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Chapter 5: Problem 8

Find the particular solution determined by the given condition. $$y^{\prime}=3 x^{2}-x+5 ; \quad y=6 \text { when } x=0$$

Short Answer

Expert verified
The particular solution is \(y = x^3 - \frac{x^2}{2} + 5x + 6\).

Step by step solution

01

- Integrate the Differential Equation

To find the particular solution, start by integrating the given differential equation. The differential equation is given as: \(y' = 3x^2 - x + 5\) Integrate both sides with respect to \(x\): \(\frac{dy}{dx} = 3x^2 - x + 5\) \(\rightarrow y = \int (3x^2 - x + 5) \, dx\)
02

- Perform the Integration

Integrate the right-hand side term by term: \(y = \int 3x^2 \, dx - \int x \, dx + \int 5 \, dx\) Compute each integral: \(\int 3x^2 \, dx = x^3\), \(\int x \, dx = \frac{x^2}{2}\), and \(\int 5 \, dx = 5x\) Thus, \(y = x^3 - \frac{x^2}{2} + 5x + C\) where \(C\) is the integration constant.
03

- Substitute Initial Condition

Use the initial condition \(y(0) = 6\) to determine the constant \(C\). Substituting \(x = 0\) and \(y = 6\) into the integrated equation: \(6 = 0^3 - \frac{0^2}{2} + 5(0) + C\) This simplifies to \(6 = C\).
04

- Write the Particular Solution

Substitute the value of \(C = 6\) back into the general solution: \(y = x^3 - \frac{x^2}{2} + 5x + 6\) This gives the particular solution.

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Key Concepts

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

Initial Condition
In many problems involving differential equations, you鈥檒l encounter an 'initial condition'. This information allows you to determine a specific solution, known as a particular solution, out of the infinite possibilities provided by a general solution.
For example, if you have a differential equation and are given that y equals a certain value when x is 0, this is your initial condition. In this exercise, it is given as `y=6` when `x=0`. The initial condition helps to uniquely determine the integration constant, which you will learn about shortly.
Integration
Integration is a fundamental process in calculus used to find functions when given their derivatives. In the context of solving differential equations, integration helps us find the general solution.
In our exercise, we start with the differential equation\(y' = 3x^2 - x + 5\). To find the function for y, we need to integrate the right-hand side with respect to x. This means we compute \[\frac{dy}{dx} = 3x^2 - x + 5 \ \rightarrow y = \int (3x^2 - x + 5) \, \ dx \].
Each term of the equation is integrated separately:
  • \(\int 3x^2 \, dx = x^3\)
  • \(\int x \, dx = \frac{x^2}{2}\)
  • \(\int 5 \, dx = 5x\)
General Solution
The general solution of a differential equation is found after performing the integration. This solution includes the integration constant, as it represents the family of all possible solutions.
Once we have integrated the terms, we get:
\(y = x^3 - \frac{x^2}{2} + 5x + C\)
This expression is the general solution of the differential equation. It reflects an entire set of functions that satisfy the differential equation. To pinpoint the exact function, we need to use the initial condition to find the specific value of 'C'.
Integration Constant
After integration, you will always find an unknown constant, often referred to as the integration constant 'C'. This constant accounts for any vertical shift in the graph of the function. Without it, you wouldn't be able to fully describe the family of possible solutions to your differential equation.
In our exercise, we determine 'C' by employing the initial condition. Given `y=6` when `x=0`, substituting these values into our general solution:
\(6 = 0^3 - \frac{0^2}{2} + 5(0) + C\), which simplifies to \(6 = C\).
Therefore, 'C' equals 6. Now, by substituting this value back into the general solution, we get the particular solution:
\(y = x^3 - \frac{x^2}{2} + 5x + 6\).

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