Quiz 1 Practice

Friday, September 11, 2026 — Lectures 1 and 2. Confirmed on the course page, checked September 9.

Homework 1 · Homework 2

Original practice problems modeled on the assigned homework, not a prediction of the quiz. Work each problem before opening its solution. Problems 21–22 are mixed checks; 19–20 cover Lecture 2’s assigned piecewise reading.

Methods available

Lecture 1 - Introduction: differentiate to verify; recover coefficients/forcing from a known function; direct and successive integration; autonomous equations, direction fields, and equilibria.

Lecture 2 - Linear First-Order Differential Equations: standard linear form; homogeneous exponential/power-law solutions; integrating factor; variation of parameter; constant shift to equilibrium. Its assigned pp. 25–26 add solving and matching on successive intervals. HW 2 also assigns the linear-drag model on pp. 78–79.

Use product/chain rules, the fundamental theorem of calculus, substitution, and exponential/logarithm algebra. No separation of variables, general superposition theorem, Euler method, Laplace transforms, or nonlinear stability derivative test is needed here.

Formula check

$$y^{\prime }=v(t)⟹y(t)=y_0+{\int }_{t_0}^{t}v(s)ds.$$ $$P(t)={\int }_{t_0}^{t}p(s)ds,y^{\prime }+py=g⟹y(t)=e^{-P(t)}\left[y_0+{\int }_{t_0}^t{e}^{P(s)}g(s)ds\right].$$

Variation of parameter: $y=C(t)e^{-P(t)}$ gives $C^{\prime }=e^{P}g$.

$$y^{\prime }=ay+b,a\ne 0:y_{\ast }=-b/a,y(t)=y_{\ast }+(y_0-y_{\ast })e^{a(t-t_0)}.$$

For $t>0$, $y^{\prime }=(a/t)y$ gives $y=C{t}^a$.

1. Classification

Classify each as linear or nonlinear. For linear equations, give $p(t),g(t)$ and homogeneous/nonhomogeneous status.

$$\begin{array}{rl}\text{(a)}& (1+t^2)y^{\prime }+2ty=\cos t,\\ \text{(b)}& y{y}^{\prime }+y=t,\\ \text{(c)}& e^t{y}^{\prime }+t^{2}y=0,\\ \text{(d)}& \frac{y^{\prime }}{y}=t+\frac{1}{y}(y\ne 0).\end{array}$$

Also give the order of $(y^{\prime \prime }{)}^3+\sin (y^{\prime })=t$.

Solution

(a) Divide by $1+t^2$:

$$p=\frac{2t}{1+t^2},g=\frac{\cos t}{1+t^2}.$$

Linear, nonhomogeneous.

(b) Nonlinear: $y$ multiplies $y^{\prime }$. For $y\ne 0$, $y^{\prime }=t/y-1$ still has nonlinear dependence on $y$.

(c) Divide by $e^t$: $p=t^2{e}^{-t}$, $g=0$. Linear, homogeneous.

(d) Multiply by $y$: $y^{\prime }-ty=1$. Linear, nonhomogeneous on the original domain $y\ne 0$; $p=-t$, $g=1$. Do not discard the domain restriction.

The last equation is second order; the cube does not change the derivative order.

2. Verify a proposed solution

Find $k$ so that $y=e^{k\sin 3t}$ solves $y^{\prime }-6(\cos 3t)y=0$. Then find $C$ so that $y=C{e}^{k\sin 3t}$ satisfies $y(\pi /6)=5$.

Solution

Differentiate by the chain rule:

$$y^{\prime }=3k\cos 3t{e}^{k\sin 3t}.$$

The residual is $(3k-6)\cos 3t{e}^{k\sin 3t}$, which vanishes identically only for $k=2$. At $t=\pi /6$, $\sin 3t=1$, so $C{e}^2=5$.

$$\boxed{C=5e^{-2},y(t)=5e^{2(\sin 3t-1)}.}$$

This is verification by substitution; no nonlinear solution technique is needed.

3. Recover the equation

The function $y=2e^{-3t}+\cos t$ solves $y^{\prime }+3y=g(t)$, $y(0)=y_0$. Find $g,y_0$.

