Introduction

In Problems 9-13, determine whether the given relation is an implicit solution to the given differential equation. Assume that the relationship implicitly defines y as a function of x and use implicit differentiation.

${\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{sin}\left(x+y\right)\mathbf{=}\mathbf{1}$, $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{2}\mathbf{xsec}\left(x+y\right)\mathbf{-}\mathbf{1}$

Use the convolution theorem to find the inverse Laplace transform of the given function.

$\frac{\mathbf{s}\mathbf{+}\mathbf{1}}{{\mathbf{(}{\mathbf{s}}^{\mathbf{2}}\mathbf{+}\mathbf{1}\mathbf{)}}^{\mathbf{2}}}$

In Problems 9–20, determine whether the equation is exact.

If it is, then solve it.

 $\mathbf{(}\mathbf{cos}\mathbf{ }\mathbf{x}\mathbf{ }\mathbf{cos}\mathbf{ }\mathbf{y}\mathbf{+}\mathbf{2}\mathbf{x}\mathbf{)}\mathbf{dx}\mathbf{-}\mathbf{(}\mathbf{sin}\mathbf{ }\mathbf{x}\mathbf{ }\mathbf{sin}\mathbf{ }\mathbf{y}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{)}\mathbf{dy}\mathbf{=}\mathbf{0}$

Question: In Problem, find the first three nonzero terms in the power series expansion for the product f(x) g(x).

In Problems 9–20, determine whether the equation is exact.

If it is, then solve it.

${\mathbf{\text{e}}}^{\mathbf{\text{t}}}\left(\text{y - t}\right)\mathbf{\text{dt + }}\left({\text{1 + e}}^{\text{t}}\right)\mathbf{\text{dy = 0}}$

Question:In Problem find the first three nonzero terms in the power series expansion for the product f(x)g(x).

In Problems 9–20, determine whether the equation is exact.

If it is, then solve it.

$\left(\frac{t}{y}\right)\mathbf{dy}\mathbf{+}\mathbf{(}\mathbf{1}\mathbf{+}\mathbf{ln}\mathbf{ }\mathbf{y}\mathbf{)}\mathbf{dt}\mathbf{=}\mathbf{0}$

In Problems 13 and 14, find an integrating factor of the form and solve the equation.

$\left(\mathbf{12}\mathbf{+}\mathbf{5}\mathbf{xy}\right)\mathbf{dx}\mathbf{+}\left(\mathbf{6}{\mathbf{xy}}^{\mathbf{-}\mathbf{1}}\mathbf{+}\mathbf{3}{\mathbf{x}}^{\mathbf{2}}\right)\mathbf{dy}\mathbf{=}\mathbf{0}$

Verify that $\mathbf{\varphi }\left(x\right)\mathbf{=}\frac{\mathbf{2}}{\left(1-{\mathrm{ce}}^x\right)}\mathbf{,}$ where c is an arbitrary constant, it is a one-parameter family of solutions to $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\frac{\mathbf{y}\left(y-2\right)}{\mathbf{2}}$. Graph the solution curves corresponding to $\mathbf{c}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{\pm }\mathbf{1}\mathbf{,}\mathbf{\pm }\mathbf{2}$ using the same coordinate axes.

In Problems 13-19, find at least the first four nonzero terms in a power series expansion of the solution to the given initial value problem.

$\left(x^2+1\right)\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}{\mathbf{e}}^{\mathbf{x}}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$

Newton’s law of cooling states that the rate of change in the temperature T(t) of a body is proportional to the difference between the temperature of the medium M(t) and the temperature of the body. That is, $\frac{\mathbf{dT}}{\mathbf{dt}}\mathbf{=}\mathbf{K}\left[\mathbf{M}\left(\mathbf{t}\right)\mathbf{-}\mathbf{T}\left(\mathbf{t}\right)\right]$ where K is a constant. Let $\mathbf{K}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{04}{\left(\mathbf{min}\right)}^{\mathbf{-}\mathbf{1}}$ and the temperature of the medium be constant, $\mathbf{M}\left(\mathbf{t}\right)\mathbf{=}\mathbf{293} \mathbf{kelvins}$ . If the body is initially at 360 kelvins, use Euler’s method with h = 3.0 min to approximate the temperature of the body after 

(a) 30 minutes. 

