Chapter 7 · Introduction to Differential Equations

7.1 Basics of Differential Equations

Equations whose unknown is a function: what they are, how to check a solution, and how one initial condition picks a single curve out of a whole family.

A freshly poured cup of coffee cools quickly at first and then more and more slowly. A bacterial colony grows faster the bigger it already is. A painkiller leaves the bloodstream at a rate that depends on how much of it is still there. In each case we do not know a formula for the quantity itself — but we do know a rule for how fast it changes. Written in symbols, such a rule is an equation that involves an unknown function and its derivatives: a differential equation.

Differential equations are the language in which physics, engineering, biology, medicine and economics state their laws. In this section we learn to read that language: to recognise a differential equation and its order, to check whether a proposed function really is a solution, and to see why a differential equation usually has infinitely many solutions — and how one extra piece of information (an initial condition) singles out the one we want. We finish with the simplest and most important model of all, $y^\prime=ky$.

What Is a Differential Equation?

In algebra the unknown in an equation is a number: $x^2-5x+6=0$ asks “which numbers $x$ make this true?”. In a differential equation the unknown is a function $y=y(x)$, and the equation links $y$ to one or more of its derivatives. For example $$y^\prime=2x,\qquad \frac{dy}{dt}=-0.25\,y,\qquad y^{\prime\prime}+9y=0,\qquad \frac{dT}{dt}=-k\left(T-22\right).$$ The first says “the slope of $y$ at every $x$ equals $2x$”; the second says “$y$ decreases at a rate equal to a quarter of its current value”.

📘 Definition 7.1 — Differential equation; order

A differential equation is an equation involving an unknown function $y=y(x)$ and one or more of its derivatives. (When the unknown depends on a single variable, as throughout this chapter, it is called an ordinary differential equation.)

The order of a differential equation is the order of the highest derivative that appears in it. So $y^\prime=2x$ is first-order, $y^{\prime\prime}+9y=0$ is second-order, and $y^{\prime\prime\prime}-4y^\prime=x$ is third-order.

✏️ Example 1 — Finding the order

State the order of each differential equation.

a $y^\prime=3x^2-1$
b $y^{\prime\prime}+4y=0$
c $\left(\dfrac{dy}{dx}\right)^3+y=x$
d $\dfrac{d^3y}{dx^3}-x\dfrac{dy}{dx}=e^x$
e $t^2y^{\prime\prime}+\left(y^\prime\right)^4=y$

💡 Show solution▶

a The highest derivative is $y^\prime$: first-order.

b The highest derivative is $y^{\prime\prime}$: second-order.

c Only the first derivative appears — it is cubed, but a power of a derivative does not raise the order. First-order.

d The highest derivative is $\dfrac{d^3y}{dx^3}$: third-order.

e $y^{\prime\prime}$ is the highest derivative (the fourth power of $y^\prime$ is irrelevant): second-order.

⚠️ Order is not the same as power

The order counts how many times $y$ has been differentiated, not the exponent on a derivative. $\left(y^\prime\right)^5=y$ is first-order; $y^{\prime\prime}=1$ is second-order. Also, “first-order” tells you how many constants to expect in the general solution (one), as we shall see below.

Verifying a Solution

📘 Definition 7.2 — Solution of a differential equation

A solution of a differential equation on an interval $I$ is a function $y=f(x)$ that is differentiable enough times on $I$ (as many times as the order) and that turns the equation into an identity — a statement true for every $x$ in $I$ — when $f$ and its derivatives are substituted for $y$ and its derivatives.

Finding solutions can be hard, but checking one is always routine: differentiate the candidate, substitute, and simplify. If the left side minus the right side (the residual) simplifies to $0$ for every $x$, the candidate is a solution. If the residual is non-zero for even one $x$ in the interval, it is not.

🧭 Problem-solving strategy — verifying a solution

Compute every derivative of the candidate that appears in the equation.
Substitute the candidate and its derivatives into the left side and simplify.
Do the same for the right side.
If the two sides agree for every $x$ in the interval, it is a solution; otherwise it is not. (If an initial condition is given, check that too.)

✏️ Example 2 — Verifying a first-order solution

Verify that $y=e^{-3x}+2x+3$ is a solution of $y^\prime+3y=6x+11$, and check whether it also satisfies the initial condition $y(0)=4$.

💡 Show solution▶

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

Substitute $$\begin{aligned}y^\prime+3y&=\left(-3e^{-3x}+2\right)+3\left(e^{-3x}+2x+3\right)\\&=-3e^{-3x}+2+3e^{-3x}+6x+9\\&=6x+11.\end{aligned}$$ The left side equals the right side for every real $x$, so $y$ is a solution on $(-\infty,\infty)$.