Solution

$$y^{\prime }=-6e^{-3t}-\sin t,$$

so

$$\boxed{g(t)=y^{\prime }+3y=3\cos t-\sin t,y_0=3.}$$

4. Successive integration

Solve $y^{\prime \prime \prime }=6$ with $y(0)=1$, $y^{\prime }(0)=-2$, $y^{\prime \prime }(0)=0$. How many constants occur in the general solution?

Solution

Integrate successively:

$$\begin{array}{rl}{y}^{\prime \prime }& =6t+C_1,\\ y^{\prime }& =3t^2+C_{1}t+C_2,\\ y& =t^3+\frac{C_1}{2}{t}^2+C_{2}t+C_3.\end{array}$$

There are three constants. The data give $C_1=0$, $C_2=-2$, $C_3=1$:

$$\boxed{y=t^3-2t+1.}$$

5. Direction field

For $y^{\prime }=y(3-y)$, determine whether it is autonomous, find the equilibria, make a slope-sign table, and sketch the trajectory through $y(0)=1$. Do not solve the nonlinear equation explicitly.

Solution

Autonomous; equilibria at $y=0,3$.

Region	$y^{\prime }$	Motion as $t$ increases
$y<0$	$-$	Down
$0<y<3$	$+$	Up
$y>3$	$-$	Down

For a field sketch, repeat the same slopes along each horizontal line. Example slopes: $f(-1)=-4$, $f(1)=2$, $f(2)=2$, $f(4)=-4$.

The trajectory through $(0,1)$ rises toward $y=3$ and flattens there. The field points toward $3$ from either side and away from $0$. This uses slopes and equilibria, without separation of variables.

6. Equilibrium or zero slope?

Find all constant solutions of (a) $y^{\prime }=t(y+2)$ and (b) $y^{\prime }=ty+2$. Then construct an autonomous equation with equilibria $-1,2$ whose slopes are positive between them and negative outside.

Solution

(a) $y=-2$ makes the right-hand side zero for every $t$.

(b) No constant works for every $t$. The curve $y=-2/t$ has zero field slope at its points, but its derivative is $2/t^2\ne 0$; it is not an equilibrium solution.

One construction is

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

Its roots and factor signs give exactly the requested behavior.

7. Homogeneous IVPs

Solve (a) $y^{\prime }=-3y$, $y(2)=-4$; (b) $t{y}^{\prime }-3y=0$, $y(2)=4$, $t>0$; (c) $y^{\prime }+(1+t^2)y=0$, $y(0)=0$.

Solution

(a) Use the exponential form from Lecture 2:

$$\boxed{y=-4e^{-3(t-2)}.}$$

The value increases toward zero; its magnitude decreases.

(b) $y^{\prime }=(3/t)y$, so use the power-law form $y=C{t}^3$. Since $8C=4$,

$$\boxed{y=\frac{1}{2}{t}^3,t>0.}$$

(c) $P=t+t^3/3$ and $y=C{e}^{-P}$. At zero, $C=0$, so $\boxed{y\equiv 0}$.

8. Integrating factor with a nonzero initial time

Solve

$$y^{\prime }-\frac{2t}{1+t^2}y=4t,y(1)=2.$$

Use an integrating factor normalized at $t_0=1$.

Solution

$$P(t)={\int }_1^t\frac{-2s}{1+s^2}ds=\ln \frac{2}{1+t^2},e^P=\frac{2}{1+t^2}.$$

Then

$$\begin{array}{rl}y(t)& =\frac{1+t^2}{2}\left[2+{\int }_1^t\frac{8s}{1+s^2}ds\right]\\ & =\boxed{\frac{1+t^2}{2}\left[2+4\ln \frac{1+t^2}{2}\right]}.\end{array}$$

The logarithm vanishes at $t=1$, giving $y(1)=2$. Differentiating the final expression gives $y^{\prime }-2ty/(1+t^2)=4t$.

9. Variation of parameter

Solve $2y^{\prime }+4y=e^t$, $y(0)=1$ by variation of parameter.