(b) 60 minutes.

Verify that ${\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{cy}}^{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{,}$ where c is an arbitrary non-zero constant, is a one-parameter family of implicit solutions to $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\frac{\mathbf{xy}}{{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}}$ and graph several of the solution curves using the same coordinate axes.

Stefan’s law of radiation states that the rate of change in the temperature of a body at T (t) kelvins in a medium at M (t) kelvins is proportional to ${\mathbf{M}}^{\mathbf{4}}\mathbf{-}{\mathbf{T}}^{\mathbf{4}}$ . That is, $\frac{\mathbf{dT}}{\mathbf{dt}}\mathbf{=}\mathbf{K}\left[\mathbf{M}{\left(\mathbf{t}\right)}^{\mathbf{4}}\mathbf{-}\mathbf{T}{\left(\mathbf{t}\right)}^{\mathbf{4}}\right]$  where K is a constant. Let  $\mathbf{K}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{9}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}{\left(\mathbf{min}\right)}^{\mathbf{-}\mathbf{1}}$ and assume that the medium temperature is constant, M (t) = 293 kelvins. If T (0) = 360 kelvins, use Euler’s method with h = 3.0 min to approximate the temperature of the body after 

(a) 30 minutes. 

(b) 60 minutes.

Show that $\mathbf{\varphi }\left(x\right)\mathbf{=}{\mathbf{Ce}}^{\mathbf{3}\mathbf{x}}\mathbf{+}\mathbf{1}$ is a solution to $\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{-}\mathbf{3}\mathbf{y}\mathbf{=}\mathbf{-}\mathbf{3}$ for any choice of the constant C. Thus, ${\mathbf{Ce}}^{\mathbf{3}\mathbf{x}}\mathbf{+}\mathbf{1}$ is a one-parameter family of solutions to the differential equation. Graph several of the solution curves using the same coordinate axes.

In the electrical circuit of Figure\({\bf{5}}{\bf{.52}}\), take\({{\bf{R}}_{\bf{1}}}{\bf{ = }}{{\bf{R}}_{\bf{2}}}{\bf{ = 1\Omega ,C = 1\;F}}\), and\({\bf{L = 1H}}\). Derive three equations for the unknown currents\({{\bf{I}}_{\bf{1}}}{\bf{,}}{{\bf{I}}_{\bf{2}}}\), and \({{\bf{I}}_{\bf{3}}}\) by writing Kirchhoff's voltage law for loops \({\bf{1}}\)and\({\bf{2}}\), and Kirchhoff's current law for the top juncture. Find the general solution.

In Problem 19, solve the given initial value problem 

 $\begin{array}{l}y\text{'}\text{'}\text{'}-y\text{'}\text{'}-4y\text{'}+4y=0\\ y\left(0\right)=-4\\ y\text{'}\left(0\right)=-1\\ y\text{'}\text{'}\left(0\right)=-19\end{array}$

In Problems 9–20, determine whether the equation is exact.

If it is, then solve it.

 $\left(\mathbf{2}\mathbf{x}\mathbf{+}\frac{\mathbf{y}}{\mathbf{1}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}{\mathbf{y}}^{\mathbf{2}}}\right)\mathbf{dx}\mathbf{+}\left(\frac{\mathbf{x}}{\mathbf{1}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}{\mathbf{y}}^{\mathbf{2}}}\mathbf{-}\mathbf{2}\mathbf{y}\right)\mathbf{dy}\mathbf{=}\mathbf{0}$

In Problems $\mathbf{15}-\mathbf{24}$ , solve for $\mathbf{Y}\mathbf{\left(}\mathbf{s}\mathbf{\right)}$, the Laplace transform of the solution  $\mathbf{y}\left(\mathbf{t}\right)$ to the given initial value problem.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{3}\mathbf{y}\mathbf{=}{\mathbf{t}}^{\mathbf{3}}\mathbf{;} \mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{0}\mathbf{,} \mathbf{y}\mathbf{\text{'}}\left(\mathbf{0}\right)\mathbf{=}\mathbf{0}$

In Problems 11–20, determine the partial fraction expansions for the given rational function.

