Detailed solutions

Compare reasoning, not handwriting

Use these solutions after an honest attempt. Put your work beside the solution and locate the first line where the decisions differ. A different valid method is not an error; explain why it works and compare its cost.

Chapter 1

Solution 1.1

Rain contributes $+12$ m³/hour. Leakage contributes $-0.04W$ m³/hour because $0.04$ has units 1/hour and $W$ has units m³. Therefore

$$W'=12-0.04W.$$

Both right-side terms and $W'$ have units m³/hour. The zero-rate volume is $W=300$ m³, which also gives a behavioral check.

Solution 1.2

For $y=2+3e^{-t}$, differentiate: $y'=-3e^{-t}$. Substitute into the right side: $2-y=2-(2+3e^{-t})=-3e^{-t}=y'$. The condition is $y(0)=2+3=5$. The curve decreases toward the stable level 2, matching the sign of $2-y$ above 2.

Solution 1.3

The rate factors as $0.4N(1-N/50)$. At $N=20$, both factors are positive, so the state rises. At $N=50$, the second factor is zero, so 50 is an equilibrium. At $N=80$, the second factor is negative while $N$ is positive, so the state falls. The equation points positive populations toward 50.

Solution 1.4

Salt enters at $(3\text{ L/min})(0.08\text{ kg/L})=0.24$ kg/min. It leaves at $(3\text{ L/min})(S/200\text{ kg/L})=0.015S$ kg/min. Thus

$$S'=0.24-0.015S,\qquad S(0)=10.$$

At zero rate, $0.24=0.015S$, so the model suggests a long-run amount of $S=16$ kg.

Solution 1.5

1. The coefficient $0.2$ has units $1/\text{hour}$, so $D'(0)=-0.2(10)=-2$ g/hour.
2. Differentiate the proposed function: $D'=-2e^{-0.2t}=-0.2D$. Also $D(0)=10$, and its positive decreasing graph matches removal.
3. After one hour, $D(1)=10e^{-0.2}\approx8.1873$ g. The fraction lost is $1-e^{-0.2}\approx0.18127$, or about 18.13%, not 20%.
4. Exact 20% loss at each hourly checkpoint instead gives $D_n=10(0.8)^n$. A continuous exponential matching those checkpoints would use $D'=(\ln0.8)D\approx-0.22314D$. An instantaneous percentage rate and a finite-interval percentage loss are different statements.

Solution 1.6

1. Net volume inflow is $2-1=1$ L/min, so $V(t)=100+t$ L for $t\ge0$, as long as the tank does not overflow.
2. Salt enters at $(2\text{ L/min})(3\text{ g/L})=6$ g/min.
3. Well-mixed outflow has concentration $S(t)/(100+t)$ g/L. At 1 L/min, it removes $S(t)/(100+t)$ g/min. Therefore

$$S'=6-\frac{S}{100+t},\qquad S(0)=0.$$

Initially, $S'(0)=6$ g/min because there is no salt to remove yet. Both the amount and volume vary, so replacing the outflow concentration by $S/100$ would use the wrong volume. Exercise 5.6 solves this model.

Chapter 2

Solution 2.1

The equilibrium is $y=3$. Below 3, $y-3\lt 0$, so solutions fall; above 3, they rise. Arrows point away from 3, so it is unstable. The initial values 1 and 5 move down and up, respectively, while 3 remains constant.

Solution 2.2

For $y'=\cos t$, each vertical column has a common slope because only time matters. For $y'=\cos y$, each horizontal row has a common slope because only state matters. For $y'=\cos(t+y)$, equal slopes lie on diagonal bands where $t+y$ is constant.

Solution 2.3

Between 0 and 1, $y'=y(1-y)\gt 0$, so the solution rises toward the stable equilibrium 1. Differentiate along a solution:

$$y''=y'(1-2y).$$

While $0\lt y\lt 1/2$, both factors are positive, so the graph is concave up. After $y$ passes $1/2$, $1-2y\lt 0$, so it is concave down. The growth rate peaks at the inflection state $y=1/2$.