Solution

First divide by $2$: $y^{\prime }+2y=\frac{1}{2}{e}^t$. A nonzero homogeneous solution is $e^{-2t}$. Set $y=C(t)e^{-2t}$:

$$C^{\prime }{e}^{-2t}=\frac{1}{2}{e}^t⟹C^{\prime }=\frac{1}{2}{e}^{3t}⟹C=\frac{1}{6}{e}^{3t}+K.$$

Hence $y=\frac{1}{6}{e}^t+K{e}^{-2t}$. The initial condition gives $K=5/6$:

$$\boxed{y=\frac{1}{6}{e}^t+\frac{5}{6}{e}^{-2t}.}$$

10. Normalize before integrating

Find the general solution of $t{y}^{\prime }+2y=t^3$ on $t>0$, then impose $y(1)=0$.

Solution

$$y^{\prime }+\frac{2}{t}y=t^2,P=2\ln t,\mu =t^2.$$

Thus

$$(t^{2}y{)}^{\prime }=t^4⟹t^{2}y=\frac{t^5}{5}+C.$$

The general solution is $y=t^3/5+C/t^2$. Since $C=-1/5$,

$$\boxed{y=\frac{t^3-t^{-2}}{5},t>0.}$$

11. A forcing matched to the coefficient

Solve $y^{\prime }+2ty=6t$, $y(0)=-1$, and find its limit as $t\to \infty$.

Solution

Here $P=t^2$ and the integrating-factor integral is

$${\int }_0^{t}6s{e}^{s^2}ds=3(e^{t^2}-1).$$

Therefore

$$y=e^{-t^2}\left[-1+3(e^{t^2}-1)\right]=\boxed{3-4e^{-t^2}}.$$

The limit is $3$. The substitution is $u=s^2$, $du=2sds$.

12. Read the coefficient from a curve

A solution of $t{y}^{\prime }-\alpha y=0$ on $t>0$ passes through $(1,2)$ and $(3,18)$. Find $\alpha$ and $y(t)$.

Solution

The power-law form gives $y=C{t}^{\alpha }$. At $t=1$, $C=2$. At $t=3$,

$$18=2\cdot 3^{\alpha }⟹3^{\alpha }=9.$$

Thus $\boxed{\alpha =2,y=2t^2}$.

13. Match qualitative behavior

All four solutions start at $y(0)=2$. Match each equation to its behavior: exponential decay; periodic variation; exponential growth; increasing with recurring horizontal tangents.

$$\begin{array}{rl}\text{(a)}& y^{\prime }+2y=0,\\ \text{(b)}& y^{\prime }+(\cos t)y=0,\\ \text{(c)}& y^{\prime }-(1-\cos t)y=0,\\ \text{(d)}& y^{\prime }-y=0.\end{array}$$

Solution

Use $y=2\exp (-{\int }_0^{t}p(s)ds)$:

Equation	Solution	Behavior
(a)	$2e^{-2t}$	Exponential decay
(b)	$2e^{-\sin t}$	Periodic variation
(c)	$2e^{t-\sin t}$	Increasing, horizontal tangents at $t=2\pi n$
(d)	$2e^t$	Exponential growth

For (c), $y^{\prime }=(1-\cos t)y\ge 0$. The slope vanishes at isolated times, so the curve has repeated flattening without decreasing.

14. Long-term behavior

Solve and find the limit:

$$y^{\prime }+(1+\cos t)y=2(1+\cos t),y(0)=5.$$

Solution

$$P=t+\sin t,{\int }_0^{t}2(1+\cos s)e^{s+\sin s}ds=2(e^{t+\sin t}-1).$$

So

$$\boxed{y=2+3e^{-t-\sin t}}.$$

Since $\sin t$ stays bounded, $t+\sin t\to \infty$ and $y\to 2$.

15. What controls the limit?

(a) A nonconstant solution satisfies $y^{\prime }+\lambda y=4$. For which real $\lambda$ does it have a finite limit as $t\to \infty$?

(b) Does $p(t)>0$ alone force every solution of $y^{\prime }+p(t)y=0$ to tend to zero? Test $p(t)=e^{-t}$, $y(0)=2$.

