3.2E: Exercises

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Exercises

Exercise $\PageIndex{1}$

Find the general solution.

(a) $y'+ay=0$ ($a$=constant)

(b) $y'+3x^2y=0$

(d) $xy'+3y=0$

(e) $x^2y'+y=0$

Answer

(a) Since $y'/y=-a $, $y=e^{-ax+c}$.

(b)Since $y'/y=-3x^2 $, $y=e^{-x^3+c}$.

(c)Since $y'/y=\frac{-\ln(x)}{x} $, $y=e^{\frac{-1}{2}\ln^2(x)+c}$.

(d)Since $y'/y=\frac{-3}{x} $, $y=\frac{c}{x^3}$.

(e)Since $y'/y=\frac{-1}{x^2} $, $ y=e^{\frac{-1}{x}+c}$.

Exercise $\PageIndex{2}$

Solve the initial value problem.

(a) ${y'+\left({1+x\over x}\right)y=0,\quad y(1)=1}$

(b) ${xy'+\left(1+{1\over\ln x}\right)y=0,\quad y(e)=1}$

(d) ${y'-\left({2x\over 1+x^2}\right)y=0,\quad y(0)=2}$

(e) ${y'+{k\over x}y=0,\quad y(1)=3 \quad {(k=constant)}}$

(f) $ y'+(\tan kx)y=0,\quad y(0)=2 \quad {(k=constant)}$

Answer

(a) Since $y=\frac{e^{-x+c}}{x}$ and $ y(1)=1$, then $c=1$.

(b) Since $y=\frac{c}{x\ln(x)}$ and $ y(e)=1$, then $c=1$.

(d) Since $y=(1+x^2)e^c$ and $y(0)=2$, $c=\ln(2)$.

(e) Since $y=\frac{c}{x^k}$ and $y(1)=3$, $c=3$. Compare treatment of $c$ in (d) and (e). In (e) $c=e^{c_0}=3$ versus in (d) $e^c=2$.

(f) Since $y=e^c\cos^{\frac{1}{k}}(kx)$ and $y(0)=2$, $c=\ln(2)$.

Exercise $\PageIndex{3}$

Find the general solution. Also, plot a direction field and some integral curves on the rectangular region $\{-2\le x\le2,\ -2\le y\le2$\}.}

(a) $y'+3y=1$

(b) ${y'+\left({1\over x}-1\right)y=-{2\over x}}$

(d) ${y'+{2x\over1+x^2}y={e^{-x}\over1+x^2}}$

Answer

(a) $ y=\frac{1}{3}+\frac{c}{e^{3x}}$

$Fig E3.2.3a.png$

(b) $ y = \frac{2}{x}+\frac{ce^x}{x} $

$clipboard_e017c4699e9f186fe5b5aaf76718c6a4b.png$
Figure $\PageIndex{1}$: $y' + (\frac{1}{x}-1)y =-\frac{2}{x}$

Figure $\PageIndex{1}$: $y'+2xy=xe^{-x^2}$

(d) $ y= \frac{-1}{e^x(1+x^2)} + \frac{c}{1+x^2} $

$clipboard_eef9ae1ae06e674a55c0ee6b914d8353f.png$
Figure $\PageIndex{1}$: $y'+\frac{2x}{1+x^2}y=\frac{e^{-x}}{1+x^2}$

Exercise $\PageIndex{4}$

Find the general solution.

(a) ${y'+{1\over x}y={7\over x^2}+3}$

(b) ${y'+{4\over x-1}y = {1\over (x-1)^5}+{\sin x\over (x-1)^4}}$

(d) ${xy'+2y={2\over x^2}+1}$

(e) $y'+(\tan x)y=\cos x$

(f) ${(1+x)y'+2y={\sin x \over 1 + x}}$

(g) $(x-2)(x-1)y'-(4x-3)y=(x-2)^3$

(h) $y'+(2\sin x\cos x) y=e^{-\sin^2x}$

(i) $x^2y'+3xy=e^x$

Answer

(a) $ y = \frac{7\ln(x)}{x}+\frac{3x}{2}+\frac{c}{x}$

(e) $y = x\cos(x)+c\cos(x)$

(g) $y= \frac{-(x-2)^3}{4(x-1)}+\frac{c(x-2)^5}{x-1}$ Hint: Use partial fractions.