Solution 2.4

For $y=0$, both $y'$ and $\sqrt y$ are zero. For $y=t^2/4$ with $t\ge0$, $y'=t/2$ and $\sqrt y=t/2$. Both satisfy $y(0)=0$. The usual no-crossing/unique-future rule fails because $\sqrt y$ is not Lipschitz in $y$ at zero. In fact, solutions may wait at zero before departing.

Solution 2.5

The slope vanishes where $t-y=0$, or $y=t$. Below this line the rate is positive; above it the rate is negative. But the candidate $y=t$ has derivative 1 while the equation demands 0 along the line, so the zero-slope line is not a solution.

At $(0,0)$, $y'(0)=0$. Differentiating the equation along the solution gives $y''=1-y'$, hence $y''(0)=1$. The curve has a local minimum at zero and rises for sufficiently small positive time. As an optional check, $y=t-1+e^{-t}$ has the correct initial condition and derivative, with expansion $y=t^2/2+O(t^3)$.

Solution 2.6

Differentiate along a solution, remembering that $y$ also depends on $t$:

$$y''=2t-y'=2t-t^2+y.$$

At $(1,1)$, $y'=0$ and $y''=2\gt 0$, so this solution has a strict local minimum at $t=1$. At $(-1,1)$, $y'=0$ and $y''=-2\lt 0$, so that solution has a strict local maximum at $t=-1$. The smooth rate rule guarantees local solutions through both points. A horizontal field mark reports only the first derivative; the second derivative distinguishes these two turning behaviors.

Chapter 3

Solution 3.1

(a) $y'=t^2/y$ is separable: $y,dy=t^2dt$. (b) $y'+e^t y=1$ is first-order linear with $p=e^t$. (c) $y'=y\sin y$ is autonomous and separable; draw the phase line first. (d) $y''+9y=0$ is homogeneous second-order constant coefficient, so use characteristic roots.

Solution 3.2

Factor $y'=ty+t=t(y+1)$, so $dy/(y+1)=t,dt$: it is separable. The equation $y'=ty+1$ gives $y'-ty=1$, but the right side $ty+1$ does not factor into a time-only and state-only product. Both equations are first-order linear; separation is the shorter option for the first.

Solution 3.3

On $t\gt 0$, $y'=y/t$ is separable because $dy/y=dt/t$. It is also homogeneous first-order linear: $y'-(1/t)y=0$. Separation uses less setup. Both produce $y=Ct$.

Solution 3.4

The sum $y^2+t$ cannot be factored as $a(t)b(y)$, so it is not separable. The squared state violates the linear form $y'+p(t)y=q(t)$. Useful next steps include drawing a direction field, computing a numerical solution for specified initial data, deriving local Taylor coefficients, and proving qualitative bounds or monotonicity.

Solution 3.5

For $t\ne2$, divide by $t-2$:

$$y'+\frac{3}{t-2}y=t-2.$$

This is first-order linear with $p(t)=3/(t-2)$ and $q(t)=t-2$. The maximal coefficient-continuity interval containing 3 is $(2,\infty)$; the other interval is $(-\infty,2)$. The next move would be an integrating factor, $\mu=|t-2|^3$.

The original equation still makes an algebraic demand at $t=2$: $3y(2)=0$, so a differentiable extension must have $y(2)=0$. This is a necessary condition, not a guarantee that every branch extends. Dividing by a coefficient that can vanish requires an interval check.

Solution 3.6

The state factor is $y(1+ty)$, whose second factor mixes time and state; the $y^2$ term also rules out linearity in $y$. First record the equilibrium $y=0$. On a branch where $y\ne0$, set $v=1/y$. The chain rule gives

$$v'=-\frac{y'}{y^2}=-\frac{y+ty^2}{y^2}=-v-t,$$

so $v'+v=-t$, a first-order linear equation with integrating factor $e^t$. Linearity depends on the unknown: the transformed equation is linear in $v$, while the original remains nonlinear in $y$. Recovering $y=1/v$ requires $v\ne0$, and the separate equilibrium $y=0$ cannot be recovered by taking a finite reciprocal.