Solution

(a) If $\lambda \ne 0$, the equilibrium shift gives

$$y=\frac{4}{\lambda }+C{e}^{-\lambda t},C\ne 0.$$

For $\lambda >0$, the limit is $4/\lambda$. For $\lambda <0$, the nonconstant exponential term is unbounded. For $\lambda =0$, $y=4t+C$, also unbounded. Thus $\boxed{\lambda >0}$ is necessary and sufficient here.

(b)

$$P(t)={\int }_0^t{e}^{-s}ds=1-e^{-t},y(t)=2e^{e^{-t}-1}⟶\boxed{2/e}.$$

Positive $p$ decreases the magnitude. Decay to zero requires the accumulated integral to diverge to $+\infty$ for nonzero initial data.

16. Approach to equilibrium

Solve $y^{\prime }=-2y+12$, $y(1)=3$. Find the equilibrium and the first time $t\ge 1$ when $y$ has completed 90% of the change from its initial value toward equilibrium.

Solution

The equilibrium is $y_{\ast }=6$. Set $z=y-6$, giving $z^{\prime }=-2z$, $z(1)=-3$:

$$\boxed{y=6-3e^{-2(t-1)}}.$$

After 90% of the change, the remaining gap is $0.1(6-3)=0.3$:

$$3e^{-2(t-1)}=0.3⟹\boxed{t=1+\frac{\ln 10}{2}}.$$

17. Cooling

(a) A metal sample has temperature $T(t)=18+72e^{-t/5}$ in degrees Celsius. Under Newton’s cooling law, find its initial temperature, room temperature, and rate constant.

(b) Another sample starts at $80^{\circ }\mathrm{C}$ in a $20^{\circ }\mathrm{C}$ room and cools to $50^{\circ }\mathrm{C}$ in 10 min. Derive its temperature using the equilibrium shift. When does it reach $35^{\circ }\mathrm{C}$?

Solution

(a) $T(0)=90^{\circ }\mathrm{C}$, the room is $18^{\circ }\mathrm{C}$, and $k=1/5$ per time unit.

(b) $T^{\prime }=-k(T-20)$. Set $z=T-20$:

$$z^{\prime }=-kz,z(0)=60,T=20+60e^{-kt}.$$

At 10 min, $30=60e^{-10k}$, so $k=\ln 2/10$ per minute. Thus

$$\boxed{T(t)=20+602^{-t/10}}.$$

At $35^{\circ }\mathrm{C}$, the excess temperature is $15=60/4$, so $\boxed{t=20\text{ min}}$ from the start.

18. Free fall followed by linear drag

Use upward-positive velocity and $g=10m/{\mathrm{s}}^2$. A body is released from rest; after 3 s, a parachute opens. After opening, $v^{\prime }=-10-2v$. Find the velocity at opening and 2 s after opening. Derive the second phase by shifting to equilibrium.

Solution

Before opening, $v^{\prime }=-10$, $v(0)=0$, so $v=-10t$. At opening, $v=-30m/s$ (speed $30m/s$).

Reset time to $\tau =0$ at opening. The equilibrium velocity is $v_{\ast }=-5$. Set $z=v+5$:

$$z^{\prime }=-2z,z(0)=-25,v(\tau )=-5-25e^{-2\tau }.$$

Two seconds later,

$$\boxed{v(2)=-5-25e^{-4}m/s}.$$

The velocity moves upward toward $-5$ while the body continues downward; its speed decreases.

19. Piecewise forcing

Find a continuous solution on $[0,2]$:

$$y^{\prime }+y=g(t),y(0)=0,g(t)=\{\begin{array}{ll}2,& 0\le t<1,\\ 0,& 1\le t\le 2.\end{array}$$

Must the derivative be continuous at $t=1$?

Solution

On $0\le t<1$, shift to equilibrium $2$:

$$y=2-2e^{-t}.$$

The matching value is $y(1)=2(1-e^{-1})$. On the second interval solve $y^{\prime }+y=0$ with that initial value:

$$\boxed{y(t)=\{\begin{array}{ll}2(1-e^{-t}),& 0\le t<1,\\ 2(1-e^{-1})e^{-(t-1)},& 1\le t\le 2.\end{array}}$$

The one-sided derivatives are $2e^{-1}$ and $-2(1-e^{-1})$, so there is a corner. The function is continuous and solves the DE on each open piece; it has no two-sided derivative at the switch.

