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1. 1.Computation · 6 min

   Every verification below must substitute into the original equation, not into a rearranged one.

   1. (a)State the order of each equation and whether it is linear, with a one-line reason each: $x^2{y}^{\prime \prime }+e^{x}y=1,y{y}^{\prime \prime }=1,y^{\prime \prime \prime }+(\sin x)y=y^{\prime }.$
   2. (b)Show that every member of the family $y=C{e}^{x^2}-\frac{1}{2}$ is a solution of $y^{\prime }=x(2y+1)$, and find the member through the point $(0,1)$.
   3. (c)Find every value of $r$ for which $y=e^{rx}$ is a solution of $y^{\prime \prime }-2y^{\prime }-8y=0$, and write down a two-parameter family of solutions, saying what justifies combining them.
   Show the solution to question 1

   (a) Order and linearity. The first equation, $x^2{y}^{\prime \prime }+e^{x}y=1$, is second order (the highest derivative is $y^{\prime \prime }$) and linear: $y$ and $y^{\prime \prime }$ appear to the first power, each multiplied by a function of $x$ alone, and the right side $1$ involves no $y$. The second, $y{y}^{\prime \prime }=1$, is second order and nonlinear, because $y$ multiplies $y^{\prime \prime }$. The third, $y^{\prime \prime \prime }+(\sin x)y=y^{\prime }$, is third order and linear: rearranged as $y^{\prime \prime \prime }-y^{\prime }+(\sin x)y=0$, every term is a derivative of $y$ times a function of $x$. The $\sin$ acts on $x$, not on $y$ — linearity is a statement about how $y$ enters, not about $x$.

   (b) Verify the family, then pick the member. For $y=C{e}^{x^2}-\frac{1}{2}$ the chain rule gives $y^{\prime }=2Cx{e}^{x^2}$. On the other side, $2y+1=2C{e}^{x^2}$, so

   $$x(2y+1)=2Cx{e}^{x^2}=y^{\prime }$$

   for every $x$ and every $C$: the whole family solves the equation. Through $(0,1)$ we need $y(0)=C-\frac{1}{2}=1$, so $C=\frac{3}{2}$ and

   $$y=\frac{3}{2}{e}^{x^2}-\frac{1}{2}.$$

   Check: $y(0)=\frac{3}{2}-\frac{1}{2}=1$, and $y^{\prime }=3x{e}^{x^2}$ while $x(2y+1)=x\cdot 3e^{x^2}$.

   (c) Exponential trial solution. With $y=e^{rx}$, $y^{\prime }=r{e}^{rx}$, $y^{\prime \prime }=r^2{e}^{rx}$:

   $$y^{\prime \prime }-2y^{\prime }-8y=e^{rx}\left(r^2-2r-8\right)=e^{rx}(r-4)(r+2).$$

   Since $e^{rx}$ is never zero this vanishes exactly when $r=4$ or $r=-2$. So $e^{4x}$ and $e^{-2x}$ are solutions, and because the equation is linear and homogeneous any combination $y=c_1{e}^{4x}+c_2{e}^{-2x}$ is a solution too (superposition). Check on $e^{-2x}$: $4e^{-2x}-2(-2e^{-2x})-8e^{-2x}=(4+4-8)e^{-2x}=0$.

2. 2.Computation · 7 min

   Consider the equation $\frac{dy}{dx}=2\sqrt{y}$.

   1. (a)Show that $y=0$ and $y=\{\begin{array}{ll}0,& x\le 0,\\ x^2,& x>0\end{array}$ are both solutions on the whole real line with $y(0)=0$. Check differentiability at $x=0$ explicitly.
   2. (b)State the existence–uniqueness theorem for first-order initial-value problems. Identify which hypothesis fails at $(0,0)$, and explain why the theorem does apply at $(0,1)$.
   3. (c)Solve the equation with $y(0)=1$ by separating variables, then determine the largest interval containing $0$ on which your formula actually satisfies the equation, and say what went wrong outside it.
   Show the solution to question 2

   (a) Two solutions through the origin. The constant $y=0$ gives $y^{\prime }=0$ and $2\sqrt{0}=0$, so it is a solution everywhere, with $y(0)=0$.

   For the piecewise function: on $x<0$ it is $0$, and both sides of the equation are $0$. On $x>0$ it is $x^2$, so $y^{\prime }=2x$ while $2\sqrt{y}=2\sqrt{x^2}=2|x|=2x$ because $x>0$. At $x=0$ the derivative must be checked from the definition: the left-hand difference quotient is $0$, and the right-hand one is $\frac{h^2-0}{h}=h\to 0$. So $y^{\prime }(0)=0$, which equals $2\sqrt{y(0)}=0$. The function is differentiable everywhere and satisfies the equation at every $x$, with $y(0)=0$. Two different solutions pass through the same point.

   (b) Which hypothesis fails. The existence–uniqueness theorem says: if $f(x,y)$ and $\partial f/\partial y$ are continuous on some open rectangle containing $(x_0,y_0)$, then $y^{\prime }=f(x,y)$, $y(x_0)=y_0$ has a solution on some open interval about $x_0$, and only one. Here $f(x,y)=2\sqrt{y}$ and

   $$\frac{\partial f}{\partial y}=\frac{1}{\sqrt{y}},$$

   which is unbounded as $y\to 0^{+}$ and undefined for $y\le 0$. No open rectangle around $(0,0)$ avoids the line $y=0$ (indeed $f$ itself is not even defined below it), so the uniqueness hypothesis fails there and the theorem is silent — and (a) shows uniqueness genuinely fails. Around $(0,1)$ the rectangle $|x|<1$, $\frac{1}{2}<y<\frac{3}{2}$ stays inside $y>0$, where both $f$ and $1/\sqrt{y}$ are continuous, so the theorem promises exactly one solution near $x=0$ through $(0,1)$.

   (c) Separate, then be honest about the square root. For $y>0$ divide by $\sqrt{y}$:

   $$\frac{dy}{\sqrt{y}}=2dx⟹2\sqrt{y}=2x+C.$$

   The condition $y(0)=1$ gives $C=2$, so $\sqrt{y}=x+1$ and $y=(x+1{)}^2$. Check: $y^{\prime }=2(x+1)$ and $2\sqrt{y}=2|x+1|$. These agree exactly when $x+1\ge 0$. So the formula is a solution on $(-1,\infty )$ and not for $x<-1$: there $y^{\prime }<0$ while the right side is positive. What went wrong is the squaring of $\sqrt{y}=x+1$, which silently discards the requirement $x+1\ge 0$ that the square root carried. The largest interval containing $0$ on which the formula solves the equation is $(-1,\infty )$ (it also works at the endpoint $x=-1$ itself).