\(\frac{s}{{(s - 1)\left( {{s^2} - 1} \right)}}\)

Chemical Reactions. The reaction between nitrous oxide and oxygen to form nitrogen dioxide is given by the balanced chemical equation\({\bf{2NO + }}{{\bf{O}}_{\bf{2}}}{\bf{ = 2N}}{{\bf{O}}_{\bf{2}}}\). At high temperatures the dependence of the rate of this reaction on the concentrations of NO,\({{\bf{O}}_{\bf{2}}}\), and \({\bf{N}}{{\bf{O}}_{\bf{2}}}\) is complicated. However, at \({\bf{2}}{{\bf{5}}^{\bf{o}}}\)C the rate at which NO2 is formed obeys the law of mass action and is given by the rate equation\(\frac{{{\bf{dx}}}}{{{\bf{dt}}}}{\bf{ = k(\alpha  - x}}{{\bf{)}}^{\bf{2}}}\left( {{\bf{\beta   - }}\frac{{\bf{x}}}{{\bf{2}}}} \right)\), where \({{\bf{x}}_{\bf{1}}}{{\bf{t}}_{\bf{2}}}\) denotes the concentration of \({\bf{N}}{{\bf{o}}_{\bf{2}}}\) at time t, k is the rate constant, a is the initial concentration of NO, and b is the initial concentration of \({{\bf{O}}_{\bf{2}}}\). At \({\bf{2}}{{\bf{5}}^{\bf{o}}}\)C, the constant k is \({\bf{7}}{\bf{.13 \times 1}}{{\bf{0}}^{\bf{3}}}{{\bf{(litre)}}^{\bf{2}}}{\bf{/(mole}}{{\bf{)}}^{\bf{2}}}\)  (second).Let \({\bf{\alpha   = 0}}{\bf{.0010}}\) mole/L, \({\bf{\beta  = 0}}{\bf{.0041}}\) mole/L, and (\({\bf{x(0) = 0}}\) mole>L. Use the fourth-order Runge–Kutta algorithm to approximate \({\bf{x}}\left( {{\bf{10}}} \right)\). For a tolerance of\({\bf{\varepsilon  = 0}}{\bf{.000001}}\), use a stopping procedure based on the relative error.

Determine which values of m the function $\mathit{\varphi }\left(x\right)\mathbf{=}{\mathbf{x}}^{\mathbf{m}}$  is a solution to the given equation.

(a) $\mathbf{3}{\mathbf{x}}^{\mathbf{2}}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dx}}^{\mathbf{2}}}\mathbf{+}\mathbf{11}\mathbf{x}\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{-}\mathbf{3}\mathbf{y}\mathbf{=}\mathbf{0}$

(b) ${\mathbf{x}}^{\mathbf{2}}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dx}}^{\mathbf{2}}}\mathbf{-}\mathbf{x}\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{-}\mathbf{5}\mathbf{y}\mathbf{=}\mathbf{0}$

Verify that the function $\mathbf{\varphi }\left(x\right)\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{x}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{x}}$ is a solution to the linear equation $\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dx}}^{\mathbf{2}}}\mathbf{+}\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{=}\mathbf{0}$ for any choice of the constants ${\mathbf{c}}_{\mathbf{1}}$ and ${\mathbf{c}}_{\mathbf{2}}$. Determine ${\mathbf{c}}_{\mathbf{1}}$ and ${\mathbf{c}}_{\mathbf{2}}$ so that each of the following initial conditions is satisfied.