Chapter 4

Solution 4.1

Separate and integrate:

$$y,dy=3t^2dt,\qquad \frac{y^2}{2}=t^3+C.$$

The condition gives $C=2$, so $y^2=2t^3+4$. Since $y(0)=2\gt 0$, select

$$y=\sqrt{2t^3+4}.$$

Differentiation gives $y'=3t^2/y$. The real solution interval containing zero requires $2t^3+4\gt 0$, so $t\gt -\sqrt[3]{2}$.

Solution 4.2

The division safeguard records equilibria $y=0$ and $y=2$. For other solutions,

$$\int\frac{dy}{y(y-2)}=t+C \quad\Rightarrow\quad \frac12\ln\left|\frac{y-2}{y}\right|=t+C.$$

Exponentiating with a signed nonzero constant gives $(y-2)/y=Ce^{2t}$, hence the non-equilibrium family $y=2/(1-Ce^{2t})$, $C\ne0$, on any interval where the denominator is nonzero. The equilibria $y=0$ and $y=2$ complete the solution set. Using $y(0)=1$ gives $C=-1$, hence

$$y=\frac{2}{1+e^{2t}}.$$

This solution remains between the two equilibria and decreases toward zero, matching the phase line.

Solution 4.3

The equilibria are 0 and 500. Partial fractions or the standard logistic integration gives

$$P=\frac{500}{1+Ae^{-0.6t}}.$$

From $P(0)=50$, $50=500/(1+A)$, so $A=9$:

$$P(t)=\frac{500}{1+9e^{-0.6t}}.$$

The rate is positive between 0 and 500 and negative above 500, so positive nonzero solutions approach 500.

Solution 4.4

Integrating $dy/(1+y^2)=dt$ gives $\arctan y=t+C$. The condition $y(0)=1$ gives $C=\pi/4$, so

$$y=\tan(t+\pi/4).$$

The nearest tangent poles occur when $t+\pi/4=-\pi/2$ and $t+\pi/4=\pi/2$. The maximal interval containing zero is

$$(-3\pi/4,\ \pi/4).$$

Solution 4.5

Record $y=0$ before division; it does not satisfy the initial condition. For the negative branch, separate and integrate:

$$\frac{dy}{y^2}=2t,dt,\qquad -\frac1y=t^2+C.$$

Using $y(0)=-2$ gives $C=1/2$, hence

$$y=-\frac{2}{1+2t^2}.$$

The denominator is positive for every real $t$, so the maximal interval is $(-\infty,\infty)$. Differentiation gives $y'=8t/(1+2t^2)^2=2ty^2$, and $y(0)=-2$. The curve decreases before zero and increases after zero, but never crosses $y=0$: a change in derivative sign changes direction, not necessarily the sign of the state.

Solution 4.6

The equilibrium $y=0$ is excluded by $y(0)=4$. On the positive branch,

$$\int y^{-3/2},dy=\int dt \quad\Rightarrow\quad -\frac2{\sqrt y}=t+C.$$

The initial condition gives $C=-1$, so $2/\sqrt y=1-t$. The left side is positive, which requires $t\lt 1$. Therefore

$$y=\frac4{(1-t)^2},\qquad -\infty<t<1.$$

On this interval, $y'=8/(1-t)^3=y^{3/2}$, and $y(0)=4$. The solution blows up as $t\to1^-$. Although the rational expression is also finite for $t\gt 1$, there its derivative is negative while $y^{3/2}$ is positive. Squaring without preserving the branch inequality would introduce a false continuation.

Chapter 5

Solution 5.1

The zero-rate state in $y'=3y+6=3(y+2)$ is $-2$, and arrows point away, so it is unstable. The integrating factor is $e^{-3t}$, giving

$$y=-2+Ce^{3t}.$$

From $y(0)=1$, $C=3$, so $y=-2+3e^{3t}$. It grows away from $-2$, as predicted.