(i) $y=\frac{(x-1)e^x}{x^3}+\frac{c}{x^3}$

Exercise $\PageIndex{5}$

Solve the initial value problem and sketch the graph of the solution.

(a) $y'+7y=e^{3x},\quad y(0)=0$

(b) ${(1+x^2)y'+4xy={2\over 1+x^2},\quad y(0)=1}$

(d) ${y'+ (\cot x)y=\cos x,\quad y\left({\pi\over 2}\right)=1}$

(e) ${y'+{1\over x}y={2\over x^2}+1,\quad y(-1)=0}$

Answer

(a) $y=\frac{e^{3x}}{10}-\frac{1}{10e^{3x}} $

$Fig E3.2.5a.png$

$Fig E3.2.5c.png$

(e) $y=\frac{\ln(x^2)}{x}+1+\frac{1}{x}$

$Fig E3.2.5e.png$

Exercise $\PageIndex{6}$

Solve the initial value problem.

(a) ${(x-1)y'+3y={1\over (x-1)^3} + {\sin x\over (x-1)^2},\quad y(0)=1}$

(b) $xy'+2y=8x^2,\quad y(1)=3$

(d) $y'+2xy=x,\quad y(0)=3$

(e) ${(x-1)y'+3y={1+(x-1)\sec^2x\over (x-1)^3},\quad y(0)=-1}$

(f) ${(x+2)y'+4y={1+2x^2\over x(x+2)^3},\quad y(-1)=2}$

(g) $(x^2-1)y'-2xy=x(x^2-1),\quad y(0)=4$

(h) $(x^2-5)y'-2xy=-2x(x^2-5),\quad y(2)=7$

Answer

(a) $y=\frac{ln|x-1|}{(x-1)^3}-\frac{\cos(x)}{(x-1)^3}+\frac{c}{(x-1)^3}$ thus $c=0$.

(e) $y= \frac{x-\tan(x)+x\tan(x)+\ln|\cos(x)|+c}{(x-1)^3}$ thus $c=0$.

(f) $y= \frac{\ln|x|+x^2+c}{(x+2)^4}$ thus $c=1$.

(h) $y= (-\ln(x^2-5)+c)(x^2-5)$ thus $c= -7$.

Exercise $\PageIndex{7}$

Solve the initial value problem and leave the answer in a form involving a definite integral.

You can solve these problems numerically by methods discussed in Chapter~3.

(a) $y'+2xy=x^2,\quad y(0)=3$

(b) ${y'+{1\over x}y={\sin x\over x^2},\quad y(1)=2}$

(d) ${y'+{2x\over 1+x^2}y={e^x\over (1+x^2)^2}, \quad y(0)=1}$

(e) $xy'+(x+1)y=e^{x^2},\quad y(1)=2$

Answer: (a) $y=(\frac{xe^{x^2}}{2}-\frac{1}{2}\int_0^{x^2} {e^t \, dt}+c)e^{x^{-2}}$, thus $c=4$.

Exercise $\PageIndex{8}$

Experiments indicate that glucose is absorbed by the body at a rate proportional to the amount of glucose present in the bloodstream. Let $\lambda$ denote the (positive) constant of proportionality. Now suppose glucose is injected into a patient's bloodstream at a constant rate of $r$ units per unit of time. Let $G=G(t)$ be the number of units in the patient's bloodstream at time $t>0$. Then $G'=-\lambda G+r,$ where the first term on the right is due to the absorption of the glucose by the patient's body and the second term is due to the injection. Determine $G$ for $t>0$, given that $G(0)=G_0$. Also, find $\lim_{t\to\infty}G(t)$.

Answer: Add texts here. Do not delete this text first.

Exercise $\PageIndex{9}$

(a) Plot a direction field and some integral curves for equation A: $xy'-2y=-1$ on the rectangular region $\{-1\le x\le 1, -.5\le y\le 1.5\}$. What do all the integral curves have in common?

(b) Show that the general solution of (A) on $(-\infty,0)$ and $(0,\infty)$ is $y={1\over2}+cx^2.$

$y=1 \over 2+c_1x^2, x\ge 0$ and $1\over 2+c_2x^2, x < 0,$.

where $c_1$ and $c_2$ are arbitrary constants.

(d) Conclude from \part{c} that all solutions of (A) on $(-\infty,\infty)$ are solutions of the initial value problem $xy'-2y=-1,\quad y(0)={1\over2}.$

(e) Use part b to show that if $x_0\ne0$ and $y_0$ is arbitrary, then the initial value problem $xy'-2y=-1,\quad y(x_0)=y_0$ has infinitely many solutions on $(-\infty,\infty)$. Explain why this doesn't contradict Theorem~\ref{thmtype:3.3.1} \part{b}.

Answer: Add texts here. Do not delete this text first.

Exercise $\PageIndex{10}$

Suppose $f$ is continuous on an open interval $(a,b)$ and $\alpha$ is a constant.

(a) Derive a formula for the solution of the initial value problem $y'+\alpha y=f(x),\quad y(x_0)=y_0, $ where $x_0$ is in $(a,b)$ and $y_0$ is an arbitrary real number.

(b) Suppose $(a,b)=(a,\infty)$, $\alpha > 0$ and $\displaystyle{\lim_{x\to\infty} f(x)=L}$. Show that if $y$ is the solution of (a), then

$\displaystyle{\lim_{x\to \infty} y(x)=L/\alpha}$.

Answer

Add texts here. Do not delete this text first.

Exercise $\PageIndex{11}$

Assume that all functions in this exercise are defined on a common interval $(a,b)$

(a) Prove: If $y_1$ and $y_2$ are solutions of $y'+p(x)y=f_1(x)$ and $y'+p(x)y=f_2(x)$ respectively, and $c_1$ and $c_2$ are constants, then $y=c_1y_1+c_2y_2$ is a solution of $y'+p(x)y=c_1f_1(x)+c_2f_2(x).$

(This is the principle of superposition)

(b) Use (a) to show that if $y_1$ and $y_2$ are solutions of the nonhomogeneous equation $y'+p(x)y=f(x),$ then $y_1-y_2$ is a solution of the homogeneous equation B: $y'+p(x)y=0$.

(c) Use (a) to show that if $y_1$ is a solution of (A) and $y_2$ is a solution of (B), then $y_1+y_2$ is a solution of (A).

Answer

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Exercise $\PageIndex{12}$

Some nonlinear equations can be transformed into linear equations by changing the dependent variable. Show that if $g'(y)y'+p(x)g(y)=f(x)$ where $y$ is a function of $x$ and $g$ is a function of $y$, then the new dependent variable $z=g(y)$ satisfies the linear equation $z'+p(x)z=f(x).$

Answer: Add texts here. Do not delete this text first.

Exercise $\PageIndex{13}$

We've shown that if $p$ and $f$ are continuous on $(a,b)$ then every solution of equation A: $y'+p(x)y=f(x)$ on $(a,b)$ can be written as $y=uy_1$, where $y_1$ is a nontrivial solution of the complementary equation for (A) and $u'=f/y_1$. Now suppose $f$, $f'$, \dots, $f^{(m)}$ and $p$, $p'$, \dots, $p^{(m-1)}$ are continuous on $(a,b)$, where $m$ is a positive integer, and define \begin{eqnarray*} f_0&=&f,\\ f_j&=&f_{j-1}'+pf_{j-1},\quad 1\le j\le m. \end{eqnarray*}

Show that $u^{(j+1)}={f_j\over y_1},\quad 0\le j\le m.$

Answer: Add texts here. Do not delete this text first.