(a) $\mathbf{y}\left(0\right)\mathbf{=}\mathbf{2}\mathbf{,}\mathbf{ }\mathbf{y}\mathbf{\text{'}}\left(0\right)\mathbf{=}\mathbf{1}$

(b) $\mathbf{y}\left(1\right)\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{ }\mathbf{y}\mathbf{\text{'}}\left(1\right)\mathbf{=}\mathbf{0}$

A particular solution and a fundamental solution set are given for a nonhomogeneous equation and its corresponding homogeneous equation. 

(a) Find a general solution to the nonhomogeneous equation. 

(b) Find the solution that satisfies the specified initial condition.

$$\begin{gathered} {\mathbf{y}}^{\left(4\right)}\mathbf{-}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{5}\mathbf{cosx}\mathbf{;} \\ \mathbf{y}\left(0\right)\mathbf{=}\mathbf{2}\mathbf{,}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{y}\mathbf{\text{'}}\left(0\right)\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\left(0\right)\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\left(0\right)\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{;} \\ {\mathbf{y}}_{\mathbf{p}}\mathbf{=}\mathbf{cosx}\mathbf{;}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\left\{e^x\mathrm{cosx}, e^x\mathrm{sinx}, e^{-x}\mathrm{cosx}, e^{-x}\mathrm{sinx}\right\} \end{gathered}$$

In Problems 23-28, determine whether Theorem 1 implies that the given initial value problem has a unique solution.

$\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}{\mathbf{y}}^{\mathbf{4}}\mathbf{-}{\mathbf{x}}^{\mathbf{4}}\mathbf{,}\mathbf{ }\mathbf{y}\left(0\right)\mathbf{=}\mathbf{7}$

In Problems 21–26, solve the initial value problem.

$\mathbf{(}{\mathbf{e}}^{\mathbf{t}}\mathbf{y}\mathbf{+}{\mathbf{te}}^{\mathbf{t}}\mathbf{y}\mathbf{)}\mathbf{dt}\mathbf{+}\mathbf{(}\mathbf{t}{\mathbf{e}}^{\mathbf{t}}\mathbf{+}\mathbf{2}\mathbf{)}\mathbf{dy}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{(}\mathbf{0}\mathbf{)}\mathbf{=}\mathbf{-}\mathbf{1}$

Question:(a) Use the general solution given in Example 5 to solve the IVP. 4x"+e^-0.1tx=0,x(0)=1,x'(0)=$-\frac{1}{2}$. Also use J'₀(x)=-J₁ (x) and Y'₀(x)=-Y₁(x)=-Y₁(x) along withTable 6.4.1 or a CAS to evaluate coefficients.

(b) Use a CAS to graph the solution obtained in part (a) for.

Question: In Problems 23–26, express the given power series as a series

with generic term X^k .

23.$\sum _{n=1}^{\infty }n{a}_n{x}^{n-1}$

In Problems 21–26, solve the initial value problem $\mathbf{(}{\mathbf{e}}^{\mathbf{t}}\mathbf{x}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{dt}\mathbf{+}\mathbf{(}{\mathbf{e}}^{\mathbf{t}}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{dx}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{x}\mathbf{(}\mathbf{1}\mathbf{)}\mathbf{=}\mathbf{1}$

Question:Use a CAS to graph  J_3/2(x),J_-3/2(x),J_5/2(x), and J_-5/2(x).

Question: In Problems 23–26, express the given power series as a series

with generic term X^k .

24.$\sum _{n=2}^{\infty }n(n-1)a_n{x}^{n+2}$

In Problems 21–26, solve the initial value problem.

 $\mathbf{(}{\mathbf{y}}^{\mathbf{2}}\mathbf{ }\mathbf{sin}\mathbf{ }\mathbf{x}\mathbf{)}\mathbf{dx}\mathbf{+}\mathbf{(}\frac{\mathbf{1}}{\mathbf{x}}\mathbf{-}\frac{\mathbf{y}}{\mathbf{x}}\mathbf{)}\mathbf{dy}\mathbf{,}\mathbf{y}\mathbf{(}\mathit{\pi }\mathbf{)}\mathbf{=}\mathbf{1}$

Question: In Problems 23–26, express the given power series as a series with generic term X^K.

