/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 Let \(y^{\prime}+4 y=0\) a) Sh... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 11

Let \(y^{\prime}+4 y=0\) a) Show that \(y=e^{-4 x}\) is a solution of this differential equation. b) Show that \(y=C e^{-4 x}\) is a solution, where \(C\) is a constant.

Short Answer

Expert verified
Both functions, \( y = e^{-4x} \) and \( y = Ce^{-4x} \), are solutions to the differential equation.

Step by step solution

01

Differentiate the given solution

Given the potential solution \(y = e^{-4x}\), we need to find \(y'\). To do this, differentiate \(y = e^{-4x}\) with respect to \(x\). Using the chain rule: \(y' = -4e^{-4x}\).
02

Substitute into the differential equation

Substitute \(y = e^{-4x}\) and \(y' = -4e^{-4x}\) into the differential equation \(y' + 4y = 0\). This gives:\[-4e^{-4x} + 4e^{-4x} = 0.\]
03

Simplify the equation

The equation simplifies to:\[0 = 0,\]confirming that the equation holds true. This verifies that \(y = e^{-4x}\) is indeed a solution to the differential equation \(y' + 4y = 0\).
04

Differentiate the general solution with constant

Consider the general solution \(y = Ce^{-4x}\), where \(C\) is a constant. Differentiate it with respect to \(x\) to get:\(y' = C(-4)e^{-4x} = -4Ce^{-4x}.\)
05

Substitute the general solution into the differential equation

Substitute \(y = Ce^{-4x}\) and \(y' = -4Ce^{-4x}\) into the differential equation:\(-4Ce^{-4x} + 4Ce^{-4x} = 0.\)
06

Verify the solution

The equation simplifies to:\[0 = 0,\]which confirms that \(y = Ce^{-4x}\) is a valid solution to the differential equation \(y' + 4y = 0\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Solution Verification
Solution verification involves checking whether a proposed function satisfies a given differential equation. To verify a solution, you'll often follow these steps:
  • Differentiate the function: Calculate the derivative of the proposed solution. For instance, when given a function like \(y = e^{-4x}\), you find \(y' = -4e^{-4x}\) using differentiation rules.
  • Substitute into the differential equation: Insert both the function and its derivative back into the original differential equation. For our exercise, the equation is \(y' + 4y = 0\).
  • Check for equalities: Ensure that after substitution, the left-hand side and the right-hand side of the equation match. For instance, \(-4e^{-4x} + 4e^{-4x} = 0\) confirms the solution because it simplifies to \(0 = 0\).
By following these steps, you can systematically verify that a function is indeed a solution to the given differential equation.
Chain Rule
The chain rule is a fundamental concept in calculus used to differentiate compositions of functions. It's particularly handy when dealing with exponentials involving complex exponents.When differentiating a function like \(y = e^{-4x}\), the chain rule helps by acknowledging the inner function, which in this instance is \(-4x\). The chain rule states:\[\frac{d}{dx} [f(g(x))] = f'(g(x)) \cdot g'(x)\]For \(y = e^{-4x}\):
  • Identify the outer function: For \(e^{-4x}\), the outer function is the exponential function \(e^u\), where \(u = -4x\).
  • Differentiate the outer function: The derivative of \(e^u\) with respect to \(u\) is \(e^u\).
  • Multiply by the derivative of the inner function: Since \(u = -4x\), its derivative is \(-4\). Thus, \(d/dx[e^{-4x}] = (-4)e^{-4x}\).
Using the chain rule effectively allows us to quickly find that \(y' = -4e^{-4x}\), crucially aiding in both understanding and solving differential equations.
Exponential Functions
Exponential functions are pivotal in calculus, especially when solving differential equations where growth and decay are modeled. The form \(y = e^{kx}\), where \(k\) is a constant, is a typical example.Key properties of exponential functions include:
  • Continuous Growth/Decay: Depending on the sign of \(k\), the function depicts exponential growth (\(k > 0\)) or decay (\(k < 0\)). In \(y = e^{-4x}\), the negative sign indicates a steadily decreasing function as \(x\) increases.
  • Differentiation: The derivative of \(e^{kx}\) remains an exponential function, ##: \(\frac{d}{dx} [e^{kx}] = ke^{kx}\). This property is crucial for differential equations as it enables simple substitution.
  • Smooth Curve and Domain: Exponential functions are always smooth and defined for all real numbers, which makes them easy to work with in many mathematical contexts.
Understanding these properties allows better application of exponential functions in solving differential equations, making them an essential tool for analysts and engineers alike.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Approximate the integral $$\int_{-\infty}^{\infty} e^{-x^{2}} d x$$

The amount of money, \(A(t)\), in Ina's savings account after \(t\) years is modeled by the differential equation \(d A / d t=0.0418 A\) a) What is the continuous growth rate? b) Find the particular solution, \(A(t),\) if Ina's account is worth \(\$ 3479.02\) after 2 yr. c) Find the amount that Ina deposited initially.

The capitalized cost, \(c,\) of an asset over its lifetime is the total of the initial cost and the present value of all maintenance expenses that will occur in the future. It is computed with the formula $$ c=c_{0}+\int_{0}^{L} m(t) e^{-k t} d t $$ where \(c_{0}\) is the initial cost of the asset, \(L\) is the lifetime (in years), \(k\) is the interest rate (compounded continuously), and \(m(t)\) is the annual cost of maintenance. Find the capitalized cost under each set of assumptions. $$ \begin{array}{l} c_{0}=\$ 300,000, k=5 \%, m(t)=\$ 30,000+\$ 500 t, \\ L=20 \end{array} $$

For each probability density function, over the given interval, find \(\mathrm{E}(x), \mathrm{E}\left(x^{2}\right),\) the mean, the variance, and the standard deviation. $$ f(x)=\frac{1}{5}, \quad[3,8] $$

Domar's capital expansion model is $$ \frac{d I}{d t}=h k I $$ where \(I\) is the investment, \(h\) is the investment productivity (constant), \(k\) is the marginal productivity to the consumer (constant), and \(t\) is the time. a) Use separation of variables to solve the differential equation. b) Rewrite the solution in terms of the condition \(I_{0}=I(0)\)
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Geometry

Read Explanation

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation

Calculus

Read Explanation

Mechanics Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.