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

$$W^{\prime }=12-0.04W.$$

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

$$S^{\prime }=0.24-0.015S,S(0)=10.$$

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

Chapter 2

Solution 2.1

The equilibrium is (y=3). Below 3, (y-3<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)>0), so the solution rises toward the stable equilibrium 1. Differentiate along a solution:

$$y^{\prime \prime }=y^{\prime }(1-2y).$$

While (0<y<1/2), both factors are positive, so the graph is concave up. After (y) passes (1/2), (1-2y<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.

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 simply the shorter option for the first.

Solution 3.3

On (t>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.

Chapter 4

Solution 4.1

Separate and integrate:

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

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

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

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

Solution 4.2

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

$$\int \frac{dy}{y(y-2)}=t+C\Rightarrow \frac{1}{2}\ln \left|\frac{y-2}{y}\right|=t+C.$$

Using (y(0)=1) gives ((y-2)/y=-e^{2t}), 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

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

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

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

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

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

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>0), (\mu=t^2). Then

$$(t^{2}y{)}^{\prime }=t^3.$$

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

$$t=4\ln 6\approx 7.17\text{ time units}.$$

Solution 5.4

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

$$y^{\prime }+y=(2t-C{e}^{-t})+(t^2+C{e}^{-t})=t^2+2t.$$

Thus one valid equation is

$$y^{\prime }+y=t^2+2t.$$

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

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. For the outer initial states, the cubic rate also warns that unbounded growth in magnitude may occur in finite time.

Solution 6.3

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

$$\frac{d{P}^{\prime }}{dP}=r-\frac{2r}{K}P.$$

The critical state is (P=K/2), and the downward-opening parabola makes it a maximum. The maximum rate is (rK/4). The largest sustainable population is the equilibrium (K); these are different quantities.

Solution 6.4

For (a>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<0), (y'=y^2+|a|&gt0), so no equilibria exist and every solution rises.

Chapter 7

Solution 7.1

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

$$y=C_1{e}^{3t}+C_2{e}^{-2t}.$$

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

$$y=(C_1+C_{2}t)e^{-3t}.$$

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).

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

$$y=\frac{\cos t-\cos 2t}{3}.$$

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

Near resonance, the response contains a difference such as (\cos(\omega t)-\cos(\omega_0t)). The identity rewrites it as a fast oscillation with average frequency ((\omega+\omega_0)/2), multiplied by a slow envelope (\sin((\omega-\omega_0)t/2)). When the frequencies are close, the envelope varies slowly, producing beats. As they meet, the limiting envelope grows proportionally to (t).

Chapter 9

Solution 9.1

The system is already diagonal:

$$x=C_1{e}^{-t},y=C_2{e}^{-3t}.$$

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

$$\mathbf{x}=C_1{e}^{3t}\left[\begin{array}{c}1\\ 1\end{array}\right]+C_2{e}^t\left[\begin{array}{c}1\\ -1\end{array}\right].$$

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

$$\left[\begin{array}{c}{x}_1^{\prime }\\ x_2^{\prime }\end{array}\right]=\left[\begin{array}{cc}0& 1\\ -5& -2\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right].$$

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<0), eigenvalues have opposite signs: saddle. If (\Delta>0) and (D>0), the roots are real with common sign set by (\tau): stable node for (\tau<0), unstable node for (\tau>0). If (D<0), they are complex: spiral sink for (\tau<0), spiral source for (\tau>0), and center in the ideal linear case (\tau=0). The curves (\Delta=0) and (D=0) are classification boundaries.

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).

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<P_n<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

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

Decay without alternating sign requires (0\le1-ah<1), hence

$$0<h\le \frac{1}{a}.$$

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

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:

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

At (t=0,y=-1), (C=1), so

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

The quadratic denominator has discriminant (1-2<0), so it never vanishes; the solution exists for all real (t).

Solution M8

On (t>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):

$$y=\frac{t^3}{5}+\frac{4}{5t^2}.$$

Solution M9

Roots are (-1\pm i), so (y=e^{-t}(C_1\cos t+C_2\sin t)). The conditions give (C_1=1) and (-C_1+C_2=0), so

$$y=e^{-t}(\cos t+\sin t).$$

Negative real part plus rotation gives a spiral sink in the ((y,y')) phase plane.

Solution M10

Natural motion is (C_1\cos t+C_2\sin t). Try (A\sin3t+B\cos3t). Matching gives (-8A=1), (-8B=0), hence

$$y=C_1\cos t+C_2\sin t-\frac{1}{8}\sin 3t.$$

Solution M11

At (y_0=0.5), the slope is (0.5(1.5)=0.75), so (y_1=0.5+0.5(0.75)=0.875). The next slope is (0.875(1.125)=0.984375), so (y_2=1.3671875). Both values remain between the unstable equilibrium 0 and stable equilibrium 2 and move upward, matching the phase line.

Solution M12

The matrix has eigenvalue (-1) along ([1,1]^T) and (-3) along ([1,-1]^T). Both decay, so the origin is a stable node. The slower (e^{-t}) mode dominates generically, so trajectories approach tangent to ([1,1]^T).

Solution C1

If (C) is concentration in mg/L, the amount in the fixed volume is (VC). Its balance is (V C'=I-kC), so

$$C^{\prime }+\frac{k}{V}C=\frac{I}{V}.$$

Here (I) has units mg/hour, (k) has units L/hour, and every term in the amount balance has units mg/hour. The equilibrium is (C_=I/k), stable because the displacement equation is ((C-C_)'=-(k/V)(C-C_*)). From (C(0)=0),

$$C(t)=\frac{I}{k}\left(1-e^{-kt/V}\right).$$

For 90% of equilibrium, (e^{-kt/V}=0.1), so (t=(V/k)\ln10). The ratio (I/k) sets the target; (k/V) sets the approach rate, though changing (k) affects both.

Solution C2

Both models start with positive rate, but (P_1) has unlimited positive feedback while (P_2) weakens as it approaches 10. Thus (P_1=e^t) grows without bound, while (P_2) approaches 10. At (P=1), rates are 1 and 0.9, so their Taylor series already differ in the linear term:

$$P_1=1+t+O(t^2),P_2=1+0.9t+O(t^2).$$

The exponential field has horizontal-row slopes proportional to (P); the logistic field adds a zero-slope row at 10 and reverses sign above it. Numerical checks for the logistic model should preserve the interval ((0,10)).

Solution C3

Counterclockwise rotation with envelope (e^{-0.5t}) can use

$$A=\left[\begin{array}{cc}-0.5& -1\\ 1& -0.5\end{array}\right].$$

Its eigenvalues are (-0.5\pm i). Since (x'=-0.5x-y), differentiating and eliminating (y) gives

$$x^{\prime \prime }+x^{\prime }+1.25x=0.$$

The roots of this scalar equation are the same (-0.5\pm i), confirming the envelope and angular frequency.

I compared first decisions, updated my error log, and re-solved missed problems without looking.

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