20. Piecewise coefficient-matched forcing

Find a continuous solution on $[0,2\pi ]$:

$$y^{\prime }+(\sin t)y=g(t),y(0)=2,g(t)=\{\begin{array}{ll}\sin t,& 0\le t\le \pi ,\\ -\sin t,& \pi <t\le 2\pi .\end{array}$$

Solution

On the first interval, $P(t)=1-\cos t$. The integral of $e^P\sin t$ is $e^P$, so

$$y=1+e^{\cos t-1},y(\pi )=1+e^{-2}.$$

For the second interval, normalize at $\pi$:

$$P_2(t)={\int }_{\pi }^t\sin sds=-\cos t-1,$$

and

$${\int }_{\pi }^t-e^{P_2(s)}\sin sds=1-e^{P_2(t)}.$$

Thus

$$\boxed{y(t)=\{\begin{array}{ll}1+e^{\cos t-1},& 0\le t\le \pi ,\\ -1+(2+e^{-2})e^{\cos t+1},& \pi <t\le 2\pi .\end{array}}$$

Both pieces give $1+e^{-2}$ at $\pi$. Here both one-sided derivatives are zero because $\sin \pi =0$; a piecewise definition does not necessarily create a corner.

21. Mixed check A

Without looking up a method, solve

$$(1+t^2)y^{\prime }+2ty=2t,y(0)=3.$$

Give the equilibrium, limit, and whether the solution increases or decreases for $t>0$.

Solution

Normalize, then use $P=\ln (1+t^2)$ and $\mu =1+t^2$:

$$((1+t^2)y{)}^{\prime }=2t⟹(1+t^2)y=t^2+C.$$

$C=3$, so

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

The equilibrium is $1$, and $y\to 1$. Since $y^{\prime }=-4t/(1+t^2{)}^2<0$ for $t>0$, it decreases.

22. Mixed check B

Solve $y^{\prime }+2y=e^{-t}$, $y(1)=0$, by variation of parameter. Check both the equation and the initial condition.

Solution

Set $y=C(t)e^{-2t}$. Substitution gives $C^{\prime }{e}^{-2t}=e^{-t}$, so $C^{\prime }=e^t$ and $C=e^t+K$.

Thus $y=e^{-t}+K{e}^{-2t}$. At $t=1$, $K=-e$:

$$\boxed{y=e^{-t}-e^{1-2t}}.$$

Verification:

$$y^{\prime }+2y=(-e^{-t}+2e^{1-2t})+2(e^{-t}-e^{1-2t})=e^{-t},$$

and $y(1)=e^{-1}-e^{-1}=0$.

Scope and source map

The Quiz 1 announcement explicitly names Lectures 1–2. The matching regular assignments are HW 1–2; the homework page preserves the instructor’s intended order and special instructions. The syllabus says quizzes draw on homework assigned at least one day earlier; the website does not establish the exact in-class completion date of every topic.

Practice	Homework models
1	1.2.3; 2.1.1, 3, 5, 6, 9
2–3	1.2.5, 7, 8, 9, 13, 14
4	1.2.10, 23
5–6	1.3.5, 1, 2, 4, 10
7–11	2.2.1, 3, 4, 7, 11, 14, 20, 21
12–15	2.2.25, 27, 28, 36, 37, 39
16–18	2.2.29(c); 2.3.19, 21; 2.9.18(a,b)
19–20	2.2.41 and assigned pp. 25–26
21–22	Mixed linear-IVP practice

Reviewed: Lecture 1, direction-field supplement, Lecture 2, linked homework scans and answer sheets, piecewise reading, linear-drag reading, background sheet, and homework approach.

The linked Test 1 preparation sheet concerns a later assessment and does not narrow Quiz 1. Its calculator/formula-sheet rules are not assumed to apply to this quiz. Lecture 3 and later methods, and the separately labeled extra-credit tasks, are not used in this set.