   At $x=-1$ the solution reaches $y=0$, the line where uniqueness breaks down, and there it may be continued to the left as the equilibrium $y=0$: the function equal to $0$ for $x\le -1$ and to $(x+1{)}^2$ for $x>-1$ solves the equation on the whole line, by the same check as in (a).

3. 3.Sketch · 8 min

   Let $y^{\prime }=2x-y$. A sketch with no reasoning beside it earns nothing here: every feature you draw must be justified from the equation.

   1. (a)Compute the slope at the nine points with $x,y\in \{-1,0,1\}$ and describe the isoclines. Where are the segments horizontal?
   2. (b)Find the one solution that is a straight line, and verify it by substitution.
   3. (c)Let $y$ be any solution and put $u(x)=y(x)-(2x-2)$. Show that $u^{\prime }=-u$, and deduce how every solution behaves relative to the line as $x\to \infty$ and as $x\to -\infty$.
   4. (d)Sketch the direction field on the window $-2\le x\le 2$, $-4\le y\le 4$ together with the solutions through $(0,0)$ and through $(0,-3)$. Mark every point where either curve has a horizontal tangent and say whether it is a maximum or a minimum.
   Show the solution to question 3

   (a) Tabulate the slope $2x-y$. Reading each row at $x=-1$, $x=0$, $x=1$ in turn:
   along $y=1$ the slopes are $-3,-1,1$;
   along $y=0$ they are $-2,0,2$;
   along $y=-1$ they are $-1,1,3$.

   The isoclines are the curves $2x-y=c$, that is the parallel lines $y=2x-c$, all of slope $2$. Segments are horizontal on the isocline $c=0$, the line $y=2x$; above that line ($y>2x$) every slope is negative, below it every slope is positive.

   (b) The straight-line solution. A line $y=mx+b$ is a solution only if the field's slope along it equals $m$ everywhere. On the isocline $c=2$, the line $y=2x-2$, every segment has slope $2$, which is the slope of the line itself. Substituting: $y^{\prime }=2$ and $2x-y=2x-(2x-2)=2$, so $y=2x-2$ is a solution. No other line works, since the field's slope along $y=mx+b$ is $(2-m)x-b$, constant only when $m=2$, and then equal to $m=2$ only when $b=-2$.

   (c) The deviation from the line decays. If $y$ is any solution and $u=y-(2x-2)$, then

   $$u^{\prime }=y^{\prime }-2=(2x-y)-2=-(y-2x+2)=-u.$$

   By the natural-growth theorem (with $k=-1$) $u=u_0{e}^{-x}$, where $u_0=u(0)$. As $x\to \infty$, $u\to 0$: every solution approaches the line $y=2x-2$, from above if $u_0>0$ and from below if $u_0<0$, and never crosses it, since $u$ never changes sign. As $x\to -\infty$, $|u|\to \infty$: the solutions fan out away from the line in both directions.

   (d) The sketch. Draw the segments from the table (and the horizontal ones along $y=2x$), the solution line $y=2x-2$ through $(0,-2)$ and $(1,0)$, and two curves.

   Through $(0,0)$. Here $u_0=0-(-2)=2>0$, so the curve lies above the line for all $x$ and settles onto it from above as $x$ increases (it passes about $(2,2.3)$ against the line's $(2,2)$). At the origin the slope is $0$; just to the left the point is above $y=2x$, so the curve is falling, and just to the right it is below $y=2x$, so it is rising: the origin is a local minimum, and it is the only place this curve meets $y=2x$, because $y-2x=-2+2e^{-x}$ vanishes only at $x=0$. Going left the curve climbs steeply ($u=2e^{-x}$ grows), passing about $(-1,1.4)$ and leaving the window at the top.

   Through $(0,-3)$. Here $u_0=-3+2=-1<0$: the curve stays below the line, so it is also below $y=2x$ and its slope is always positive (it is $3$ at the start). It rises toward the line from below without ever reaching it (about $(1,-0.4)$ and $(2,1.9)$ against the line's $(1,0)$ and $(2,2)$), and to the left it plunges downward, since $u=-e^{-x}\to -\infty$, leaving the window at the bottom near $x=-1$.

   Mark the horizontal tangent at the origin, the dashed line $y=2x$ of horizontal segments, and the solution line $y=2x-2$ drawn solid.

4. 4.Computation · 7 min

   Consider the initial-value problem $y^{\prime }=2xy$, $y(0)=1$. Keep every number exact — as a fraction or a terminating decimal — until the final comparison.

   1. (a)Use Euler's method with $h=0.5$ to estimate $y(1)$. Show the table of $x_n$, $y_n$ and the slope used at each step.
   2. (b)Repeat with $h=0.25$.
   3. (c)Solve the problem exactly and compute the exact value of $y(1)$. Explain from the shape of the solution curve why both estimates err in the direction they do, and why the second is closer.
   Show the solution to question 4

   (a) Two steps of size $0.5$. The slope function is $f(x,y)=2xy$, and each step uses the slope at the current point: $y_{n+1}=y_n+0.5\cdot 2x_n{y}_n$.
   Step 0: at $(x_0,y_0)=(0,1)$ the slope is $2\cdot 0\cdot 1=0$, so $y_1=1+0.5\cdot 0=1$.
   Step 1: at $(x_1,y_1)=(0.5,1)$ the slope is $2\cdot 0.5\cdot 1=1$, so $y_2=1+0.5\cdot 1=1.5$.

   So $y(1)\approx 1.5$. Note the first step moves nowhere: at $x=0$ the field is flat whatever $y$ is.

   (b) Four steps of size $0.25$. Now $y_{n+1}=y_n+0.25\cdot 2x_n{y}_n$.
   Step 0: at $(0,1)$ the slope is $0$, so $y_1=1$.
   Step 1: at $(0.25,1)$ the slope is $0.5$, so $y_2=1+0.125=1.125$.
   Step 2: at $(0.5,1.125)$ the slope is $1.125$, so $y_3=1.125+0.28125=1.40625$.
   Step 3: at $(0.75,1.40625)$ the slope is $2.109375$, so $y_4=1.40625+0.52734375=1.93359375$.

