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Chapter 6: Problem 12

Find the solution of the initial value problem. $$\frac{d y}{d x}=\sin x, \quad y(0)=3$$

Short Answer

Expert verified
The solution is \(y = -\cos x + 4\).

Step by step solution

01

Recognize the Differential Equation

Identify that the problem is asking to solve a first-order differential equation given by \(\frac{dy}{dx} = \sin x\). Also note, we have an initial condition \(y(0) = 3\).
02

Integrate the Differential Equation

To solve \(\frac{dy}{dx} = \sin x\), the first step is to integrate both sides with respect to \(x\). The integration of \(\sin x\) with respect to \(x\) is \(-\cos x\), so we have \(y = -\cos x + C\), where \(C\) is the integration constant.
03

Apply Initial Condition

Use the initial condition \(y(0) = 3\) to find the constant \(C\). Substitute \(x = 0\) into the equation \(y(0) = -\cos (0) + C\). Since \(\cos(0) = 1\), we have \(-1 + C = 3\). Solving for \(C\), we get \(C = 4\).
04

Write the Final Solution

Substitute the constant \(C = 4\) back into the equation \(y = -\cos x + C\). The solution to the differential equation with the initial value is therefore \(y = -\cos x + 4\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integration
Integration is a fundamental concept in calculus, playing a crucial role in solving differential equations. It's like the reverse of differentiation.
When you have a derivative, and you want to find the original function, you perform integration. In this context, we need to solve \( \frac{dy}{dx} = \sin x \) by integrating both sides.
  • Definite Integration: When we integrate with specific limits, we find the area under the curve between those two points.
  • Indefinite Integration: Here, we integrate without limits, resulting in a family of functions that include an arbitrary constant \(C\).
In our problem, we perform indefinite integration on \( \sin x \). This results in the solution \( y = -\cos x + C \). Understanding how integration reverses differentiation is key to grasping these mathematical concepts.
Initial Value Problem
An initial value problem is a type of differential equation that includes conditions specified for the solution at a particular point. It tells us where exactly our curve should pass.
In the exercise we're working on, our initial value is \( y(0) = 3 \). This means we want our solution curve to pass through the point where \( x = 0 \) and \( y = 3 \).
  • By introducing this condition, it helps us determine the exact values that our solution function should take at certain points.
  • It is essential to incorporate these conditions to ensure our solution is not just a general solution but the precise one that satisfies given constraints.
For our specific problem, this means after integrating, we don鈥檛 just stop there. We use the condition to solve for the constant \(C\), ensuring our solution fits perfectly where required.
Integration Constant
The integration constant \(C\) is a reminder of the family nature of indefinite integration. Without specific limits, the integral \( \int f(x) \, dx \) could lead to multiple solutions, forming a family of curves.
Whenever you integrate without limits, the constant \(C\) is essential.
  • Why \(C\): It accounts for all the vertical shifts (or translations) any two curves might have in relation to each other.
  • When solving an initial value problem, the \(C\) must be found using the initial condition provided.
In this exercise, substituting \( y(0) = 3 \) into the integrated equation \( y = -\cos x + C \) allows us to solve for \( C \). We discover that \( C = 4 \), ensuring our specific curve passes through the defined initial point.

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Most popular questions from this chapter

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