Problem 13 - Calculus: Graphical, Numerical, Algebraic

Question

In Exercises \(11-20,\) solve the initial value problem explicitly. $$\frac{d u}{d x}=7 x^{6}-3 x^{2}+5\( and \)u=1\( when \)x=1$$

Step-by-Step Solution

Answer

The solution to the given initial value problem is \( u(x) = x^7 - x^3 + 5x - 4 \)

Step 1: Integration

To find \( u(x) \), integrate the given derivative equation with respect to \( x \). The integral of \( \frac{d u}{d x}=7 x^{6}-3 x^{2}+5 \) is \( u(x) = \frac{7x^7}{7} - \frac{3x^3}{3} + 5x + C \) where \( C \) is the unknown arbitrary constant. Simplify to obtain \( u(x) = x^7 - x^3 + 5x + C \)

Step 2: Apply the Initial Condition

Now apply the given initial condition \( u=1 \) when \( x=1 \) to find the value of the arbitrary constant \( C \). Substituting \( x=1, u=1 \) in the \( u(x) \) equation, we get \( 1 = 1^7 - 1^3 + 5*1 + C \), which simplifies to \( C = 1 \- (1 - 1 + 5) = -4\)

Step 3: Formulate the Solution

Substitute \( C \) back into the equation of \( u(x) \) to obtain the final solution. Therefore, the solution of the given initial value problem is \( u(x) = x^7 - x^3 + 5x - 4 \)

Key Concepts

Differential Equations

Differential equations are equations that involve an unknown function and its derivatives. Solving these equations helps us understand changes between variables. For example, in calculus, a differential equation might express how velocity changes with respect to time.
There are different types of differential equations:

• Ordinary Differential Equations (ODEs): Deal with functions of one variable and their derivatives.
• Partial Differential Equations (PDEs): Involve multiple variables and their partial derivatives.

In this exercise, we deal with an ordinary differential equation. It's given by \( \frac{d u}{d x} = 7 x^{6} - 3 x^{2} + 5 \) which ties the function \( u(x) \) to its rate of change with respect to \( x \). Solving it involves finding \( u(x) \) that satisfies this relationship.

Initial Value Problem

An initial value problem (IVP) specifies not only a differential equation but also an initial condition. This condition gives the value of the unknown function at a specific point, ensuring a unique solution.
In this case, the initial condition is \( u=1 \) when \( x=1 \). This means that when \( x \) is 1, the function \( u(x) \) must equal 1.

• The initial condition is crucial as it allows us to solve for the constant \( C \) that arises during integration.
• With the specific initial value, we determine the unique solution that fits the condition, setting this solution apart from all others that might solve the differential equation.

For the given problem, after integrating, we use \( x=1 \) and \( u=1 \) to find \( C = -4 \). This ensures our solution satisfies the initial requirements.

Integration Techniques

Integration is at the heart of solving differential equations, allowing us to find the function from its derivative.
In this exercise, we used basic integration rules like:

• The reverse power rule: Integrating \( x^n \) results in \( \frac{x^{n+1}}{n+1} \).
• The linearity of integration: We can integrate each term separately and combine the results.

Using these techniques on the equation \( \frac{d u}{d x} = 7 x^{6} - 3 x^{2} + 5 \), we integrate each term to find:\[ u(x) = x^7 - x^3 + 5x + C \]This leads to finding the function whose derivative matches the given expression. Integration also introduces the constant \( C \), representing the set of all possible solutions before applying the initial condition.