   So $y(1)\approx 1.93359375=\frac{495}{256}$.

   (c) The exact solution and the direction of the error. The equation is separable: for $y\ne 0$, $\frac{dy}{y}=2xdx$, so $\ln |y|=x^2+C$ and $y=A{e}^{x^2}$ with $A$ any constant (the equilibrium $y=0$ is $A=0$). The condition $y(0)=1$ gives $A=1$:

   $$y=e^{x^2},y(1)=e\approx \mathrm{2.718.}$$

   Check: $y^{\prime }=2x{e}^{x^2}=2xy$ and $y(0)=1$. The errors are $e-1.5\approx 1.22$ and $e-1.934\approx 0.78$.

   Both estimates are too small, and this could have been predicted. Differentiating the equation, $y^{\prime \prime }=2y+2x{y}^{\prime }=2y+4x^{2}y=(2+4x^2)y>0$ for $y>0$: the solution is concave up, so each tangent line lies below the curve and every Euler step lands under the true value. Worse, the next slope is then computed from a $y_n$ that is already too small, so the undershoot compounds. Halving $h$ reduces the error (here by the factor $0.78/1.22\approx 0.64$). The rule that the accumulated error is proportional to $h$ — so that halving $h$ halves the error — is a statement about small $h$; with steps this coarse on a solution that bends this fast it has not yet taken hold, but the trend is the right one.

5. 5.Critique · 8 min

   A student is asked to find all solutions of $\frac{dy}{dx}=4x(2-y)$ and, in particular, the solution with $y(0)=5$. The submitted work reads:

   Step 1. Separate: $\frac{dy}{2-y}=4xdx$.

   Step 2. Integrate: $-\ln (2-y)=2x^2+C$, so $2-y=e^{-2x^2-C}=A{e}^{-2x^2}$ with $A=e^{-C}>0$.

   Step 3. Hence $y=2-A{e}^{-2x^2}$ with $A>0$: every solution lies below the line $y=2$.

   Step 4. The condition $y(0)=5$ gives $5=2-A$, so $A=-3<0$, which is impossible. The initial-value problem has no solution.

   Step 5. Also, $y=2$ is not a solution, since no positive $A$ produces it.

   Steps 3, 4 and 5 are all false.

   1. (a)Find the single error from which all three false steps follow, and explain how each of Steps 3, 4 and 5 inherits it. Then say why the existence–uniqueness theorem should have made the student suspicious of Step 4 before checking any algebra.
   2. (b)Give the correct general solution, including any solution that separation discards, then the solution with $y(0)=5$, and check it in the original equation and the initial condition.
   Show the solution to question 5

   (a) One dropped absolute value. The error is in Step 2. The integral is

   $$\int \frac{dy}{2-y}=-\ln |2-y|+C,$$

   with an absolute value. Writing $\ln (2-y)$ instead silently assumes $2-y>0$, that is $y<2$, and everything downstream inherits that assumption. Step 3's "every solution lies below $y=2$" is not a discovery but the assumption read back. Step 4 fails precisely because $(0,5)$ lies above $y=2$, outside the region the student had unknowingly restricted to. Step 5 is a separate oversight of the same kind: Step 1 divides by $2-y$, which is illegal where $2-y=0$, so the constant $y=2$ has to be tested on its own — and it is a solution, since both sides are then $0$.

   The existence–uniqueness theorem should have stopped Step 4 on sight. Here $f(x,y)=4x(2-y)$ and $\partial f/\partial y=-4x$ are polynomials, continuous on the whole plane, so through every point, including $(0,5)$, there passes a solution, and only one. (The equation is also linear, $y^{\prime }+4xy=8x$, so that solution exists on all of $\mathbb{R}$.) A method that concludes "no solution" has contradicted a theorem; the method, not the theorem, is wrong.

   (b) The repaired argument. First the equilibrium: $y=2$ makes both sides zero and is a solution. For $y\ne 2$, separating and integrating correctly,

   $$-\ln |2-y|=2x^2+C⟹|2-y|=e^{-C}{e}^{-2x^2}⟹2-y=\pm e^{-C}{e}^{-2x^2}.$$

   Write $A=\pm e^{-C}$, which is any nonzero constant; allowing $A=0$ restores $y=2$. So the complete general solution is

   $$y=2-A{e}^{-2x^2},A\in \mathbb{R}.$$

   Solutions with $A<0$ lie above the line $y=2$ and are just as legitimate as those below it. For $y(0)=5$: $5=2-A$, so $A=-3$ and

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

   Check: $y^{\prime }=-12x{e}^{-2x^2}$, while $4x(2-y)=4x\cdot \left(-3e^{-2x^2}\right)=-12x{e}^{-2x^2}$; and $y(0)=2+3=5$. This solution decreases toward $2$ as $|x|\to \infty$ but never reaches it, which is uniqueness at work: it cannot cross the solution $y=2$.

6. 6.Computation · 8 min

   Solve $x\frac{dy}{dx}=y-\frac{x^3}{y^2}$ for $x>0$ with $y(1)=2$.

   1. (a)Show that the equation is homogeneous: write $y^{\prime }$ as a function of $v=y/x$ alone, or verify $f(tx,ty)=f(x,y)$.
   2. (b)Substitute $y=vx$, solve the resulting separable equation, and give $y$ explicitly.
   3. (c)Check your answer by substitution, and state the largest interval containing $x=1$ on which the solution is valid, explaining what goes wrong at each end.
   Show the solution to question 6

   (a) It depends only on $y/x$. Dividing by $x$,

   $$\frac{dy}{dx}=\frac{y}{x}-\frac{x^2}{y^2}=v-\frac{1}{v^2},v=\frac{y}{x},$$

   so the right side is $F(v)=v-v^{-2}$, a function of the ratio alone: the equation is homogeneous. Equivalently, with $f(x,y)=y/x-x^2/y^2$, replacing $(x,y)$ by $(tx,ty)$ gives $\frac{ty}{tx}-\frac{t^2{x}^2}{t^2{y}^2}=f(x,y)$. (It is not separable as it stands, since $y/x-x^2/y^2$ is a difference, not a product.)

