# 2.2E: Separable Equations (Exercises)

- Page ID
- 18247

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**In Exercises 1 -6 find all solutions.**

[exer:2.2.1] \( {y'={3x^2+2x+1\over y-2}}\)&

[exer:2.2.2] \((\sin x)(\sin y)+(\cos y)y'=0\)

[exer:2.2.3] \(xy'+y^2+y=0\) &

[exer:2.2.4] \(y' \ln |y|+x^2y= 0\)

[exer:2.2.5] \( {(3y^3+3y \cos y+1)y'+{(2x+1)y\over 1+x^2}=0}\)

[exer:2.2.6] \(x^2yy'=(y^2-1)^{3/2}\)

**In Exercises 7–10 find all solutions. Also, plot a direction field and some integral curves on the indicated rectangular region.**

[exer:2.2.7] \( {y'=x^2(1+y^2)}; \; \{-1\le x\le1,\ -1\le y\le1\}\)

[exer:2.2.8] \(y'(1+x^2)+xy=0 ; \; \{-2\le x\le2,\ -1\le y\le1\}\)

[exer:2.2.9] \(y'=(x-1)(y-1)(y-2); \; \{-2\le x\le2,\ -3\le y\le3\}\)

[exer:2.2.10] \((y-1)^2y'=2x+3; \; \{-2\le x\le2,\ -2\le y\le5\}\)

**In Exercises 11 and 12 solve the initial value problem.**

[exer:2.2.11] \( {y'={x^2+3x+2\over y-2}, \quad y(1)=4}\)

[exer:2.2.12] \(y'+x(y^2+y)=0, \quad y(2)=1\)

**In Exercises 13-16 solve the initial value problem and graph the solution.**

[exer:2.2.13] \((3y^2+4y)y'+2x+\cos x=0, \quad y(0)=1\)

[exer:2.2.14] \( {y'+{(y+1)(y-1)(y-2)\over x+1}=0, \quad y(1)=0}\)

[exer:2.2.15] \(y'+2x(y+1)=0, \quad y(0)=2\)

[exer:2.2.16] \(y'=2xy(1+y^2),\quad y(0)=1\)

**In Exercises 17–23 solve the initial value problem and find the interval of validity of the solution.**

[exer:2.2.17] \(y'(x^2+2)+ 4x(y^2+2y+1)=0, \quad y(1)=-1\)

[exer:2.2.18] \(y'=-2x(y^2-3y+2), \quad y(0)=3\)

[exer:2.2.19] \( {y'={2x\over 1+2y}, \quad y(2)=0}\) &

[exer:2.2.20] \(y'=2y-y^2, \quad y(0)=1\)

[exer:2.2.21] \(x+yy'=0, \quad y(3) =-4\)

[exer:2.2.22] \(y'+x^2(y+1)(y-2)^2=0, \quad y(4)=2\)

[exer:2.2.23] \((x+1)(x-2)y'+y=0, \quad y(1)=-3\)

[exer:2.2.24] Solve \( {y'={(1+y^2) \over (1+x^2)}}\) explicitly.

[exer:2.2.25] Solve \( {y'\sqrt{1-x^2}+\sqrt{1-y^2}=0}\) explicitly.

[exer:2.2.26] Solve \( {y'={\cos x\over \sin y},\quad y (\pi)={\pi\over2}}\) explicitly.

[exer:2.2.27] Solve the initial value problem \[y'=ay-by^2,\quad y(0)=y_0.\] Discuss the behavior of the solution if a \(y_0\ge0\); b \(y_0<0\).

[exer:2.2.28] The population \(P=P(t)\) of a species satisfies the logistic equation \[P'=aP(1-\alpha P)\] and \(P(0)=P_0>0\). Find \(P\) for \(t>0\), and find \(\lim_{t\to\infty}P(t)\).

[exer:2.2.29] 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 \[I'=rI(S-I),\] where \(r\) is a positive constant. Assuming that \(I(0)=I_0\), find \(I(t)\) for \(t>0\), and show that \(\lim_{t\to\infty}I(t)=S\).

[exer:2.2.30] The result of Exercise [exer:2.2.29} 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 [exer:2.2.29}, 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 \[I'=rI(S-I)-qI \tag{A}\] where \(q\) is a positive constant.

- Choose \(r\) and \(S\) positive. By plotting direction fields and solutions of (A) on suitable rectangular grids \[R=\{0\le t \le T,\ 0\le I \le d\}\] in the \((t,I)\)-plane, verify that if \(I\) is any solution of (A) 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 \(q\ge rS\).
- To verify the experimental results of (a), use separation of variables to solve (A) with initial condition \(I(0)=I_0>0\), and find \(\lim_{t\to\infty}I(t)\).

[exer:2.2.31] Consider the differential equation \[y'=ay-by^2-q, \tag{A} \] 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 \(y(0)=y_0\).

- Choose \(a\) and \(b\) positive and \(q<a^2/4b\). By plotting direction fields and solutions of (A) on suitable rectangular grids \[R=\{0\le t \le T,\ c\le y \le d\} \tag{B]\] 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 \(y_0=y_1\)?)
- Choose \(a\) and \(b\) positive and \(q=a^2/4b\). By plotting direction fields and solutions of (A) on suitable rectangular grids of the form (B), 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 \(t\).
- Choose positive \(a\), \(b\) and \(q>a^2/4b\). By plotting direction fields and solutions of (A) on suitable rectangular grids of the form (B), discover that no matter what \(y_0\) is, \(y(t)=-\infty\) for some finite value of \(t\).
- Verify your results experiments analytically. Start by separating variables in (A) to obtain \[{y'\over ay-by^2-q}=1.\] 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
*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 \(q>q_0\).

[exer:2.2.32] 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.

[exer:2.2.33] Suppose a disease spreads according to the model of Exercise [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 [exer:2.2.31} to describe what happens as \(t\to\infty\).

[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.

[exer:2.2.35] \( {y'+y={2xe^{-x}\over1+ye^x}}\)&

[exer:2.2.36] \( {xy'-2y={x^6\over y+x^2}}\)

[exer:2.2.37] \( {y'-y}={(x+1)e^{4x}\over(y+e^x)^2}\)&

[exer:2.2.38] \(y'-2y= {xe^{2x}\over1-ye^{-2x}}\)

[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.

- \(xy'+y=h(x)p(xy)\)
- \( {xy'-y=h(x) p\left({y\over x}\right)}\)
- \(y'+y=h(x) p(e^xy)\)
- \(xy'+ry=h(x) p(x^ry)\)
- \( {y'+{v'(x)\over v(x)}y= h(x) p\left(v(x)y\right)}\)