Solution 5.2

Here $p(t)=2/t$, so on $t\gt 0$, $\mu=t^2$. Then

$$(t^2y)'=t^3.$$

Integration gives $t^2y=t^4/4+C$, or $y=t^2/4+C/t^2$. The condition $y(1)=0$ gives $C=-1/4$:

$$y=\frac{t^2}{4}-\frac{1}{4t^2}.$$

Solution 5.3

Rewrite as $T'+0.25T=5$, or read the equilibrium directly as 20. The solution is

$$T=20+60e^{-0.25t}.$$

Set $T=30$: $10=60e^{-0.25t}$, so $e^{-0.25t}=1/6$ and

$$t=4\ln6\approx7.17\text{ time units}.$$

Solution 5.4

The transient $Ce^{-t}$ is annihilated by the operator $D+1$. Apply it to the proposed family:

$$y'+y=(2t-Ce^{-t})+(t^2+Ce^{-t})=t^2+2t.$$

Thus one valid equation is

$$y'+y=t^2+2t.$$

The polynomial $t^2$ is the forced response; $Ce^{-t}$ is the decaying transient.

Solution 5.5

The integrating factor is $e^{2t}$, so

$$(e^{2t}y)'=e^t,\qquad e^{2t}y=e^t+C.$$

Thus $y=e^{-t}+Ce^{-2t}$. The initial condition gives $C=-1$, hence $y=e^{-t}-e^{-2t}$. Its derivative is $-e^{-t}+2e^{-2t}$, which is positive before $t=\ln2$ and negative afterward. On $t\ge0$, the maximum is therefore

$$y(\ln2)=\frac12-\frac14=\frac14.$$

Finally, $y'+2y=e^{-t}$, and $y(0)=0$. The solution first builds up under the input, then falls to zero as the input fades.

Solution 5.6

On $t\ge0$, the integrating factor is $\mu=100+t$. Multiplying every term gives

$$[(100+t)S]'=6(100+t).$$

Integration yields $(100+t)S=3(100+t)^2+C$. At $t=0$, $S=0$, so $C=-30000$, and

$$S=3(100+t)-\frac{30000}{100+t}.$$

Consequently the concentration is

$$\frac{S}{100+t}=3-\frac{30000}{(100+t)^2}\longrightarrow3\text{ g/L}.$$

The concentration approaches that of the inflow, but volume grows without bound in this idealized no-overflow model, so the amount grows without a finite limit. As a check, $S'=3+30000/(100+t)^2=6-S/(100+t)$, and $S'(0)=6$ g/min, as predicted in Exercise 1.6.

Chapter 6

Solution 6.1

The equilibria are 0, 1, and 3. The sign pattern of $y(y-1)(y-3)$, from left to right, is $-,+,-,+$. Arrows point away from 0, toward 1, and away from 3. Thus 0 and 3 are unstable, while 1 is stable.

Solution 6.2

The phase line gives: $y(0)=-1$ decreases away below 0; $y(0)=1/2$ rises toward 1; $y(0)=2$ falls toward 1; and $y(0)=4$ rises away above 3. The two inner solutions tend to 1 as $t\to\infty$. The outer solutions instead become unbounded in finite positive time: separation expresses the time to reach a state as an integral of $1/[y(y-1)(y-3)]$, and the integral to either infinite endpoint converges because its magnitude behaves like $1/|y|^3$. Thus their arrows must not be interpreted as solutions that remain finite for every positive time.

Solution 6.3

Treat $P'=rP-(r/K)P^2$ as a function of state. Differentiate with respect to $P$:

$$\frac{dP'}{dP}=r-\frac{2r}{K}P.$$

With $r,K\gt 0$, the critical state is $P=K/2$, and the downward-opening parabola makes it a maximum. The maximum rate is $rK/4$. The carrying-capacity population is the equilibrium $K$; these have different units. A population may start above $K$, but then its rate is negative, so $K$ is not an absolute upper bound on every initial state.

Solution 6.4

For $a\gt 0$, equilibria are $\pm\sqrt a$. The rate is positive outside them and negative between them, so $-\sqrt a$ is stable and $+\sqrt a$ is unstable. For $a=0$, $y'=y^2\ge0$; zero is semistable, attracting from below and repelling above. For $a\lt 0$, $y'=y^2+|a|\gt 0$, so no equilibria exist and every solution rises on its interval of existence.

Solution 6.5

The only equilibrium is $y=2$. Below it, $(y-2)^3\lt 0$, so the rate is positive. Above it, the rate is negative. Both arrows point inward: 2 is asymptotically stable.

