Problem 89

Question

Which of the following graphs shows the solution of the initial value problem $$ \frac{d y}{d x}=2 x, \quad y=4 \text { when } x=1 ? $$ Give reasons for your answer. (GRAPHS CAN'T COPY)

Step-by-Step Solution

Verified

Answer

The graph of $ y=x^2+3 $, passing through $ (1, 4) $, is the solution.

1Step 1: Understand the Differential Equation

The initial value problem given is $ \frac{dy}{dx} = 2x $ with the initial condition $ y = 4 \text{ when } x=1 $. This indicates a first-order linear differential equation where the rate of change of $ y $ with respect to $ x $ is proportional to $ 2x $.

2Step 2: Solve the Differential Equation

Integrate both sides of the equation $ \frac{dy}{dx} = 2x $ with respect to $ x $ to find $ y(x) $. The integral of $ 2x $ with respect to $ x $ is $ x^2 $, so the solution becomes: \[ y(x) = x^2 + C \] where $ C $ is the constant of integration.

3Step 3: Apply Initial Conditions

Use the initial condition $ y = 4 $ when $ x = 1 $ to find the value of $ C $. Substitute $ x = 1 $ and $ y = 4 $ into $ y(x) = x^2 + C $: $ 4 = 1^2 + C $. Thus, $ 4 = 1 + C $, which simplifies to $ C = 3 $.

4Step 4: Final Solution for y(x)

Substitute $ C = 3 $ back into the equation to get the final solution: $ y(x) = x^2 + 3 $. This is the function that needs to be plotted.

5Step 5: Match the Graph with the Solution

The graph that represents the solution will show a parabola opening upwards, passing through the point $ (1, 4) $, because the function is $ y = x^2 + 3 $. Check each of the provided graphs to identify which one meets these criteria.

Key Concepts

Initial Value ProblemIntegrationConstant of IntegrationGraphical Solution

Initial Value Problem

An initial value problem in differential equations is where you are given a differential equation along with specific values of the function at certain points, called initial conditions. This kind of problem answers the question of how a function behaves over time or space, starting from a specific point. In this exercise, the initial value problem is defined with

An equation: $ \frac{dy}{dx} = 2x $
An initial condition: $ y = 4 $ when $ x = 1 $

This means that initially, when $ x = 1 $, the value of the function $ y $ is 4. Solving the initial value problem involves finding a specific solution that satisfies both the differential equation and the initial condition.

Integration

Integration is a fundamental concept in calculus that is used to solve differential equations, among other things. In this problem, integration is used to find the antiderivative of the derivative $ \frac{dy}{dx} = 2x $. This involves finding what the function $ y $ must be such that its derivative with respect to $ x $ is $ 2x $. By integrating $ 2x $ with respect to $ x $, we get $ x^2 $. However, because integration can result in a family of functions, we also include a constant of integration, $ C $, to account for any vertical shift in the graph of the function. Therefore, the solution becomes\[ y(x) = x^2 + C \].Finding this solution is crucial as it allows us to apply the initial condition to determine the specific $ y(x) $ that satisfies the problem.

Constant of Integration

When you integrate a function, such as $ 2x $, you end up with a solution plus a constant, known as the constant of integration, $ C $. This constant is important because integration can yield many solutions, each differing by a constant. The constant of integration is determined using the initial condition provided.To find $ C $, you substitute the given initial values of $ x $ and $ y $ into the integrated equation. In this problem, by plugging in $ x = 1 $ and $ y = 4 $ into the equation $ y(x) = x^2 + C $, we find:

$ 4 = 1^2 + C $
Solving for $ C $, we get $ C = 3 $

Substituting $ C = 3 $ back, the unique solution becomes $ y(x) = x^2 + 3 $. This specific value ensures that the function meets the initial condition.

Graphical Solution

A graphical solution gives a visual representation of the function obtained from solving the differential equation. For this initial value problem, the solution is $ y(x) = x^2 + 3 $. This is a parabola that opens upwards due to the $ x^2 $ term.

The graph of this parabola will pass through the point $ (1, 4) $ because $ y(1) = 1^2 + 3 = 4 $.
It indicates the initial condition has been satisfied.

Therefore, to identify which graph represents this solution, you need to look for one that shows this parabola shape with a vertex shifted up by 3 units from the origin. Seeing where it crosses $ x = 1 $ at $ y = 4 $ confirms it is the correct representation of the solution.

Problem 89

Other exercises in this chapter

Problem 88

$$\text { Find } f^{\prime}(0) \text { for } f(x)=\left\\{\begin{array}{ll}e^{-1 / x^{2}}, & x \neq 0 \\\0, & x=0\end{array}\right.$$.

View solution

Problem 89

Then find the extreme values of the function on the interval and say where they occur. $$h(x)=|x+2|-|x-3|, \quad-\infty

View solution

Problem 89

The continuous extension of $(\sin x)^{x}$ to $[0, \pi]$ a. Graph $f(x)=(\sin x)^{x}$ on the interval $0 \leq x \leq \pi .$ What value would you assign

View solution

Problem 90

Then find the extreme values of the function on the interval and say where they occur. $$k(x)=|x+1|+|x-3|, \quad-\infty

View solution