Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Finding general solutions Find the general solution of each differential equation. Use \(C, C_{1}, C_{2}, \ldots\) to denote arbitrary constants. $$y^{\prime}(x)=4 \tan 2 x-3 \cos x$$

Short Answer

Expert verified
Question: Find the general solution of the first-order differential equation given by $$y'(x) = 4\tan(2x) - 3\cos(x).$$ Answer: The general solution of the given differential equation is $$y(x) = 2\ln|\sec(2x)| + 3\sin(x) + C,$$ where \(C\) is an arbitrary constant.

Step by step solution

01

Integrate the given equation with respect to x

We will integrate the given equation with respect to x: $$y(x) = \int(4\tan(2x) - 3\cos(x))\,dx.$$
02

Break the integration into two parts

Now, we can break the integration into two separate parts: $$y(x) = \int4\tan(2x)\,dx - \int3\cos(x)\,dx.$$
03

Integrate the first part: \(\int4\tan(2x)\,dx\)

We can integrate \(4\tan(2x)\) with respect to \(x\) by using the substitution method. Let: $$u = 2x \Rightarrow \frac{du}{dx} = 2 \Rightarrow dx = \frac{1}{2}du.$$ Now we substitute into the integration: $$\int4\tan(2x)\,dx= 4\int\tan(u)\, \frac{1}{2}\,du= 2\int\tan(u)\,du.$$ Now, integrating \(\tan(u)\), we have: $$2\int\tan(u)\,du = 2(\ln|\sec(u)|+C_1) = 2\ln|\sec(2x)|+C_1$$
04

Integrate the second part: \(\int3\cos(x)\,dx\)

The integration of the second part is straightforward: $$\int 3\cos(x)\,dx = 3 \sin(x) + C_2.$$
05

Combine the results of Step 3 and Step 4

Now, we will combine the results of Step 3 and Step 4: $$y(x) = 2\ln|\sec(2x)|+C_1 + 3 \sin(x) + C_2.$$
06

Write the general solution

Finally, we will write the general solution of the given differential equation by combining the arbitrary constants \(C_1\) and \(C_2\) into a single constant \(C\): $$y(x) = 2\ln|\sec(2x)| + 3\sin(x) + C.$$

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 Vaia!

Key Concepts

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

Integration
Integration is a fundamental concept in calculus, crucial for solving differential equations. It involves finding a function, called the antiderivative, whose derivative is the integrand (the function being integrated).
When we find the integral of a function, we're essentially performing the reverse of differentiation.
This process allows us to accumulate quantities, such as areas under curves or solutions to differential equations. There are several integration methods, and selecting the most suitable one is often crucial:
  • Indefinite Integrals: These integrals don't have specified limits and include an arbitrary constant of integration, often denoted by 'C'.
  • Definite Integrals: These include limits of integration, providing a numerical value representing the accumulated change between the limits.
Breaking larger integrals into manageable parts, as seen in the original exercise, is a common technique to simplify the process.
General Solution
In the context of differential equations, a general solution encompasses all possible solutions through the inclusion of arbitrary constants.
These solutions reveal the behavior of a system or phenomenon described by the differential equation. General solutions are composed of two components:
  • Particular Solution: A specific solution that satisfies the differential equation without arbitrary constants.
  • Arbitrary Constants: Symbols like 'C', 'C_1', and 'C_2' that account for the initial conditions or specific circumstances of the problem.
The arbitrary constants in the general solution provide flexibility, allowing the user to tailor it according to initial conditions or further constraints.
This adaptability is key when applying theoretical solutions to practical scenarios.
Substitution Method
The substitution method is a powerful tool used in calculus to simplify integration, making it more manageable.
This technique involves introducing a new variable to substitute a part of the original integrand, thus transforming the integral into a simpler form.Here's how it typically works:
  • Choose a Substitution: Identify a part of the integrand that complicates direct integration.
  • Define a New Variable: Let this part of the integrand be equal to a new variable, such as 'u'.
    Express 'dx' in terms of 'du' to facilitate substitution.
  • Transform and Integrate: Rewrite the integral in terms of the new variable, integrate, then revert to the original variable.
In our example, substituting 'u = 2x' simplifies the integral of '4\( \tan(2x) \)', enabling us to integrate more easily, focusing on constant factors like '2'.
Arbitrary Constants
Arbitrary constants are a crucial part of the solution in differential equations and integrals.
These constants, denoted typically by 'C', provide the means to incorporate different particular solutions under one general framework. The need for arbitrary constants arises from the nature of integration:
  • Integration's Nature: Each indefinite integral contains a constant of integration because differentiating a constant yields zero.
    This means any constant added during integration yields the same derivative.
  • Defining Initial Conditions: Arbitrary constants are pivotal in tailoring the solution to satisfy initial conditions.
    This allows us to determine a particular solution from the general one.
In the context of differential equations, merging multiple constants (like 'C_1' and 'C_2') into a single 'C' still represents all valid solutions, preserving generality while simplifying the expression.

