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Chapter 9: Problem 1

Verify that the function \(y\) satisfies the given differential equation. $$ \begin{array}{l} y=e^{2 x}-3 e^{x}+2 \\ y^{\prime \prime}-3 y^{\prime}+2 y=4 \end{array} $$

Short Answer

Expert verified
The function \( y = e^{2x} - 3e^x + 2 \) satisfies the differential equation \( y'' - 3y' + 2y = 4 \).

Step by step solution

01

Differentiate y to find y prime

First, we need to find the first derivative of the given function \( y = e^{2x} - 3e^x + 2 \). Applying the derivative rules, we get: \( y' = \frac{d}{dx}[e^{2x}] - \frac{d}{dx}[3e^x] + \frac{d}{dx}[2] = 2e^{2x} - 3e^x + 0 \).Thus, \( y' = 2e^{2x} - 3e^x \).
02

Differentiate y prime to find y double prime

Next, we differentiate the first derivative \( y' = 2e^{2x} - 3e^x \) to find the second derivative:\( y'' = \frac{d}{dx}[2e^{2x}] - \frac{d}{dx}[3e^x] = 4e^{2x} - 3e^x \).Hence, \( y'' = 4e^{2x} - 3e^x \).
03

Substitute into the differential equation

Substitute \( y'' = 4e^{2x} - 3e^x \), \( y' = 2e^{2x} - 3e^x \), and \( y = e^{2x} - 3e^x + 2 \) into the differential equation \( y'' - 3y' + 2y \):\( (4e^{2x} - 3e^x) - 3(2e^{2x} - 3e^x) + 2(e^{2x} - 3e^x + 2) \).
04

Simplify the expression

Simplify each part of the substituted expression:- For \( y'' = 4e^{2x} - 3e^x \), we get that part.- Simplify \( -3y' = -3(2e^{2x} - 3e^x) = -6e^{2x} + 9e^x \).- Simplify \( 2y = 2(e^{2x} - 3e^x + 2) = 2e^{2x} - 6e^x + 4 \).Then combine all the terms together:\( 4e^{2x} - 3e^x - 6e^{2x} + 9e^x + 2e^{2x} - 6e^x + 4. \)
05

Further combine and check equality

Combine all terms with the same base:- Collect the \( e^{2x} \) terms: \( 4e^{2x} - 6e^{2x} + 2e^{2x} = 0 \).- Collect the \( e^x \) terms: \( -3e^x + 9e^x - 6e^x = 0 \).- Lastly, the constant term remains: \( + 4 \).This results in \( 4 \), confirming that the expression simplifies correctly to \( 4 \), matching the right side of the differential equation \( y'' - 3y' + 2y = 4 \).
06

Conclusion

Since substituting \( y'' \), \( y' \), and \( y \) into \( y'' - 3y' + 2y \) results in \( 4 \), which equals the right-hand side of the given differential equation, we have verified that the function \( y = e^{2x} - 3e^x + 2 \) satisfies the differential equation \( y'' - 3y' + 2y = 4 \).

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Key Concepts

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

First Derivative
The first derivative of a function gives you the rate of change or the slope of the function at any point. This is crucial in understanding how the function behaves and moves. When finding the first derivative of a function like
  • \( y = e^{2x} - 3e^x + 2 \)
apply derivative rules:
  • For \( e^{2x} \), use the rule \((\frac{d}{dx} e^{2x} = 2e^{2x})\).
  • For \( -3e^x \), use \((\frac{d}{dx} [-3e^x] = -3e^x)\).
  • For the constant \( 2 \), the derivative is zero.
Thus, the result is \( y' = 2e^{2x} - 3e^x \). Differentiating helps gather essential insights into the function's growth and contraction behaviors.
Second Derivative
The second derivative of a function represents the acceleration or the rate at which the first derivative itself is changing. It provides information about the concavity of the function and whether it is curving upwards or downwards. By differentiating the first derivative
  • \( y' = 2e^{2x} - 3e^x \)
we apply similar differentiation rules:
  • For \( 2e^{2x} \), differentiate to get \( \frac{d}{dx}[2e^{2x}] = 4e^{2x} \).
  • For \( -3e^x \), it remains \( \frac{d}{dx}[-3e^x] = -3e^x \).
So we find \( y'' = 4e^{2x} - 3e^x \). The second derivative can tell if the function is reaching a maximum or minimum point.
Function Verification
Verifying a function against a differential equation ensures that the function satisfies the equation under given conditions. In our example, we want to verify that
  • \( y = e^{2x} - 3e^x + 2 \)
satisfies
  • \( y'' - 3y' + 2y = 4 \).
Substitute the derivatives \( y'' \) and \( y' \) and the original function \( y \) into the equation. After simplification, it results in
  • \( 4 = 4 \),
proving that the function does indeed satisfy the equation. This process involves setting both sides of the equation equal and confirming they match, ensuring mathematical consistency.
Exponential Functions
Exponential functions involve terms where the variable is in the exponent, such as
  • \( e^{2x} \) and \( e^x \).
These functions are significant in various fields because of their continuous growth properties. Exponential functions have the unique feature where the rate of change of the function is proportional to the function's current value. They are crucial in solving differential equations because their derivatives and integrals result in the same functional form. Understanding how to manipulate and differentiate exponential functions is key in calculus, providing essential tools for many mathematical models.

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Most popular questions from this chapter

The following exercises require the use of a slope field program. For each differential equation and initial condition: a. Use a graphing calculator slope field program to graph the slope field for the differential equation on the window [-5,5] by [-5,5] b. Sketch the slope field on a piece of paper and draw a solution curve that follows the slopes and that passes through the point (0,2) c. Solve the differential equation and initial condition. d. Use your slope field program to graph the slope field and the solution that you found in part (c). How good was the sketch that you made in part (b) compared with the solution graphed in part (d)? $$ \left\\{\begin{array}{l} \frac{d y}{d x}=\frac{x^{2}}{y^{2}} \\ y(0)=2 \end{array}\right. $$

For each initial value problem, use an Euler's method graphing calculator program to find the approximate solution at the stated \(x\) -value, using 50 segments. [Hint: Use an interval that begins at the initial \(x\) -value and ends at the stated \(x\) -value. \(y^{\prime}=x e^{-y}\) \(y(1)=0.5\) Approximate the solution at \(x=3\)

The water in a 100,000 -gallon reservoir contains 0.1 gram of pesticide per gallon. Each hour, 2000 gallons of water (containing 0.01 gram of pesticide per gallon) is added and mixed into the reservoir, and an equal volume of water is drained off. a. Write a differential equation and initial condition that describe the amount \(y(t)\) of pesticide in the reservoir after \(t\) hours. b. Solve this differential equation and initial condition. c. Graph your solution on a graphing calculator and find when the amount of pesticide will reach 0.02 gram per gallon, at which time the water is safe to drink. d. Use your solution to find the "long-run" amount of pesticide in the reservoir.

A cell receives nutrients through its surface, and its surface area is proportional to the two-thirds power of its weight. Therefore, if \(w(t)\) is the cell's weight at time \(t,\) then \(w(t)\) satisfies \(w^{\prime}=a w^{2 / 3}\) where \(a\) is a positive constant. Solve this differential equation with the initial condition \(w(0)=1\) (initial weight 1 unit).

Solve each differential equation with the given initial condition. $$ \begin{array}{l} x y^{\prime}+4 y=10 x \\ y(1)=0 \end{array} $$
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