, dx \] The integration yields:\[ y = \left( \frac{9}{3}x^3 - \frac{4}{2}x^2 + 5x + C \right) = 3x^3 - 2x^2 + 5x + C \] where \( C \) is the constant of integration.

Step 3: Apply the Initial Condition

Substitute the initial condition \( y(-1) = 0 \) into the integrated function to find \( C \):\[ 0 = 3(-1)^3 - 2(-1)^2 + 5(-1) + C \]Simplifying:\[ 0 = -3 - 2 - 5 + C = -10 + C \]Therefore, \( C = 10 \).

Step 4: Write the Particular Solution

Substitute \( C = 10 \) back into the general solution to find the particular solution:\[ y = 3x^3 - 2x^2 + 5x + 10 \]

Key Concepts

Ordinary Differential Equations

An ordinary differential equation (ODE) is a type of equation that involves functions and their derivatives. In the given problem, we deal with a first-order ODE: \( \frac{dy}{dx} = 9x^2 - 4x + 5 \). This equation represents the rate of change of the function \( y \) with respect to variable \( x \). By solving this equation, we want to find the function \( y \) that satisfies this rate of change for different values of \( x \).

• An ODE is called "ordinary" because it contains derivatives with respect to only one independent variable (in this case, \( x \)).
• It is "first-order" because the highest derivative (\( \frac{dy}{dx} \)) is the first derivative.

Recognizing these components is crucial in understanding the kind of solution process one needs to undertake.

Integration

Integration is the mathematical process used to solve differential equations. It is essentially the reverse of differentiation, allowing us to find the original function given its rate of change.In solving the problem \( \frac{dy}{dx} = 9x^2 - 4x + 5 \), we find the function \( y \) by integrating:\[ y = \int (9x^2 - 4x + 5) \, dx \].The integration process involves finding the antiderivative:

• The term \( 9x^2 \) integrates to \( 3x^3 \), since the integral of \( x^n \) is \( \frac{x^{n+1}}{n+1} \).
• The term \( -4x \) integrates to \( -2x^2 \).
• The term \( 5 \) integrates to \( 5x \).

After integrating each term, we include a constant of integration, signified as \( C \), because integration can yield multiple possible original functions that differ by a constant.

Initial Value Problem

An initial value problem (IVP) is a type of differential equation problem that, besides the equation, provides additional conditions called initial conditions. These are used to find the particular solution from the general solution.In our problem, the initial condition is given as \( y(-1) = 0 \). This tells us the exact state of function \( y \) when \( x = -1 \), helping us find the value of \( C \) in our integrated function:\[ y = 3x^3 - 2x^2 + 5x + C \]. Substituting \( x = -1 \) and \( y = 0 \):\[ 0 = 3(-1)^3 - 2(-1)^2 + 5(-1) + C \] simplifies to \( C = 10 \).

• This step narrows down the many possible solutions to the specific one that passes through the point (-1, 0).

This demonstrates the importance of the initial condition in obtaining a unique solution to the differential equation.

Constant of Integration

When you integrate an expression, you always add a constant of integration, \( C \), because integration is the reverse of differentiation, which loses constant values.To elaborate, if \( \frac{dy}{dx} = 9x^2 - 4x + 5 \), then the integrated form becomes:\[ y = 3x^3 - 2x^2 + 5x + C \],where \( C \) could be any real number, representing numerous possible vertical shifts of the curve.

• The constant signifies that there are infinitely many antiderivatives, with the graph of each being a vertical translation of others.
• The purpose of applying the initial condition is to determine the specific value of \( C \) for which the solution fits the initial condition, pinning down one particular antiderivative.

In our case, we found \( C = 10 \), making it the specific solution that satisfies the initial condition given in the problem statement.