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

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. The equation \(u^{\prime}(x)=\left(x^{2} u^{7}\right)^{-1}\) is separable. b. The general solution of the separable equation \(y^{\prime}(t)=\frac{t}{y^{7}+10 y^{4}}\) can be expressed explicitly with \(y\) in terms of \(t\) c. The general solution of the equation \(y y^{\prime}(x)=x e^{-y}\) can be found using integration by parts.

Consider the general first-order linear equation \(y^{\prime}(t)+a(t) y(t)=f(t) .\) This equation can be solved, in principle, by defining the integrating factor \(p(t)=\exp \left(\int a(t) d t\right) .\) Here is how the integrating factor works. Multiply both sides of the equation by \(p\) (which is always positive) and show that the left side becomes an exact derivative. Therefore, the equation becomes $$p(t)\left(y^{\prime}(t)+a(t) y(t)\right)=\frac{d}{d t}(p(t) y(t))=p(t) f(t)$$ Now integrate both sides of the equation with respect to t to obtain the solution. Use this method to solve the following initial value problems. Begin by computing the required integrating factor. $$y^{\prime}(t)+\frac{3}{t} y(t)=1-2 t, y(2)=0$$

Equilibrium solutions \(A\) differential equation of the form \(y^{\prime}(t)=f(y)\) is said to be autonomous (the function \(f\) depends only on y. The constant function \(y=y_{0}\) is an equilibrium solution of the equation provided \(f\left(y_{0}\right)=0\) (because then \(y^{\prime}(t)=0\) and the solution remains constant for all \(t\) ). Note that equilibrium solutions correspond to horizontal lines in the direction field. Note also that for autonomous equations, the direction field is independent of t. Carry out the following analysis on the given equations. a. Find the equilibrium solutions. b. Sketch the direction field, for \(t \geq 0\). c. Sketch the solution curve that corresponds to the initial condition \(y(0)=1\). $$y^{\prime}(t)=y(y-3)(y+2)$$

In this section, several models are presented and the solution of the associated differential equation is given. Later in the chapter, we present methods for solving these differential equations. One possible model that describes the free fall of an object in a gravitational field subject to air resistance uses the equation \(v^{\prime}(t)=g-b v,\) where \(v(t)\) is the velocity of the object for \(t \geq 0, g=9.8 \mathrm{m} / \mathrm{s}^{2}\) is the acceleration due to gravity, and \(b>0\) is a constant that involves the mass of the object and the air resistance. a. Verify by substitution that a solution of the equation, subject to the initial condition \(v(0)=0,\) is \(v(t)=\frac{g}{b}\left(1-e^{-b t}\right)\) b. Graph the solution with \(b=0.1 \mathrm{s}^{-1}\) c. Using the graph in part (b), estimate the terminal velocity \(\lim _{t \rightarrow \infty} v(t)\)

The following initial value problems model the payoff of a loan. In each case, solve the initial value problem, for \(t \geq 0,\) graph the solution, and determine the first month in which the loan balance is zero. $$B^{\prime}(t)=0.0075 B-1500, B(0)=100,000$$

See all solutions

Recommended explanations on Math Textbooks

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.

Sign-up for free