25.$\sum _{n=0}^{\infty }{a}_n{x}^{n+1}$ 

Question: In Problems 23–26, express the given power series as a series with generic term .

26. $\sum _{n=1}^{\infty }\frac{a_n}{n+3}{x}^{n+3}$

Question: Show that,

$X^2.\sum _{n-0}^{\infty }n(n+1)a_n{x}^n=\sum _{n-2}^{\infty }(n-2)(n-1)a_{n-2}{x}^n$ 

Question: Show that,$28. \sum _{n=0}^{\infty }{a}_n{x}^{n+1}+\sum _{n=1}^{\infty }n{b}_n{x}^{n-1}=b_1+\sum _{n-1}^{\infty }[2a_{n-1}+(n+1\left)b_{n+1}\right]x^n$

(a) For the initial value problem (12) of Example 9. Show that ${\mathbf{\varphi }}_{\mathbf{1}}\left(x\right)\mathbf{=}\mathbf{0}$ and ${\mathbf{\varphi }}_{\mathbf{2}}\left(x\right)\mathbf{=}{\left(x-2\right)}^{\mathbf{3}}$ are solutions. Hence, this initial value problem has multiple solutions. (See also Project G in Chapter 2.)

(b) Does the initial value problem $\mathit{y}\mathbf{\text{'}}\mathbf{=}\mathbf{3}{\mathit{y}}^{\frac{\mathbf{2}}{\mathbf{3}}}\mathbf{,}$ $\mathbf{y}\left(0\right)\mathbf{=}{\mathbf{10}}^{\mathbf{-}\mathbf{7}}$ have a unique solution in a neighbourhood of $\mathbf{x}\mathbf{=}\mathbf{0}$?

Implicit Function Theorem. Let $\mathbf{G}\left(x,y\right)$ have continuous first partial derivatives in the rectangle $\mathbf{R}\mathbf{=}\left\{\left(x,y\right):a<x<b,c<y<d\right\}$ containing the point $(x_0,y_0)$. If $\mathbf{G}\left(x_0,y_0\right)\mathbf{=}\mathbf{0}$ and the partial derivative ${\mathbf{G}}_{\mathbf{y}}\left(x_0,y_0\right)\mathbf{\ne }\mathbf{0}$, then there exists a differentiable function $\mathbf{y}\mathbf{=}\mathbf{\varphi }\left(x\right)$, defined in some interval $\mathbf{I}\mathbf{=}\left(x_0-\delta ,y_0+\delta \right)$, that satisfies G for all for $\mathbf{G}\left(x,\varphi \left(x\right)\right)$ all $\mathbf{x}\mathbf{\in }\mathbf{I}$.

The implicit function theorem gives conditions under which the relationship $\mathbf{G}\left(x,y\right)\mathbf{=}\mathbf{0}$ implicitly defines y as a function of x. Use the implicit function theorem to show that the relationship $\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{+}{\mathbf{e}}^{\mathbf{xy}}\mathbf{=}\mathbf{0}$ given in Example 4, defines y implicitly as a function of x near the point $\left(0,-1\right)$.

Aging spring. As a spring ages, its “spring constant” decreases on value. One such model for a mass-spring system with an aging spring is

mx"(t)+bx'(t)+ke^-ηt x(t)=0

where m is the mass, the damping constant, k and η positive constants and x(t) displacement of the spring from equilibrium position. Let m=1 kg, b=2 N sec/m, k=1 N/m, η =1 sec^-1. The system is set in motion by displacing the mass 1m from it equilibrium position and releasing it (x(0)=1, x'(0)=0). Find at least the first four nonzero terms in a power series expansion of about t=0 of displacement. 

Question: In Problems 29–34, determine the Taylor series about the point X₀  for the given functions and values of X₀.