Yet $f'(y)=-3(y-2)^2$, so $f'(2)=0$. A zero derivative makes the first-derivative test inconclusive, not unstable. Starting at 1, the solution rises toward 2; starting at 3, it falls toward 2. Smoothness prevents crossing the equilibrium. Each trajectory is monotone and bounded, and any limit different from 2 would retain a nonzero rate, so both limits must equal 2.

Solution 6.6

Complete the square in the rate:

$$P'=2-H-\frac{(P-4)^2}{8}.$$

For $0\le H\lt 2$, the two equilibria are

$$P_-=4-\sqrt{16-8H},\qquad P_+=4+\sqrt{16-8H}.$$

The rate is positive between them and negative outside. Thus $P_-$ is unstable and $P_+$ is asymptotically stable. For $0\lt H\lt 2$, a starting population between 0 and $P_-$ declines to extinction, while a starting population above $P_-$ approaches $P_+$. The lower equilibrium is a survival threshold. At $H=0$, it is zero itself, and positive populations instead approach 8.

At $H=2$, $P'=-(P-4)^2/8$: the double equilibrium 4 attracts from above and repels from below, so it is semistable. At $H\gt 2$, the rate is strictly negative everywhere and there are no equilibria; positive populations reach extinction. Because a constant harvest would otherwise drive $P$ below zero, the physical model stops at extinction. The threshold $H=2$ equals the unharvested model's maximum growth rate.

Chapter 7

Solution 7.1

The characteristic equation $r^2-r-6=(r-3)(r+2)=0$ gives roots 3 and $-2$. Thus

$$y=C_1e^{3t}+C_2e^{-2t}.$$

The conditions give $C_1+C_2=1$ and $3C_1-2C_2=0$, hence $C_1=2/5$, $C_2=3/5$. The growing $e^{3t}$ mode dominates for large positive time.

Solution 7.2

The characteristic polynomial is $(r+3)^2$, so

$$y=(C_1+C_2t)e^{-3t}.$$

The first condition gives $C_1=0$. Differentiating then evaluating at zero gives $C_2=2$. Therefore $y=2te^{-3t}$.

Solution 7.3

The roots are $-\zeta\pm\sqrt{\zeta^2-1}$. At $\zeta=0$, they are pure imaginary and the phase portrait is a center. At $1/2$, they are complex with negative real part, producing a spiral sink. At 1, the negative root repeats, giving critical damping and a degenerate stable node. At 2, both roots are real and negative, giving an overdamped stable node.

Solution 7.4

Envelope $e^{-t}$ requires real part $-1$. Period $\pi$ requires angular frequency $2$, since $2\pi/\beta=\pi$. Desired roots are $-1\pm2i$. Their polynomial is

$$(r+1)^2+4=r^2+2r+5.$$

Choose $a=2$, $b=5$.

Solution 7.5

The roots are $\pm2i$, so $x=C_1\cos2t+C_2\sin2t$. Position gives $C_1=1$, and velocity gives $2C_2=2$, hence

$$x=\cos2t+\sin2t=\sqrt2\cos(2t-\pi/4).$$

The amplitude is $\sqrt2$ and the period is $2\pi/2=\pi$. For $E=(x')^2/2+2x^2$, differentiate:

$$E'=x'x''+4xx'=x'(x''+4x)=0.$$

Initially $E=2^2/2+2(1)^2=4$, so $E=4$ at every time. This gives an independent physical check: the undamped oscillator exchanges kinetic and potential energy without losing their sum.

Solution 7.6

The characteristic equation is $r(r+2)=0$. Therefore

$$y=C_1+C_2e^{-2t},\qquad y'=-2C_2e^{-2t}.$$

Velocity gives $-2C_2=-4$, so $C_2=2$. Position then gives $C_1=-1$, hence $y=-1+2e^{-2t}$ and $\lim_{t\to\infty}y=-1$. Direct substitution gives $y''+2y'=8e^{-2t}-8e^{-2t}=0$.

The root 0 supplies the constant mode $e^{0t}=1$, while the root $-2$ supplies a decaying mode. Undamped oscillation needs a nonzero imaginary pair; a real zero root alone gives no rotation or periodic motion.