   (b) Substitute $y=vx$. Then $y^{\prime }=v+x{v}^{\prime }$ by the product rule, and the equation becomes

   $$v+x{v}^{\prime }=v-\frac{1}{v^2}⟹x\frac{dv}{dx}=-\frac{1}{v^2}⟹v^{2}dv=-\frac{dx}{x}.$$

   Integrating (with $x>0$, so $\ln |x|=\ln x$): $\frac{v^3}{3}=-\ln x+C$, that is $v^3=C_1-3\ln x$ with $C_1=3C$. Nothing was divided by a function of $v$ here, so no solution was lost. Replacing $v=y/x$ and multiplying by $x^3$,

   $$y^3=x^3\left(C_1-3\ln x\right).$$

   The condition $y(1)=2$ gives $8=C_1$, so

   $$y=x(8-3\ln x{)}^{1/3}.$$

   (c) Check, and the interval of validity. Write $y=xu$ with $u=(8-3\ln x{)}^{1/3}$, so $u^3=8-3\ln x$. Differentiating that relation, $3u^2{u}^{\prime }=-3/x$, hence $u^{\prime }=-\frac{1}{x{u}^2}$. Then

   $$y^{\prime }=u+x{u}^{\prime }=u-\frac{1}{u^2},\text{while}\frac{y}{x}-\frac{x^2}{y^2}=u-\frac{x^2}{x^2{u}^2}=u-\frac{1}{u^2}.$$

   They agree, and $y(1)=1\cdot 8^{1/3}=2$.

   The cube root is defined for every real number, so the formula makes sense for all $x>0$ — but it stops being a solution where $y=0$, since the original equation has $y^2$ in a denominator. That happens when $8-3\ln x=0$, i.e. at $x=e^{8/3}$, and there $u^{\prime }=-1/(x{u}^2)$ blows up: the curve arrives at the $x$-axis with a vertical tangent. Going the other way, as $x\to 0^{+}$ the formula tends to $0$ but $x=0$ itself is excluded (we divided by $x$). So the largest interval containing $x=1$ on which the solution is valid is

   $$0<x<e^{8/3}.$$

   The equation's own singularities, at $x=0$ and at $y=0$, are exactly what ends the interval at each side.

7. 7.Proof · 9 min

   Throughout, $P$ and $Q$ are continuous on an open interval $I$, and $\mu (x)=e^{\int P(x)dx}$ is an integrating factor for $y^{\prime }+P(x)y=Q(x)$. Do not quote the solution formula; argue from the product rule, in the manner of the proof that every solution of $y^{\prime }=ky$ is $C{e}^{kt}$.

   1. (a)Let $y_1$ and $y_2$ be any two solutions on $I$ and put $w=y_1-y_2$. Prove that $w^{\prime }+P(x)w=0$, and then, by considering $u=\mu w$, that $w=C/\mu (x)$ for some constant $C$.
   2. (b)Deduce that two solutions which agree at a single point $x_0\in I$ agree on all of $I$, and that if $y_p$ is any one solution then the general solution is exactly $y_p+C/\mu (x)$.
   3. (c)Verify that $y=\sin x$ solves $y^{\prime }+(\cot x)y=2\cos x$ on $(0,\pi )$, and use (b) — not the integrating-factor recipe — to write down the general solution on $(0,\pi )$.
   Show the solution to question 7

   (a) The difference of two solutions. By linearity of differentiation,

   $$w^{\prime }+Pw=\left(y_1^{\prime }+P{y}_1\right)-\left(y_2^{\prime }+P{y}_2\right)=Q-Q=0$$

   on $I$. Next, ${\mu }^{\prime }=P\mu$ by the chain rule, since the derivative of $e^{\int P}$ is $e^{\int P}$ times the derivative of the exponent. Put $u=\mu w$. By the product rule,

   $$u^{\prime }={\mu }^{\prime }w+\mu w^{\prime }=P\mu w+\mu w^{\prime }=\mu \left(w^{\prime }+Pw\right)=0$$

   on $I$. A function whose derivative vanishes on an interval is constant there (a consequence of the Mean Value Theorem), so $u=C$ for some constant $C$, and since $\mu =e^{\int P}>0$ is never zero, $w=C/\mu (x)$. $■$

   (b) Uniqueness, and the shape of the general solution. If $y_1(x_0)=y_2(x_0)$ then $w(x_0)=0$, so $C=\mu (x_0)w(x_0)=0$ and $w\equiv 0$ on $I$: the two solutions coincide on the whole interval. This is the uniqueness half of the theorem for linear equations, and note that it holds on all of $I$, not merely near $x_0$.

   For the general solution, let $y_p$ be any one solution. By (a), every solution $y$ satisfies $y-y_p=C/\mu$ for some $C$, so every solution has the form $y_p+C/\mu$. Conversely each such function is a solution: since $(1/\mu {)}^{\prime }=-{\mu }^{\prime }/{\mu }^2=-P/\mu$,

   $${\left(y_p+\frac{C}{\mu }\right)}^{\prime }+P\left(y_p+\frac{C}{\mu }\right)=\left(y_p^{\prime }+P{y}_p\right)+C\left(-\frac{P}{\mu }+\frac{P}{\mu }\right)=Q.$$

   So the solutions are exactly the one-parameter family $y=y_p+C/\mu (x)$. $■$

   (c) One known solution is enough. For $y=\sin x$: $y^{\prime }=\cos x$ and $(\cot x)\sin x=\cos x$, so $y^{\prime }+(\cot x)y=2\cos x$ on $(0,\pi )$, as required. Here $P=\cot x$, and on $(0,\pi )$ we have $\sin x>0$, so $\int \cot xdx=\ln \sin x$ and $\mu =\sin x$, $1/\mu =\csc x$. By (b) the general solution is

   $$y=\sin x+C\csc x,0<x<\pi .$$

   Check of the new term: $(C\csc x{)}^{\prime }=-C\csc x\cot x$, and $(\cot x)(C\csc x)$ cancels it exactly, so $C\csc x$ contributes $0$ to the left side. No integral of $\mu Q$ was needed. Notice where the family breaks down: $\csc x$ blows up at $0$ and $\pi$, precisely the points where $P=\cot x$ stops being continuous.

