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Chapter 5: Problem 4

$$\text { Find the general solution and three particular solutions.}$$ $$y^{\prime}=e^{4 x}-x+2$$

Short Answer

Expert verified
General solution: \[ y = \frac{1}{4}e^{4x} - \frac{x^2}{2} + 2x + C \]Particular solutions: When C=0, C=1, and C=-1.

Step by step solution

01

- Integrate the Differential Equation

To find the general solution of the differential equation, integrate both sides with respect to x. The equation is given as \[ y' = e^{4x} - x + 2 \]Integrating both sides, \[ y = \frac{1}{4}e^{4x} - \frac{x^2}{2} + 2x + C \] where C is the constant of integration.
02

- Write the General Solution

The general solution includes the constant of integration C. Thus, the general solution to the differential equation is: \[ y = \frac{1}{4}e^{4x} - \frac{x^2}{2} + 2x + C \]
03

- Find Particular Solutions

To find three particular solutions, select three different values for the constant C and substitute them into the general solution.
04

- First Particular Solution

Choose C = 0. Substitute C into the general solution: \[ y = \frac{1}{4}e^{4x} - \frac{x^2}{2} + 2x \]
05

- Second Particular Solution

Choose C = 1. Substitute C into the general solution: \[ y = \frac{1}{4}e^{4x} - \frac{x^2}{2} + 2x + 1 \]
06

- Third Particular Solution

Choose C = -1. Substitute C into the general solution: \[ y = \frac{1}{4}e^{4x} - \frac{x^2}{2} + 2x - 1 \]

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Key Concepts

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

General Solution
The **general solution** of a differential equation includes all possible solutions, encompassing an integral constant. For the given differential equation, we start with the equation \[ y' = e^{4x} - x + 2 \] Integrating both sides with respect to \(x\), we get \[ y = \frac{1}{4}e^{4x} - \frac{x^2}{2} + 2x + C \] Here, \(C\) is the constant of integration, representing an entire family of solutions, since it can be any real number. Including this constant gives us the general solution to the differential equation. The general solution captures all the possible behaviors of the system described by the differential equation.
Particular Solutions
Particular solutions are found by assigning specific values to the constant in the general solution. These solutions are 'particular' because they represent specific instances of the family of solutions. For instance, to find three particular solutions, we can choose different values for \(C\). Let鈥檚 see how different values of \(C\) affect our general solution:
  • For \(C = 0\): Substitute \(C\) into the general solution to get \[ y = \frac{1}{4}e^{4x} - \frac{x^2}{2} + 2x \]
  • For \(C = 1\): Substitute \(C\) into the general solution to get \[ y = \frac{1}{4}e^{4x} - \frac{x^2}{2} + 2x + 1 \]
  • For \(C = -1\): Substitute \(C\) into the general solution to get \[ y = \frac{1}{4}e^{4x} - \frac{x^2}{2} + 2x - 1 \]
These particular solutions provide specific curves on the graph, each differing based on the value of \(C\). They can be used to model different scenarios or initial conditions in a practical context.
Integration in Differential Equations
Integration is a vital tool in solving differential equations as it allows us to reverse the differentiation process. Given a first-order differential equation like \[ y' = e^{4x} - x + 2 \] we integrate both sides with respect to \(x\) to obtain the solution \(y\). The process involves integrating each term separately:
  • The integral of \(e^{4x}\) is \(\frac{1}{4}e^{4x}\).
  • The integral of \(-x\) is \(-\frac{x^2}{2}\).
  • The integral of the constant \(2\) is \(2x\).
Combining these results, and adding the constant of integration \(C\), we get: \[ y = \frac{1}{4}e^{4x} - \frac{x^2}{2} + 2x + C \] Integration thus transforms a differential equation into a functional form that we can work with to find general and particular solutions.

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