Chapter 8

Solution 8.1

Natural roots are $\pm i$, so $y_h=C_1\cos t+C_2\sin t$. For the polynomial input $t$, try $y_p=At+B$. Since $y_p''=0$, matching $At+B=t$ gives $A=1,B=0$. Thus

$$y=C_1\cos t+C_2\sin t+t.$$

Solution 8.2

The homogeneous solution is $C_1\cos2t+C_2\sin2t$. Try $y_p=A\cos t+B\sin t$; substitution gives $3A=1$, $3B=0$. Thus $y_p=\frac13\cos t$. The conditions give $C_1=-1/3$, $C_2=0$, so

$$y=\frac{\cos t-\cos2t}{3}.$$

Solution 8.3

(a) $Ae^{-t}$; it does not collide with the natural modes. (b) $t(A\cos2t+B\sin2t)$; frequency 2 is the natural frequency, so multiply once by $t$. (c) $At^2+Bt+C$; use the complete polynomial family.

Solution 8.4

For $y''+\omega_0^2y=F\cos(\omega t)$ with zero initial position and velocity, $\omega_0\gt 0$, and $\omega\ne\omega_0$, the response is

$$y=\frac{F[\cos(\omega t)-\cos(\omega_0t)]}{\omega_0^2-\omega^2}.$$

The identity rewrites the numerator as a fast oscillation with average angular frequency $(\omega+\omega_0)/2$, multiplied by $-2\sin((\omega-\omega_0)t/2)$. Close frequencies produce a slowly varying amplitude, or beats. In the limit $\omega\to\omega_0$, both numerator and denominator vanish; their ratio tends to $Ft\sin(\omega_0t)/(2\omega_0)$. It is this full response, not the numerator alone, whose resonant envelope grows linearly.

Solution 8.5

The homogeneous characteristic polynomial is $(r-1)^2$, so both $e^t$ and $te^t$ are natural modes. The forcing trial must therefore be $At^2e^t$, not $Ae^t$ or $Ate^t$.

For a direct verification, set $y=e^t u$. Then

$$y''-2y'+y=e^t u''.$$

The equation becomes $u''=1$, so $u=t^2/2+C_1+C_2t$. Since $y(0)=u(0)=0$, $C_1=0$. Also $y'(0)=u(0)+u'(0)=0$, so $C_2=0$. Thus

$$y=\frac12t^2e^t.$$

The transformed equation verifies the forcing, and evaluation verifies both conditions. The homogeneous modes grow rather than decay, so they should not be described as decaying transients.

Solution 8.6

The roots of $r^2+2r+2=0$ are $-1\pm i$, giving $y_h=e^{-t}(C_1\cos t+C_2\sin t)$. Try $y_p=A\cos t+B\sin t$. Substitution yields

$$(A+2B)\cos t+(B-2A)\sin t=\cos t.$$

Thus $A+2B=1$, $B-2A=0$, so $A=1/5$, $B=2/5$. Apply the initial conditions to the full solution: $C_1+1/5=0$ and $-C_1+C_2+2/5=0$. Therefore

$$y=e^{-t}\left(-\frac15\cos t-\frac35\sin t\right) +\frac15\cos t+\frac25\sin t.$$

The steady-state amplitude is $\sqrt{(1/5)^2+(2/5)^2}=1/\sqrt5$. The first term is the decaying transient. The difference of any two solutions satisfies the homogeneous equation, whose modes decay, so their periodic responses agree asymptotically.

Chapter 9

Solution 9.1

The system is already diagonal:

$$x=C_1e^{-t},\qquad y=C_2e^{-3t}.$$

Both modes decay, so the origin is a stable node. Unless $C_1=0$, the slower $e^{-t}$ mode dominates late, and trajectories approach tangent to the $x$-axis.

Solution 9.2

The matrix has eigenvalue 3 with eigenvector $[1,1]^T$, and eigenvalue 1 with eigenvector $[1,-1]^T$. Therefore

$$\mathbf{x}=C_1e^{3t}\begin{bmatrix}1\1\end{bmatrix}+C_2e^t\begin{bmatrix}1\-1\end{bmatrix}.$$

Both modes grow, so the origin is an unstable node; the $[1,1]^T$ direction dominates generically.