31. $f\left(x\right)= \frac{1+x}{1-x}.$x₀ = 0 ,    

Use the method in Problem 32 to find the orthogonal trajectories for each of the given families of curves, where k is a parameter.

 (a) $\mathbf{2}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{=}\mathbf{k}$

(b) $\mathbf{y}\mathbf{=}{\mathbf{kx}}^{\mathbf{4}}$

(c) $\mathbf{y}\mathbf{=}{\mathbf{e}}^{\mathbf{kx}}$

(d)  ${\mathbf{y}}^{\mathbf{2}}\mathbf{=}\mathbf{kx}$

[Hint: First express the family in the form F(x, y) = k.]

Mixing. Suppose a brine containing 0.2 kg of salt per liter runs into a tank initially filled with 500 L of water containing 5 kg of salt. The brine enters the tank at a rate of 5 L/min. The mixture, kept uniform by stirring, is flowing out at the rate of 5 L/min (see Figure 2.6).

(a)Find the concentration, in kilograms per liter, of salt in the tank after 10 min. [Hint: Let A denote the number of kilograms of salt in the tank at t minutes after the process begins and use the fact that

rate of increase in A =rate of input - rate of exit.

A further discussion of mixing problems is given in Section 3.2.]

(b) After 10 min, a leak develops in the tank and an additional liter per minute of mixture flows out of the tank (see Figure 2.7). What will be the concentration, in kilograms per liter, of salt in the tank 20 min after the leak develops? [Hint: Use the method discussed in Problems 31 and 32.]

Question: The Taylor series for f(x) =  ln (x) about x₂=0 given in equation (13) can also be obtained as follows:

(a) Starting with the expansion 1/ (1-s) =$\sum _{n=0}^{\infty }s\text{'}\text{'}$  and observing that

 '

obtain the Taylor series for 1/x  about x₀ = 1 .

(b) Since   use the result of part (a) and termwise integration to obtain the Taylor series for f (x) =lnx about x₀ = 1 .

Variation of Parameters. Here is another procedure for solving linear equations that is particularly useful for higher-order linear equations. This method is called variation of parameters. It is based on the idea that just by knowing the form of the solution, we can substitute into the given equation and solve for any unknowns. Here we illustrate the method for first-order equations (see Sections 4.6 and 6.4 for the generalization to higher-order equations).

(a) Show that the general solution to (20)$\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\mathbf{P}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{y}\mathbf{=}\mathbf{Q}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$   has the form$\mathbf{y}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{C}{\mathbf{y}}_{\mathbf{h}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{+}{\mathbf{y}}_{\mathbf{p}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$, where ${\mathbf{y}}_{\mathbf{h}}$ ( $\mathbf{\ne }\mathbf{0}$is a solution to equation (20) when  $\mathbf{Q}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{ }$, 

C is a constant, and${\mathbf{y}}_{\mathbf{p}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{v}\mathbf{\left(}\mathbf{x}\mathbf{\right)}{\mathbf{y}}_{\mathbf{h}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$ for a suitable function v(x). [Hint: Show that we can take ${\mathbf{y}}_{\mathbf{h}}\mathbf{=}{\mathbf{\mu }}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$ and then use equation (8).] We can in fact determine the unknown function   ${\mathbf{y}}_{\mathbf{h}}$by solving a separable equation. Then direct substitution of v${\mathbf{y}}_{\mathbf{h}}$ in the original equation will give a simple equation that can be solved for v.

Use this procedure to find the general solution to (21) $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\frac{\mathbf{3}}{\mathbf{x}}\mathbf{y}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}$ , x > 0 by completing the following steps:

(b) Find a nontrivial solution  ${\mathbf{y}}_{\mathbf{h}}$ to the separable equation (22) $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\frac{\mathbf{3}}{\mathbf{x}}\mathbf{y}\mathbf{=}\mathbf{0}$  , $\mathbf{x}\mathbf{>}\mathbf{0}$ .