8. 8.Computation · 9 min

   A tank of capacity $300$ L initially holds $100$ L of brine in which $5$ kg of salt is dissolved. Brine containing $0.25$ kg of salt per litre flows in at $4$ L/min, the tank is kept well mixed, and the mixture drains at $2$ L/min.

   1. (a)Write the differential equation for the amount $y(t)$ of salt in the tank, with its initial condition, and identify it as linear. State when the tank becomes full.
   2. (b)Solve for $y(t)$, and find the amount of salt in the tank at the moment it becomes full, as an exact number.
   3. (c)Compare the concentration in the tank at that moment with the inflow concentration, and explain why the comparison had to come out the way it did.
   4. (d)From the moment the tank is full, the excess spills over the rim so that the volume stays at $300$ L while the drain keeps running. Write the new equation for $y$ and find ${\lim }_{t\to \infty }y(t)$, with justification. You need not solve the new equation.
   Show the solution to question 8

   (a) The balance sheet. The volume is $V(t)=100+(4-2)t=100+2t$ litres, so the tank is full when $100+2t=300$, at $t=100$ min. Salt enters at $0.25\times 4=1$ kg/min. It leaves at concentration $y/V$ times the outflow rate:

   $$\text{rate out}=\frac{y}{100+2t}\cdot 2=\frac{y}{50+t}\text{kg/min}.$$

   Hence

   $$\frac{dy}{dt}=1-\frac{y}{50+t},y(0)=5,$$

   or in standard form $y^{\prime }+\frac{1}{50+t}y=1$: linear, with $P(t)=\frac{1}{50+t}$ and $Q(t)=1$. It is not separable, since $1-y/(50+t)$ does not factor into a function of $t$ times a function of $y$.

   (b) Integrating factor. $\mu =e^{\int dt/(50+t)}=e^{\ln (50+t)}=50+t$. Multiplying through,

   $$(50+t)y^{\prime }+y=50+t,\text{and indeed}\frac{d}{dt}[(50+t)y]=(50+t)y^{\prime }+y.$$

   Integrate: $(50+t)y=\frac{(50+t{)}^2}{2}+C$, so

   $$y=\frac{50+t}{2}+\frac{C}{50+t}.$$

   From $y(0)=5$: $25+\frac{C}{50}=5$, so $C=-1000$ and

   $$y(t)=\frac{50+t}{2}-\frac{1000}{50+t}.$$

   Check: $y^{\prime }=\frac{1}{2}+\frac{1000}{(50+t{)}^2}$, while $1-\frac{y}{50+t}=1-\frac{1}{2}+\frac{1000}{(50+t{)}^2}$; and $y(0)=25-20=5$. At the moment the tank is full,

   $$y(100)=\frac{150}{2}-\frac{1000}{150}=75-\frac{20}{3}=\frac{205}{3}\approx 68.3\text{kg}.$$

   (c) The concentration comparison. The concentration at $t=100$ is $\frac{205/3}{300}=\frac{205}{900}=\frac{41}{180}\approx 0.228$ kg/L, below the inflow's $0.25$ kg/L. It had to be: at every moment the tank holds a mixture of the original liquid, at $5/100=0.05$ kg/L, and inflow liquid at $0.25$ kg/L, and draining removes mixture without changing its concentration. A blend of two concentrations lies strictly between them, so the tank's concentration stays strictly between $0.05$ and $0.25$ for all $t>0$. (Sanity check on the amount: $300$ L of inflow brine would hold $75$ kg, and $\frac{205}{3}<75$.)

   (d) After the tank is full. For $t\ge 100$ the volume is fixed at $300$ L. Liquid still enters at $4$ L/min, so $4$ L/min must leave — $2$ through the drain and $2$ over the rim — all of it at concentration $y/300$:

   $$\frac{dy}{dt}=1-\frac{4y}{300}=1-\frac{y}{75}.$$

   This is autonomous with the single equilibrium $y=75$, and $f(y)=1-y/75$ has $f^{\prime }=-1/75<0$, so the equilibrium is stable: ${\lim }_{t\to \infty }y(t)=75$ kg, the amount that $300$ L of the incoming brine would contain. (Explicitly, $y=75-\frac{20}{3}{e}^{-(t-100)/75}$ for $t\ge 100$, rising from $\frac{205}{3}$ toward $75$.)

9. 9.Sketch · 9 min

   Let $\frac{dy}{dt}=y^2(4-y^2)$, and write $f(y)=y^2(4-y^2)$.

   1. (a)Find the equilibria and draw the phase line, showing the sign of $f$ on every interval between and beyond them.
   2. (b)Classify each equilibrium as stable, unstable or semistable. Use the derivative test wherever it decides; where it does not, say so and state what you used instead.
   3. (c)For the solution with $y(0)=1$, find every height $y$ at which the solution curve has an inflection point, by differentiating the equation.
   4. (d)On one set of $ty$-axes for $t\ge 0$, sketch the equilibrium lines and the solutions with $y(0)=-3$, $-1$, $1$ and $3$. State the long-term behaviour of each, justify its monotonicity and concavity, and mark the inflection point from (c).
   Show the solution to question 9

   (a) Equilibria and the phase line. Factor $f(y)=y^2(2-y)(2+y)$, so the equilibria are $y=-2$, $y=0$ and $y=2$. The sign of $f$ on each gap: for $y<-2$, $y^2>0$, $2-y>0$, $2+y<0$, so $f<0$. For $-2<y<0$ all three factors are positive, so $f>0$. For $0<y<2$ likewise $f>0$ (the factor $y^2$ does not change sign through $0$). For $y>2$, $2-y<0$, so $f<0$. On the phase line the arrows point down below $-2$, up on $(-2,0)$, up on $(0,2)$, and down above $2$.

   (b) Classification. $f^{\prime }(y)=8y-4y^3=4y\left(2-y^2\right)$. At $y=-2$: $f^{\prime }(-2)=-8(2-4)=16>0$, unstable, and the phase line agrees (arrows leave it on both sides). At $y=2$: $f^{\prime }(2)=8(2-4)=-16<0$, stable, and both neighbouring arrows point toward it. At $y=0$: $f^{\prime }(0)=0$, so the derivative test is silent and only the sign of $f$ can decide. Since $f>0$ on both sides of $0$, solutions below $0$ rise toward it while solutions above $0$ rise away from it: $y=0$ is semistable — approached from below, abandoned from above. The double factor $y^2$ is what makes $f$ touch the axis without crossing it.