Solution 9.3

Let $x_1=y$, $x_2=y'$. Then

$$\begin{bmatrix}x_1'\x_2'\end{bmatrix} =\begin{bmatrix}0&1\-5&-2\end{bmatrix} \begin{bmatrix}x_1\x_2\end{bmatrix}.$$

The eigenvalues solve $\lambda^2+2\lambda+5=0$, giving $-1\pm2i$. The negative real part and nonzero imaginary part produce a spiral sink.

Solution 9.4

The eigenvalue discriminant is $D=\tau^2-4\Delta$. If $\Delta\lt 0$, eigenvalues have opposite signs: saddle. If $\Delta\gt 0$ and $D\gt 0$, the roots are real with common sign set by $\tau$: stable node for $\tau\lt 0$, unstable node for $\tau\gt 0$. If $D\lt 0$, they are complex: spiral sink for $\tau\lt 0$, spiral source for $\tau\gt 0$, and center in the ideal linear case $\tau=0$. The curves $\Delta=0$ and $D=0$ are classification boundaries.

On $D=0$ with $\Delta\gt 0$, the repeated nonzero eigenvalue is $\tau/2$: its sign still determines decay or growth, but eigenvectors distinguish a scalar-matrix node from a defective node. On $\Delta=0$, at least one eigenvalue is zero and the origin is not an isolated equilibrium. If $\tau\lt 0$, trajectories approach a line of equilibria, so the origin is stable but not asymptotically stable; if $\tau\gt 0$, it is unstable. At $(\tau,\Delta)=(0,0)$, $A=0$ leaves every point fixed, whereas a nonzero nilpotent matrix permits linear drift. Trace and determinant alone cannot distinguish these last two cases.

Solution 9.5

First solve $y'=-y$, $y(0)=1$, obtaining $y=e^{-t}$. Then $x'+x=e^{-t}$, so the integrating factor $e^t$ gives $(e^tx)'=1$. Using $x(0)=0$ yields

$$x=te^{-t},\qquad y=e^{-t}.$$

Differentiate to check: $x'=(1-t)e^{-t}=-x+y$, and $y'=-e^{-t}=-y$.

The matrix $\begin{bmatrix}-1&1\0&-1\end{bmatrix}$ has repeated eigenvalue $-1$, but its eigenvectors satisfy $v_2=0$, leaving only the $x$-axis direction. An eigenvector mode has zero second component and cannot match $y(0)=1$. The additional $te^{-t}$ contribution is essential. In general, $y=C_2e^{-t}$ and $x=(C_1+C_2t)e^{-t}$; every solution tends to zero, so the origin is a defective stable node.

Solution 9.6

Differentiate $x'=-4y$ to get $x''=-4y'=-4x$. Since $x(0)=2$ and $x'(0)=0$,

$$x=2\cos2t,\qquad y=-x'/4=\sin2t.$$

You can also prove the invariant without solving:

$$\frac{d}{dt}(x^2+4y^2)=2x(-4y)+8yx=0.$$

Its initial value is 4, so the orbit is the ellipse $x^2+4y^2=4$, or $x^2/4+y^2=1$. At $(2,0)$, the velocity is $(0,2)$, pointing upward, so rotation is counterclockwise. The period is $\pi$. The origin is a center: nearby trajectories remain nearby, but nonzero invariant level sets never approach it. Thus it is stable, not attracting or asymptotically stable.

Chapter 10

Solution 10.1

With $h=0.2$, use $y_{n+1}=y_n+0.2(y_n-t_n)$:

Thus $y(0.6)\approx1.6$.

In fact, $y=1+t$ satisfies both the equation and initial condition, so Euler is exact at these grid points in exact arithmetic. This special straight-line solution is not evidence that Euler generally has zero error.

Solution 10.2

Euler gives $y_{n+1}=(1+h)y_n$, so from $y_0=1$, $y_n=(1+h)^n$. At $t=1$, $n=1/h$:

• $h=1$: $2$;
• $h=1/2$: $1.5^2=2.25$;
• $h=1/4$: $1.25^4\approx2.4414$.