(c) Assuming (21) has a solution of the form${\mathbf{y}}_{\mathbf{p}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{v}\mathbf{\left(}\mathbf{x}\mathbf{\right)}{\mathbf{y}}_{\mathbf{h}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$  , substitute this into equation (21), and simplify to obtain  $\mathbf{v}\mathbf{\text{'}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{{\mathbf{x}}^{\mathbf{2}}}{{\mathbf{y}}_{\mathbf{h}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}$. 

d) Now integrate to get $\mathbf{v}\left(\mathbf{x}\right)\mathbf{ }$ 

(e) Verify that $\mathbf{y}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{C}{\mathbf{y}}_{\mathbf{h}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{+}\mathbf{v}\mathbf{\left(}\mathbf{x}\mathbf{\right)}{\mathbf{y}}_{\mathbf{h}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$  is a general solution to (21).

Question: Let f(x) and g(x)  be analytic at x₀. Determine whether the following statements are always true or sometimes false:

(a) 3f(x)+g(x) is analytic at x₀ .

(b) f(x)/g(x) is analytic at x₀ .

(c) f'(x) is analytic at x₀ .

(d)  ${\left[f\left(x\right)\right]}^3-{\int }_{x_0}^{x}g\left(t\right)dt$ is analytic at x₀ .

The temperature T (in units of 100 F) of a university classroom on a cold winter day varies with time t (in hours) as $$\begin{gathered} \frac{\mathbf{dT}}{\mathbf{dt}}\mathbf{=}\left\{\mathbf{1}\mathbf{-}\mathbf{T}\mathbf{,}\mathbf{if} \mathbf{heating} \mathbf{units} \text{is} \mathbf{On} \\ \mathbf{-}\mathbf{T}\mathbf{,}\mathbf{if} \mathbf{heating} \mathbf{units} \mathbf{is} \mathbf{OFF}\mathbf{.} \\ \right. \end{gathered}$$

$\mathbf{T}\mathbf{=}\mathbf{0}\mathbf{ }$Suppose   at 9:00 a.m., the heating unit is ON from 9-10 a.m., OFF from 10-11 a.m., ON again from 11 a.m.–noon, and so on for the rest of the day. How warm will the classroom be at noon? At 5:00 p.m.?

Use Heaviside's expansion formula derived in Problem 40 to determine the inverse Laplace transform of

$F\left(s\right)=\frac{3s^2-16s+5}{(s+1)(s-3)(s-2)}$

In Problems 13-16, write a differential equation that fits the physical description. The rate of change in the temperature T of coffee at time t is proportional to the difference between the temperature M of the air at time t and the temperature of the coffee at time t.

In Problems 13-16, write a differential equation that fits the physical description. The rate of change of the mass A of salt at time t is proportional to the square of the mass of salt present at time t.

(a) Show that ${\mathbf{y}}^{\mathbf{2}}\mathbf{+}\mathbf{x}\mathbf{-}\mathbf{3}\mathbf{=}\mathbf{0}$ is an implicit solution to $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}\mathbf{y}}$ on the interval $\mathbf{(}\mathbf{-}\mathbf{\infty }\mathbf{,}\mathbf{3}\mathbf{)}$.

(b) Show that ${\mathbf{xy}}^{\mathbf{3}}\mathbf{-}{\mathbf{xy}}^{\mathbf{3}}\mathbf{sin} \mathbf{x}\mathbf{=}\mathbf{1}$ is an implicit solution to $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\frac{\left(\mathbf{x} \mathbf{cos} \mathbf{x}\mathbf{+}\mathbf{sin} \mathbf{x}\mathbf{-}\mathbf{1}\right)\mathbf{y}}{\mathbf{3}\left(\mathbf{x}\mathbf{-}\mathbf{x} \mathbf{sin} \mathbf{x}\right)}$ on the interval $\left(0,\frac{\pi }{2}\right)$.

In Problems 3-8, determine whether the given function is a solution to the given differential equation.

$\mathbf{y}\mathbf{=}\mathbf{sin} \mathbf{x}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}$, $\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dx}}^{\mathbf{2}}}\mathbf{+}\mathbf{y}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{2}$