   (c) Inflection point of the solution through $y(0)=1$. This solution starts strictly between the equilibria $0$ and $2$, cannot cross either of them by uniqueness, and increases (since $f>0$ there), so it tends to $2$. By the chain rule,

   $$y^{\prime \prime }=f^{\prime }(y)y^{\prime }=f^{\prime }(y)f(y)=4y\left(2-y^2\right)\cdot y^2\left(4-y^2\right).$$

   On $0<y<2$ the factors $4y$ and $y^2(4-y^2)$ are positive, so the sign of $y^{\prime \prime }$ is the sign of $2-y^2$: concave up while $y<\sqrt{2}$, concave down once $y>\sqrt{2}$. The solution starts at $1<\sqrt{2}$ and rises through $\sqrt{2}$ exactly once, so it has exactly one inflection point, at the height $y=\sqrt{2}$, where its slope is $f(\sqrt{2})=2\cdot 2=4$, the steepest it ever is.

   (d) The sketch. Draw the three horizontal equilibrium lines $y=-2$, $y=0$, $y=2$ for $t\ge 0$. Then:

   $y(0)=3$. Above $2$, so decreasing ($f(3)=-45$, a steep start) and trapped above $2$: it falls toward $y=2$ and flattens onto it. Concavity: for $y>2$, $4y>0$, $2-y^2<0$ and $y^2(4-y^2)<0$, so $y^{\prime \prime }>0$ — concave up throughout, approaching from above.

   $y(0)=1$. The S-curve: increasing from $1$ to $2$, concave up until it reaches height $\sqrt{2}\approx 1.41$, then concave down, levelling off just under $y=2$. Mark the inflection point at $y=\sqrt{2}$.

   $y(0)=-1$. On $(-2,0)$ the solution increases toward $0$ and never reaches it. Here $4y<0$ and $2-y^2>0$ (since $|y|<\sqrt{2}$) while $y^2(4-y^2)>0$, so $y^{\prime \prime }<0$: it rises concave down, flattening onto $y=0$ from below.

   $y(0)=-3$. Below $-2$, where $f<0$: decreasing, moving away from the unstable equilibrium and never turning back, ever more steeply as $|y|$ grows (for large $|y|$, $f\approx -y^4$). It leaves the picture through the bottom. (In fact it reaches $-\infty$ in finite time, but the sketch only needs "decreasing without bound".)

   The picture should show curves that never cross the equilibrium lines, arrows on the right edge consistent with the phase line, and the two behaviours at $y=0$ — a curve arriving from below, none arriving from above.

10. 10.Critique · 9 min

    A student is asked to solve $y^{\prime \prime }+4y^{\prime }+4y=6e^{-2x}$ with $y(0)=1$, $y^{\prime }(0)=0$, and submits the following.

    Step 1. The characteristic equation $r^2+4r+4=(r+2{)}^2=0$ gives $r=-2$, so $y_c=c_1{e}^{-2x}+c_2{e}^{-2x}$.

    Step 2. The forcing term $6e^{-2x}$ solves the complementary equation, so by the modification rule the trial $A{e}^{-2x}$ is multiplied by $x$: $y_p=Ax{e}^{-2x}$.

    Step 3. Then $y_p^{\prime }=A(1-2x)e^{-2x}$ and $y_p^{\prime \prime }=A(4x-4)e^{-2x}$, so $y_p^{\prime \prime }+4y_p^{\prime }+4y_p=A{e}^{-2x}[(4x-4)+(4-8x)+4x]=0$.

    Step 4. The left side is $0$ for every $A$, so it can never equal $6e^{-2x}$. The equation has no particular solution, and the initial-value problem has no solution.

    The algebra in Step 3 is correct.

    1. (a)Identify the two errors, in Steps 1 and 2, and state precisely what each should have been. Then name the theorem that makes the conclusion of Step 4 impossible, and say what a result of $0$ from a trial form actually tells you.
    2. (b)Solve the initial-value problem correctly, and check the equation and both initial conditions.
    Show the solution to question 10

    (a) Two errors and a theorem. Step 1 mishandles the repeated root. The sum $c_1{e}^{-2x}+c_2{e}^{-2x}$ is just $(c_1+c_2)e^{-2x}$, a one-parameter family that could never meet two initial conditions. When the characteristic root $r$ is repeated the second independent solution is $x{e}^{rx}$, so

    $$y_c=\left(c_1+c_{2}x\right)e^{-2x}.$$

    Step 2 applies the modification rule only once. Because the root is repeated, $x{e}^{-2x}$ is itself a complementary solution, so the trial $Ax{e}^{-2x}$ is annihilated by the left side — which is exactly what Step 3 discovered. The rule says: if the trial form still solves the complementary equation after multiplying by $x$, multiply by $x$ again. The correct trial is $y_p=A{x}^2{e}^{-2x}$.

    Step 4 is impossible by the existence–uniqueness theorem for linear equations: a constant-coefficient equation $a{y}^{\prime \prime }+b{y}^{\prime }+cy=G(x)$ with $G$ continuous has exactly one solution for any prescribed $y(x_0)$, $y^{\prime }(x_0)$, and it exists on the whole line. (Variation of parameters, which needs only $y_1,y_2$, would also have produced a particular solution.) A trial form that gives $0$ is telling you the trial form was wrong, never that the equation is unsolvable.

    (b) The correct solution. With $y_p=A{x}^2{e}^{-2x}$,

    $$y_p^{\prime }=A\left(2x-2x^2\right)e^{-2x},y_p^{\prime \prime }=A\left(2-4x-4x+4x^2\right)e^{-2x}=A\left(2-8x+4x^2\right)e^{-2x}.$$

    Substituting,

    $$y_p^{\prime \prime }+4y_p^{\prime }+4y_p=A{e}^{-2x}[(2-8x+4x^2)+(8x-8x^2)+4x^2]=2A{e}^{-2x}.$$

    Setting this equal to $6e^{-2x}$ gives $A=3$, so $y_p=3x^2{e}^{-2x}$ and the general solution is

    $$y=\left(c_1+c_{2}x+3x^2\right)e^{-2x}.$$

    Now the initial conditions. $y(0)=c_1=1$. Differentiating, $y^{\prime }=\left(c_2+6x\right)e^{-2x}-2\left(c_1+c_{2}x+3x^2\right)e^{-2x}$, so $y^{\prime }(0)=c_2-2c_1=c_2-2=0$ and $c_2=2$:

    $$y=\left(1+2x+3x^2\right)e^{-2x}.$$

    Check. Write $y=v{e}^{-2x}$ with $v=1+2x+3x^2$. Then $y^{\prime }=(v^{\prime }-2v)e^{-2x}$ and $y^{\prime \prime }=(v^{\prime \prime }-4v^{\prime }+4v)e^{-2x}$, so

    $$y^{\prime \prime }+4y^{\prime }+4y=[(v^{\prime \prime }-4v^{\prime }+4v)+4(v^{\prime }-2v)+4v]e^{-2x}=v^{\prime \prime }{e}^{-2x}=6e^{-2x}.$$

    Also $y(0)=v(0)=1$ and $y^{\prime }(0)=v^{\prime }(0)-2v(0)=2-2=0$. The identity $y^{\prime \prime }+4y^{\prime }+4y=v^{\prime \prime }{e}^{-2x}$ explains the whole episode: any $v$ of degree at most $1$ gives zero (those are the complementary solutions), which is why the particular solution had to carry an $x^2$.

11. 11.Proof · 10 min

    Let $x(t)$ be any solution of the mass–spring equation $m{x}^{\prime \prime }+c{x}^{\prime }+kx=0$ with $m,k>0$ and $c\ge 0$, and define its energy

    $$E(t)=\frac{1}{2}m(x^{\prime }(t){)}^2+\frac{1}{2}k(x(t){)}^2.$$

    Argue directly from the equation; the characteristic roots and the explicit solution formulas are not available in this question.

    1. (a)Prove that $E^{\prime }(t)=-c(x^{\prime }(t){)}^2$, and conclude that $E$ is constant when $c=0$ and nonincreasing when $c>0$.
    2. (b)For $c=0$, prove that every solution satisfies $|x(t)|\le \sqrt{2E(0)/k}$ for all $t$. Then compute $E$ for $x=R\cos (\omega t-\delta )$ with $\omega =\sqrt{k/m}$, and show that the bound is exactly the amplitude $R$.
    3. (c)Prove that if $x(0)=0$ and $x^{\prime }(0)=0$ then $x(t)=0$ for all $t\ge 0$. Deduce that two solutions with the same initial position and the same initial velocity are identical for $t\ge 0$, naming the property of the equation that lets you pass from the first statement to the second.
    Show the solution to question 11

    (a) Energy is dissipated by damping alone. By the chain rule,

    $$E^{\prime }=m{x}^{\prime }{x}^{\prime \prime }+kx{x}^{\prime }=x^{\prime }\left(m{x}^{\prime \prime }+kx\right).$$

    The equation says $m{x}^{\prime \prime }+kx=-c{x}^{\prime }$, so $E^{\prime }=x^{\prime }\cdot (-c{x}^{\prime })=-c(x^{\prime }{)}^2$. When $c=0$ this is identically $0$, so $E$ is constant. When $c>0$ it is $\le 0$ at every instant, so $E$ is nonincreasing — it can only decrease, and it does so exactly while the mass is moving. $■$

    (b) Conserved energy bounds the displacement. With $c=0$, $E(t)=E(0)$ for all $t$. Since $\frac{1}{2}m(x^{\prime }{)}^2\ge 0$,

    $$\frac{1}{2}kx(t{)}^2\le E(t)=E(0)⟹x(t{)}^2\le \frac{2E(0)}{k}⟹|x(t)|\le \sqrt{\frac{2E(0)}{k}}.$$

    Now take $x=R\cos (\omega t-\delta )$ with $\omega =\sqrt{k/m}$. Then $x^{\prime }=-R\omega \sin (\omega t-\delta )$ and

    $$E=\frac{1}{2}m{R}^2{\omega }^2{\sin }^2(\omega t-\delta )+\frac{1}{2}k{R}^2{\cos }^2(\omega t-\delta )=\frac{1}{2}k{R}^2\left({\sin }^2+{\cos }^2\right)=\frac{1}{2}k{R}^2,$$

    using $m{\omega }^2=k$. The bound becomes $\sqrt{2\cdot \frac{1}{2}k{R}^2/k}=R$: the energy inequality reproduces the amplitude exactly, and it is attained whenever the velocity vanishes. $■$

    (c) No energy, no motion — and uniqueness. Suppose $x(0)=0$ and $x^{\prime }(0)=0$. Then $E(0)=0$. Each term of $E$ is a nonnegative multiple of a square, so $E(t)\ge 0$ for all $t$; and by (a), for either $c=0$ or $c>0$, $E$ is nonincreasing, so $E(t)\le E(0)=0$ for $t\ge 0$. Hence $E(t)=0$ for all $t\ge 0$, and then $\frac{1}{2}kx(t{)}^2\le E(t)=0$ forces $x(t)=0$ for all $t\ge 0$.

    Now let $x_1$ and $x_2$ be two solutions with $x_1(0)=x_2(0)$ and $x_1^{\prime }(0)=x_2^{\prime }(0)$. The equation is linear and homogeneous, so by superposition the difference $w=x_1-x_2$ is again a solution, and it satisfies $w(0)=0$, $w^{\prime }(0)=0$. By what was just proved, $w(t)=0$ for all $t\ge 0$, that is $x_1=x_2$ on $t\ge 0$. $■$

    This is the uniqueness half of the existence–uniqueness theorem for this equation, obtained without ever writing down a solution: a mass at rest at equilibrium has no energy, energy can only leave, and without energy it cannot move. Superposition is what turns that one-solution statement into a statement about any two.

12. 12.Computation · 12 min

    Consider $\left(1-x^2\right)y^{\prime \prime }-2x{y}^{\prime }+2y=0$ near $x=0$.

    1. (a)Substitute $y=\sum _{n=0}^{\infty }{c}_n{x}^n$ and derive the recurrence relation for the coefficients.
    2. (b)Show that the recurrence forces $c_3=c_5=c_7=\cdots =0$, so that $y=x$ is a solution, and confirm this directly in the equation.
    3. (c)Find the solution with $y(0)=1$, $y^{\prime }(0)=0$: compute $c_2,c_4,c_6,c_8$, guess and prove a formula for $c_{2k}$, and find the radius of convergence of the series.
    4. (d)Identify the sum in (c) in closed form, using the Maclaurin series of $\ln (1+x)$ and $\ln (1-x)$, and verify that your closed form satisfies $y(0)=1$ and $y^{\prime }(0)=0$.
    Show the solution to question 12

    (a) The recurrence. With $y={\sum }_{n\ge 0}{c}_n{x}^n$ we have $y^{\prime }={\sum }_{n\ge 1}n{c}_n{x}^{n-1}$ and $y^{\prime \prime }={\sum }_{n\ge 2}n(n-1)c_n{x}^{n-2}$. Shift the index in $y^{\prime \prime }$ so that its power is $x^n$: $y^{\prime \prime }={\sum }_{n\ge 0}(n+2)(n+1)c_{n+2}{x}^n$. The other three pieces already carry $x^n$ after multiplying in:

    $$-x^2{y}^{\prime \prime }=-\sum _{n\ge 0}n(n-1)c_n{x}^n,-2x{y}^{\prime }=-\sum _{n\ge 0}2n{c}_n{x}^n,2y=\sum _{n\ge 0}2c_n{x}^n,$$

    where the first two sums may start at $n=0$ because the terms they add are zero. Collecting the coefficient of $x^n$ and setting it to zero,

    $$(n+2)(n+1)c_{n+2}-[n(n-1)+2n-2]c_n=0.$$

    The bracket is $n^2+n-2=(n+2)(n-1)$, so after cancelling $n+2$,

    $$c_{n+2}=\frac{n-1}{n+1}{c}_n,n\ge 0.$$

    (b) The polynomial solution. Take $n=1$: $c_3=\frac{0}{2}{c}_1=0$. Then $c_5=\frac{2}{4}{c}_3=0$, and inductively every odd coefficient beyond $c_1$ vanishes. So the odd part of any solution is just $c_{1}x$, and $y=x$ is a solution. Directly: $y=x$ gives $y^{\prime }=1$, $y^{\prime \prime }=0$, and $(1-x^2)\cdot 0-2x\cdot 1+2x=0$.

    (c) The even solution. The initial conditions give $c_0=y(0)=1$ and $c_1=y^{\prime }(0)=0$, so all odd coefficients are $0$ by (b). The even ones come from $c_{n+2}=\frac{n-1}{n+1}{c}_n$ at $n=0,2,4,6$:

    $$c_2=\frac{-1}{1}{c}_0=-1,c_4=\frac{1}{3}{c}_2=-\frac{1}{3},c_6=\frac{3}{5}{c}_4=-\frac{1}{5},c_8=\frac{5}{7}{c}_6=-\frac{1}{7}.$$

    Guess $c_{2k}=-\frac{1}{2k-1}$ for $k\ge 1$. Induction: it holds for $k=1$, and if $c_{2k}=-\frac{1}{2k-1}$ then, with $n=2k$,

    $$c_{2k+2}=\frac{2k-1}{2k+1}\cdot \left(-\frac{1}{2k-1}\right)=-\frac{1}{2k+1},$$

    which is the formula at $k+1$. Hence

    $$y=1-\sum _{k=1}^{\infty }\frac{x^{2k}}{2k-1}=1-x^2-\frac{x^4}{3}-\frac{x^6}{5}-\frac{x^8}{7}-\cdots$$

    Radius of convergence: with $a_k=x^{2k}/(2k-1)$, the ratio test gives $\left|a_{k+1}/a_k\right|=\frac{2k-1}{2k+1}{x}^2\to x^2$, so the series converges for $|x|<1$ and diverges for $|x|>1$: the radius is $1$. (At $x=\pm 1$ the terms are $1/(2k-1)$, and that series diverges by comparison with the harmonic series.) This is no accident: the leading coefficient $1-x^2$ vanishes at $x=\pm 1$, and a series solution about $0$ cannot see past the nearest such point.

    (d) The closed form. From $\ln (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots$ and $\ln (1-x)=-x-\frac{x^2}{2}-\frac{x^3}{3}-\cdots$, valid for $|x|<1$, subtracting cancels the even powers and doubles the odd ones:

    $$\ln \frac{1+x}{1-x}=2\left(x+\frac{x^3}{3}+\frac{x^5}{5}+\cdots \right)=2\sum _{k=1}^{\infty }\frac{x^{2k-1}}{2k-1}.$$

    Therefore ${\sum }_{k\ge 1}\frac{x^{2k}}{2k-1}=x{\sum }_{k\ge 1}\frac{x^{2k-1}}{2k-1}=\frac{x}{2}\ln \frac{1+x}{1-x}$, and

    $$y=1-\frac{x}{2}\ln \frac{1+x}{1-x},|x|<1.$$

    Check of the initial conditions: $y(0)=1-0=1$. Differentiating, with $\frac{d}{dx}\ln \frac{1+x}{1-x}=\frac{1}{1+x}+\frac{1}{1-x}=\frac{2}{1-x^2}$,

    $$y^{\prime }=-\frac{1}{2}\ln \frac{1+x}{1-x}-\frac{x}{1-x^2},y^{\prime }(0)=0-0=0.$$

    For a full check, differentiate once more: $y^{\prime \prime }=-\frac{1}{1-x^2}-\frac{1+x^2}{(1-x^2{)}^2}$. Then $(1-x^2)y^{\prime \prime }=-1-\frac{1+x^2}{1-x^2}$, $-2x{y}^{\prime }=x\ln \frac{1+x}{1-x}+\frac{2x^2}{1-x^2}$, and $2y=2-x\ln \frac{1+x}{1-x}$. The logarithms cancel and what remains is

    $$-1+2+\frac{-1-x^2+2x^2}{1-x^2}=1+\frac{x^2-1}{1-x^2}=1-1=0.$$

    So the general solution near $0$ is $c_0\left(1-\frac{x}{2}\ln \frac{1+x}{1-x}\right)+c_{1}x$: one solution is a polynomial, the other is not elementary until the series is recognised — exactly the situation the power-series method exists for.