The values rise toward $e\approx2.7183$ as the step shrinks.

Solution 10.3

The phase line shows that solutions starting between 0 and 10 rise toward 10 without crossing it. A useful numerical test should therefore: assert $0\lt P_n\lt 10$, rerun with half the step, compare the two trajectories over the same times, and reject a method or step size that crosses the invariant equilibrium. A direction-field overlay supplies an independent visual check.

Solution 10.4

Euler gives $y_{n+1}=(1-ah)y_n$. Decay in magnitude requires

$$|1-ah|<1\quad\Rightarrow\quad 0<h<\frac{2}{a}.$$

Decay without alternating sign requires $0\le1-ah\lt 1$, hence

$$0<h\le\frac1a.$$

At $h=1/a$, the numerical state reaches zero in one step.

Solution 10.5

For $f(t,y)=y-t$, the midpoint stages are:

Euler instead gives $y_1=0$ and $y_2=0+0.5(0-0.5)=-0.25$. The exact value is $y(1)=2-e\approx-0.718282$, making the absolute errors about 0.077657 for midpoint and 0.468282 for Euler.

The midpoint result is closer in this comparison because it samples the rate inside each step. It uses four rate evaluations rather than Euler's two, so the comparison fixes step size, not computational cost. One example does not prove a convergence order; repeated step-halving tests provide that evidence.

Solution 10.6

Write $a=hr$. For $0\lt z_n\lt1$ and $0\lt a\le1$, the update $z_{n+1}=z_n+az_n(1-z_n)$ is positive and greater than $z_n$. For its upper bound, factor

$$1-z_{n+1}=(1-z_n)(1-az_n).$$

Both factors are positive because $z_n\lt 1$ and $az_n\lt 1$. Thus $z_{n+1}\lt 1$, and induction preserves the interval for every finite step in exact arithmetic.

For $r=1$, $h=2$, and $z_0=0.75$,

$$z_1=0.75+2(0.75)(0.25)=1.125,$$

which crosses the equilibrium even though the exact solution cannot. The bound $hr\le1$ is sufficient to preserve the interval for every starting state within it; it does not estimate the time-trajectory error. Step-halving comparisons are still needed for accuracy, and floating-point implementations should allow roundoff-sized tolerances.

Mixed studio

Solution M1

Rewrite as $y'-ty=\sin t$. It is first-order linear with $p(t)=-t$, so $\mu=e^{-t^2/2}$. A direction field or substitution can check the eventual solution.

Solution M2

The factors separate: $(1+y^2)dy=e^t dt$. Integration gives $y+y^3/3=e^t+C$, generally best left implicit unless an initial condition selects a useful form.

Solution M3

Use homogeneous plus particular response. The characteristic root $-3$ repeats, so $y_h=(C_1+C_2t)e^{-3t}$. For the polynomial input, try $y_p=At+B$.

Solution M4

The matrix has eigenvalues $\pm i$. The system is $x'=-y, y'=x$, so $d(x^2+y^2)/dt=0$. Trajectories are counterclockwise circles centered at the origin, an ideal center.

Solution M5

Use an autonomous phase line before separation. Equilibria are $-2$ and 1. The sign pattern is positive outside and negative between, so $-2$ is stable and 1 unstable.

Solution M6

The condition holds, but the equation does not. The derivative is $y'=1/(1-t)^2=y^2$, not $y$. The claimed function solves $y'=y^2$ instead.

Solution M7

The equilibrium $y=0$ does not match the condition. Separate:

$$\int y^{-2}dy=\int(1+t)dt \quad\Rightarrow\quad -\frac1y=t+\frac{t^2}{2}+C.$$

At $t=0,y=-1$, $C=1$, so

$$y=-\frac{1}{1+t+t^2/2}.$$

The quadratic denominator has discriminant $1-2\lt 0$, so it never vanishes; the solution exists for all real $t$.

Solution M8

On $t\gt 0$, $\mu=t^2$. Then $(t^2y)'=t^4$, so $t^2y=t^5/5+C$. The condition gives $1=1/5+C$, hence $C=4/5$: