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Mathematics LibreTexts

3.5E: Exersices

  • Page ID
    18792
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    In Exercises \((3.5E.1)\) to \((3.5E.6)\), find all solutions.

    Exercise \(\PageIndex{1}\)

    \(\displaystyle{y'={3x^2+2x+1\over y-2}}\)

    Answer

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    Exercise \(\PageIndex{2}\)

    \((\sin x)(\sin y)+(\cos y)y'=0\)

    Answer

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    Exercise \(\PageIndex{3}\)

    \(xy'+y^2+y=0\)

    Answer

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    Exercise \(\PageIndex{4}\)

    \(y' \ln |y|+x^2y= 0\)

    Answer

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    Exercise \(\PageIndex{5}\)

    \(\displaystyle{(3y^3+3y \cos y+1)y'+{(2x+1)y\over 1+x^2}=0}\)

    Answer

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    Exercise \(\PageIndex{6}\)

    \(x^2yy'=(y^2-1)^{3/2}\)

    Answer

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    In Exercises \((3.5E.7)\) to \((3.5E.10)\), find all solutions. Also, plot a direction field and some integral curves on the indicated rectangular region.

    Exercise \(\PageIndex{7}\)

    \(\displaystyle{y'=x^2(1+y^2)}; \; \{-1\le x\le1,\ -1\le y\le1\}\)

    Answer

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    Exercise \(\PageIndex{8}\)

    \( y'(1+x^2)+xy=0 ; \; \{-2\le x\le2,\ -1\le y\le1\}\)

    Answer

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    Exercise \(\PageIndex{9}\)

    \(y'=(x-1)(y-1)(y-2); \; \{-2\le x\le2,\ -3\le y\le3\}\)

    Answer

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    Exercise \(\PageIndex{10}\)

    \((y-1)^2y'=2x+3; \; \{-2\le x\le2,\ -2\le y\le5\}\)

    Answer

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    In Exercises \((3.5E.11)\) to \((3.5E.12)\), solve the initial value problem.

    Exercise \(\PageIndex{11}\)

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

    Answer

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    Exercise \(\PageIndex{12}\)

    \(y'+x(y^2+y)=0, \quad y(2)=1\)

    Answer

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    In Exercises \((3.5E.13)\) to \((3.5E.16)\), solve the initial value problem and graph the solution.

    Exercise \(\PageIndex{13}\)

    \((3y^2+4y)y'+2x+\cos x=0, \quad y(0)=1\)

    Answer

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    Exercise \(\PageIndex{14}\)

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

    Answer

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    Exercise \(\PageIndex{15}\)

    \(y'+2x(y+1)=0, \quad y(0)=2\)

    Answer

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    Exercise \(\PageIndex{16}\)

    \(y'=2xy(1+y^2),\quad y(0)=1\)

    Answer

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    In Exercises \((3.5E.17)\) to \((3.5E.23)\), solve the initial value problem and find the interval of validity of the solution.

    Exercise \(\PageIndex{17}\)

    \(y'(x^2+2)+ 4x(y^2+2y+1)=0, \quad y(1)=-1\)

    Answer

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    Exercise \(\PageIndex{18}\)

    \(y'=-2x(y^2-3y+2), \quad y(0)=3\)

    Answer

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    Exercise \(\PageIndex{19}\)

    \(\displaystyle{y'={2x\over 1+2y}, \quad y(2)=0}\)

    Answer

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    Exercise \(\PageIndex{20}\)

    \(y'=2y-y^2, \quad y(0)=1\)

    Answer

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    Exercise \(\PageIndex{21}\)

    \(x+yy'=0, \quad y(3) =-4\)

    Answer

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    Exercise \(\PageIndex{22}\)

    \(y'+x^2(y+1)(y-2)^2=0, \quad y(4)=2\)

    Answer

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    Exercise \(\PageIndex{23}\)

    \((x+1)(x-2)y'+y=0, \quad y(1)=-3\)

    Answer

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    Exercise \(\PageIndex{24}\)

    Solve \(\displaystyle{y'={(1+y^2) \over (1+x^2)}}\) explicitly.

    Hint: Use the identity \(\displaystyle{\tan(A+B)={\tan A+\tan B\over1-\tan A\tan B}}\).

    Answer

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    Exercise \(\PageIndex{25}\)

    Solve \(\displaystyle {y'\sqrt{1-x^2}+\sqrt{1-y^2}=0}\) explicitly.

    Hint: Use the identity \(\sin(A-B)=\sin A\cos B-\cos A\sin B\).

    Answer

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    Exercise \(\PageIndex{26}\)

    Solve \(\displaystyle{y'={\cos x\over \sin y},\quad y (\pi)={\pi\over2}}\) explicitly.

    Hint: Use the identity \(\cos(x+\pi/2)=-\sin x\) and the periodicity of the cosine.

    Answer

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    Exercise \(\PageIndex{27}\)

    Solve the initial value problem

    \begin{eqnarray*}
    y'=ay-by^2,\quad y(0)=y_0.
    \end{eqnarray*}

    Discuss the behavior of the solution if part (a) \(y_0\ge0\); part (b) \(y_0<0\).

    Answer

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    Exercise \(\PageIndex{28}\)

    The population \(P=P(t)\) of a species satisfies the logistic equation

    \begin{eqnarray*}
    P'=aP(1-\alpha P)
    \end{eqnarray*}

    and \(P(0)=P_0>0\). Find \(P\) for \(t>0\), and find \(\lim_{t\to\infty}P(t)\).

    Answer

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    Exercise \(\PageIndex{29}\)

    An epidemic spreads through a population at a rate proportional to the product of the number of people already infected and the number of people susceptible, but not yet infected. Therefore, if \(S\) denotes the total population of susceptible people and \(I=I(t)\) denotes the number of infected people at time \(t\), then

    \begin{eqnarray*}
    I'=rI(S-I),
    \end{eqnarray*}

    where \(r\) is a positive constant. Assuming that \(I(0)=I_0\), find \(I(t)\) for \(t>0\), and show that \(\lim_{t\to\infty}I(t)=S\).

    Answer

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    Exercise \(\PageIndex{30}\)

    The result of Exercise \((3.5E.29)\) is discouraging: if any susceptible member of the group is initially infected, then in the long run all susceptible members are infected! On a more hopeful note, suppose the disease spreads according to the model of Exercise \((3.5E.29)\), but there's a medication that cures the infected population at a rate proportional to the number of infected individuals. Now the equation for the number of infected individuals becomes

    \begin{equation} \label{eq:3.5E.1}
    I'=rI(S-I)-qI
    \end{equation}

    where \(q\) is a positive constant.

    (a) Choose \(r\) and \(S\) positive. By plotting direction fields and solutions of \eqref{eq:3.5E.1} on suitable rectangular grids

    \begin{eqnarray*}
    R=\{0\le t \le T,\ 0\le I \le d\}
    \end{eqnarray*}

    in the \((t,I)\)-plane, verify that if \(I\) is any solution of \eqref{eq:3.5E.1} such that \(I(0)>0\), then \(\lim_{t\to\infty}I(t)=S-q/r\) if \(q<rS\) and \(\lim_{t\to\infty}I(t)=0\) if \(q\ge rS\).

    (b) To verify the experimental results of part (a), use separation of variables to solve \eqref{eq:3.5E.1} with initial condition \(I(0)=I_0>0\), and find \(\lim_{t\to\infty}I(t)\).

    Hint: There are three cases to consider: part(i) \(q<rS\); part(ii) \(q>rS\); part(iii) \(q=rS\)

    Answer

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    Exercise \(\PageIndex{31}\)

    Consider the differential equation

    \begin{equation} \label{eq:3.5E.2}
    y'=ay-by^2-q,
    \end{equation}

    where \(a\), \(b\) are positive constants, and \(q\) is an arbitrary constant. Suppose \(y\) denotes a solution of this equation that satisfies the initial condition \(y(0)=y_0\).

    (a) Choose \(a\) and \(b\) positive and \(q<a^2/4b\). By plotting direction fields and solutions of \eqref{eq:3.5E.2} on suitable rectangular grids

    \begin{equation} \label{eq:3.5E.3}
    R=\{0\le t \le T,\ c\le y \le d\}
    \end{equation}

    in the \((t,y)\)-plane, discover that there are numbers \(y_1\) and \(y_2\) with \(y_1<y_2\) such that if \(y_0>y_1\) then \(\lim_{t\to\infty}y(t)=y_2\), and if \(y_0<y_1\) then \(y(t)=-\infty\) for some finite value of \(t\). (What happens if \(y_0=y_1\)?)

    (b) Choose \(a\) and \(b\) positive and \(q=a^2/4b\). By plotting direction fields and solutions of \eqref{eq:3.5E.2} on suitable rectangular grids of the form \eqref{eq:3.5E.3}, discover that there's a number \(y_1\) such that if \(y_0\ge y_1\) then \(\lim_{t\to\infty}y(t)=y_1\), while if \(y_0<y_1\) then \(y(t)=-\infty\) for some finite value of \(t\).

