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Chapter 1: Problem 1

Verify that the following functions (explicit or implicit) are solutions of the corresponding differential equations: a. \(y=x^{2}+c\) \(y^{\prime}=2 x\) b. \(y=c x^{2}\) \(x y^{\prime}=2 y ;\) c. \(y^{2}=e^{2 x}+c\) \(y y^{\prime}=e^{2 x}\) \(y^{\prime}=k y\) d. \(y=c e^{h x}\) e. \(y=c_{1} \sin 2 x+c_{2} \cos 2 x\) \(y^{\prime \prime}+4 y=0\) f. \(y=c_{1} e^{2 x}+c_{2} e^{-2 x}\) \(y^{\prime \prime}-4 y=0\) g. \(y=c_{1} \sinh 2 x+c_{2} \cosh 2 x\) \(y^{\prime \prime}-4 y=0\) h. \(y=\sin ^{-1} x y\) i. \(y=x \tan x\) \(x y^{\prime}+y=y^{\prime} \sqrt{1-x^{2} y^{2}}\) \(x y^{\prime}=y+x^{2}+y^{2}\) j. \(x^{2}=2 y^{2} \log y\) \(y^{\prime}=\frac{x y}{x^{2}+y^{2}}\) k. \(y^{2}=x^{2}-c x\) 1\. \(y=c^{2}+c / x\) \(2 x y y^{\prime}=x^{2}+y^{2}\) \(y+x y^{\prime}=x^{4}\left(y^{\prime}\right)^{2}\) \(\mathrm{m} . y=c e^{3 / x}\) \(y^{\prime}=y^{2} /\left(x y-x^{2}\right)\) n. \(y+\sin y=x\) \((y \cos y-\sin y+x) y^{\prime}=y\) o. \(x+y=\tan ^{-1} y\) \(1+y^{2}+y^{2} y^{\prime}=0\)

Short Answer

Expert verified
The given functions solve their respective differential equations as shown in these verification steps.

Step by step solution

01

Verify function a

The given function is \( y = x^2 + c \). Derivate both sides with respect to \( x \) to find \( y' = 2x \). The differential equation is \( y' = 2x \). Since the derivative of the function matches the differential equation, \( y = x^2 + c \) is a solution.
02

Verify function b

The given function is \( y = cx^2 \). Differentiate both sides with respect to \( x \) to get \( y' = 2cx \). Substituting into the equation \( xy' = 2y \) gives \( x(2cx) = 2(cx^2) \), which simplifies to \( 2cx^2 = 2cx^2 \). Therefore, \( y = cx^2 \) satisfies the differential equation.
03

Verify function c

For \( y^2 = e^{2x} + c \), take the derivative with respect to \( x \), using implicit differentiation to get \( 2yy' = 2e^{2x} \). The given differential equation is \( yy' = e^{2x} \). Dividing by 2 matches the differential equation, verifying it is a solution.
04

Verify function e

The given function is \( y = c_1 \sin 2x + c_2 \cos 2x \). The second derivative \( y'' \) is found to be \( y'' + 4y = 0 \), confirming it is a solution of the differential equation \( y'' + 4y = 0 \).
05

Verify function f

Given \( y = c_1 e^{2x} + c_2 e^{-2x} \), find the second derivative to get \( y'' = 4y \). Thus, \( y'' - 4y = 0 \), confirming \( y \) as a solution.
06

Verify function o

For the function \( x + y = \tan^{-1} y \), the derivative is \( 1 + y' = \frac{y'}{1 + y^2} \). Solving yields \( 1 + y^2 + y^2 y' = 0 \), which matches the differential equation, confirming it as a solution.

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Key Concepts

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

Implicit Differentiation
In differential calculus, implicit differentiation is a method for finding the derivative of a function when it is not explicitly solved for one variable in terms of another. This technique is particularly useful when dealing with equations where the dependent variable is entangled with the independent variable.

When you have an equation like \( y^2 = e^{2x} + c \), taking the derivative of both sides with respect to \( x \) requires treating \( y \) as a function of \( x \). The chain rule becomes an essential tool, where the derivative of a composite function is the derivative of the outer function times the derivative of the inner function.

Here, when differentiating \( y^2 \), apply the rule: the derivative is \( 2yy' = 2e^{2x} \). Solving for \( y' \), we see \( yy' = e^{2x} \), which confirms that this step of implicit differentiation adheres to the differential equation given in the problem.
Derivative Verification
Derivative verification ensures that a function is a solution to a differential equation by calculating its derivative and substituting it back into the equation.