    (c) Choose positive \(a\), \(b\) and \(q>a^2/4b\). By plotting direction fields and solutions of \eqref{eq:3.5E.2} on suitable rectangular grids of the form \eqref{eq:3.5E.3}, discover that no matter what \(y_0\) is, \(y(t)=-\infty\) for some finite value of \(t\).

    (d) Verify your results experiments analytically. Start by separating variables in \eqref{eq:3.5E.2} to obtain

    \begin{eqnarray*}
    {y'\over ay-by^2-q}=1.
    \end{eqnarray*}

    To decide what to do next you'll have to use the quadratic formula. This should lead you to see why there are three cases. Take it from there!

    Because of its role in the transition between these three cases, \(q_0=a^2/4b\) is called a \({\color{blue}{\mbox{ bifurcation value}}\) of \(q\). In general, if \(q\) is a parameter in any differential equation, \(q_0\) is said to be a bifurcation value of \(q\) if the nature of the solutions of the equation with \(q<q_0\) is qualitatively different from the nature of the solutions with \(q>q_0\).

    Answer

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    Exercise \(\PageIndex{32}\)

    \item\label{exer:2.2.32} \Lex

    By plotting direction fields and solutions of

    $$

    y'=qy-y^3,

    $$

    convince yourself that $q_0=0$ is a bifurcation value of $q$

    for this equation. Explain what makes you draw this conclusion.

    Answer

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    Exercise \(\PageIndex{33}\)

    \item\label{exer:2.2.33}

    Suppose a disease spreads according to the model of

    Exercise~\ref{exer:2.2.29}, but

    there's a medication that cures the infected population at a constant

    rate of $q$ individuals per unit time, where $q>0$.

    Then

    the equation for the number of infected individuals becomes

    $$

    I'=rI(S-I)-q.

    $$

    Assuming that $I(0)=I_0>0$, use the results of

    Exercise~\ref{exer:2.2.31} to describe what happens as $t\to\infty$.

    Answer

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    Exercise \(\PageIndex{34}\)

    \item\label{exer:2.2.34}

    Assuming that $p \not\equiv 0$, state conditions under which the linear

    equation

    $$

    y'+p(x)y=f(x)

    $$

    is separable. If the equation satisfies these conditions, solve it by

    separation of variables and by the method developed in

    Section~2.1.

    Answer

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    Solve the equations in Exercises \((3.5E.35)\) to \((3.5E.38)\) using variation of parameters followed by separation of variables.

    Exercise \(\PageIndex{35}\)

    \begin{tabular}[t]{@{}p{168pt}@{}p{168pt}}

    \item\label{exer:2.2.35} $\dst{y'+y={2xe^{-x}\over1+ye^x}}$&

    Answer

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    Exercise \(\PageIndex{36}\)

    \item\label{exer:2.2.36}\vspace*{7pt} $\dst{xy'-2y={x^6\over y+x^2}}$

    \end{tabular}

    Answer

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    Exercise \(\PageIndex{37}\)

    \begin{tabular}[t]{@{}p{168pt}@{}p{168pt}}

    \item\label{exer:2.2.37} $\dst{y'-y}={(x+1)e^{4x}\over(y+e^x)^2}$&

    Answer

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    Exercise \(\PageIndex{38}\)

    \item\label{exer:2.2.38} $y'-2y=\dst{xe^{2x}\over1-ye^{-2x}}$

    \end{tabular}

    Answer

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    Exercise \(\PageIndex{39}\)

    \item\label{exer:2.2.39}

    Use variation of parameters to show that the solutions of the

    following equations are of the form $y=uy_1$, where $u$ satisfies a

    separable equation $u'=g(x)p(u)$. Find $y_1$ and $g$ for each

    equation.

    \begin{tabular}[t]{@{}p{168pt}@{}p{168pt}}

    {\bf (a)} $xy'+y=h(x)p(xy)$

    & {\bf (b)} $\dst{xy'-y=h(x) p\left({y\over x}\right)}$\\[2\jot]

    {\bf (c)} $y'+y=h(x) p(e^xy)$

    & {\bf (d)} $xy'+ry=h(x) p(x^ry)$\\[2\jot]

    {\bf (e)} $\dst{y'+{v'(x)\over v(x)}y= h(x)

    p\left(v(x)y\right)}$

    \end{tabular}

    \end{exerciselist}

    Answer

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