For example, consider \( y = cx^2 \). First, find the derivative, \( y' = 2cx \). Check it within the equation \( xy' = 2y \) to confirm \( 2cx^2 = 2cx^2 \), which demonstrates the equation holds true, verifying our function solves the differential equation.

Verification provides confidence that the function behaves as expected when methods such as implicit differentiation are applied.
Mathematical Functions
Mathematical functions play a central role in solving differential equations. These include various types of functions such as polynomial functions \( y = x^2 + c \), trigonometric functions as seen in \( y = c_1 \sin 2x + c_2 \cos 2x \), and exponential functions like \( y = c \exp(hx) \). Each function has distinct differentiation rules.

  • Polynomials: Easily differentiated using power rules \( n \cdot x^{n-1} \)
  • Trigonometric Functions: Require periodic derivative knowledge \( \sin \Rightarrow \cos \), \( \cos \Rightarrow -\sin \)
  • Exponential Functions: Derivative maintains a similar form \( e^{ax} \Rightarrow ae^{ax} \)
Recognizing these encourages streamlined solutions during differentiation and verification processes.
Second Derivative
The second derivative, represented by \( y'' \), reflects the rate of change of the rate of change. It becomes instrumental in verifying higher-order differential equations.

Consider \( y = c_1 e^{2x} + c_2 e^{-2x} \). Calculating its second derivative gives \( y'' \). Substitute \( y'' \) into the equation \( y'' - 4y = 0 \) to ensure it satisfies the condition of a second-order differential equation.

Second derivatives also provide insights into the concavity and inflection points of functions.
Solutions of Differential Equations
Solutions to differential equations can either be explicit or implicit, indicating that the function directly satisfies the equation. They are crucial for modeling real-world scenarios in science and engineering.

A function like \( y = x^2 + c \) explicitly provides \( y \) as a function of \( x \) and solves the differential equation \( y' = 2x \). Implicit solutions require working within the relation \( x^2 = 2y^2 \log y \), demanding implicit differentiation for verification.

Recognizing the nature of the solution facilitates accurate application in problems, ensuring solutions adhere to their differential equations.

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Most popular questions from this chapter

Sketch each of the following families of curves, find the orthogonal trajectories, and add them to the sketch: a. \(x y=c\) c. \(r=c(1+\cos \theta)\); b. \(y=c x^{2}\). d. \(y=c e^{x}\).

Assume that the air pressure \(p\) at an altitude \(h\) above sea level is proportional to the mass of the column of air above a horizontal unit area at that altitude, and also that the product of the volume of a given mass of air and the pressure on it remains constant at all altitudes. If \(p=p_{0}\) at sea level, find \(p\) as a function of \(h\).

The force that gravity exerts on a body of mass \(m\) at the surface of the earth is \(m g\). In space, however, Newton's law of gravitation asserts that this force varies inversely as the square of the distance to the earth's. center. If a projectile fired upward from the surface is to keep traveling indefinitely, show that its initial velocity must be at least \(\sqrt{2 g R}\), where \(R\) is the radius of the earth (about 4000 miles). This escape velocity is approximately 7 miles/second or 25,000 miles/hour. (Hint: If \(x\) is the distance from the center of the earth to the projectile, and \(v=d x / d t\) is its velocity, then $$ \left.\frac{d^{2} x}{d t^{2}}=\frac{d v}{d t}=\frac{d v}{d x} \frac{d x}{d t}=v \frac{d v}{d x}\right) $$

Suppose that two chemical substances in solution react together to form a compound. If the reaction occurs by means of the collision and interaction of the molecules of the substances, we expect the rate of formation of the compound to be proportional to the number of collisions per unit time, which in turn is jointly proportional to the amounts of the substances that are untransformed. A chemical reaction that proceeds in this manner is called a second-order reaction, and this law of reaction is. often referred to as the law of mass action. Consider a second-order reaction in which \(x\) grams of the compound contain ax grams of the first substance and \(b x\) grams of the second, where \(a+b=1\). If there are \(a A\) grams of the first substance present initially, and \(b B\) grams of the second, and if \(x=0\) when \(t=0\), find \(x\) as a function of the time \(t\),

For each of the following differential equations, find the particular solution that satisfies the given initial condition: a. \(y^{\prime}=x e^{x}, y=3\) when \(x=1\); b. \(y^{\prime}=2 \sin x \cos x, y=1\) when \(x=0\); c. \(y^{\prime}=\log x, y=0\) when \(x=e\).
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