Initial condition $y(0)=e^{0}+0+3=4$. ✓ So $y$ solves the initial-value problem $y^\prime+3y=6x+11$, $y(0)=4$.

✏️ Example 3 — Verifying a second-order solution

Show that $y=xe^x$ is a solution of $y^{\prime\prime}-2y^\prime+y=0$. Is $y=x^2e^x$ also a solution?

💡 Show solution▶

Derivatives By the product rule, $y^\prime=e^x+xe^x=(x+1)e^x$ and $y^{\prime\prime}=e^x+(x+1)e^x=(x+2)e^x$.

Substitute $$y^{\prime\prime}-2y^\prime+y=\big[(x+2)-2(x+1)+x\big]e^x=0\cdot e^x=0.$$ So $y=xe^x$ is a solution for all $x$.

Second candidate For $y=x^2e^x$: $y^\prime=\left(x^2+2x\right)e^x$, $y^{\prime\prime}=\left(x^2+4x+2\right)e^x$, so $$y^{\prime\prime}-2y^\prime+y=\big[x^2+4x+2-2x^2-4x+x^2\big]e^x=2e^x.$$ The residual $2e^x$ is never $0$, so $y=x^2e^x$ is not a solution.

Pick an equation and a candidate. The blue curve is the left side with the candidate substituted; the green dashed curve is the right side. A true solution makes them the same curve — the red gap (the residual) is zero everywhere. Press ▶ Play to sweep $x$ across the interval.

Equation and candidate:$x$:

$x$—

candidate $y(x)$—

left side—

right side—

residual (left − right)—

General Solutions, Particular Solutions and Initial-Value Problems

The simplest differential equations have the form $y^\prime=f(x)$. Solving one is exactly the antidifferentiation of Calculus I §4.1: $y=\int f(x)\,dx$. For instance $y^\prime=2x$ has solutions $y=x^2$, $y=x^2+5$, $y=x^2-\pi$, … — every function $y=x^2+C$. One differential equation, infinitely many solutions, one for each value of the arbitrary constant $C$. Their graphs form a family of solutions: vertical shifts of one parabola that fill the plane without ever crossing.

📘 Definition 7.3 — General and particular solutions

A general solution of a differential equation is an expression, containing arbitrary constants, that describes every solution (on a given interval). An $n$th-order equation typically needs $n$ independent constants: $y=x^2+C$ for $y^\prime=2x$; $y=C_1\cos3x+C_2\sin3x$ for $y^{\prime\prime}+9y=0$.

A particular solution is one specific member of the family, obtained by giving each constant a definite value. Its graph is a solution curve.

📘 Definition 7.4 — Initial-value problem

An initial-value problem (IVP) is a differential equation together with one or more initial conditions that fix the value of the unknown (and, for higher order, of its derivatives) at a single point: $$y^\prime=f(x,y),\quad y(x_0)=y_0\qquad\text{or}\qquad y^{\prime\prime}=f(x,y,y^\prime),\quad y(x_0)=y_0,\ y^\prime(x_0)=v_0.$$ A first-order equation needs one initial condition; a second-order equation needs two. Geometrically, the condition $y(x_0)=y_0$ says “the solution curve passes through the point $(x_0,y_0)$”.

🧭 Problem-solving strategy — solving an IVP

Find the general solution (with its constant(s) $C$).
Substitute the initial condition(s) and solve for the constant(s).
Write the particular solution with the numerical constants in place.
Check: it satisfies the equation and the initial condition(s).

✏️ Example 4 — Solving an initial-value problem

Solve $y^\prime=3e^x+x^2-4$, $y(0)=5$.

💡 Show solution▶

General solution Integrate both sides with respect to $x$: $$y=\int\left(3e^x+x^2-4\right)dx=3e^x+\frac{x^3}{3}-4x+C.$$

Use $y(0)=5$ $5=3e^0+0-0+C=3+C$, so $C=2$.

Particular solution $$y=3e^x+\frac{x^3}{3}-4x+2.$$

Check $y^\prime=3e^x+x^2-4$ ✓ and $y(0)=3+0-0+2=5$ ✓.

The grey curves are members of the general solution. Click or drag anywhere on the graph to choose an initial point $(x_0,y_0)$: exactly one curve of the family passes through it — the red particular solution — and its slope there (blue segment) is the slope the equation prescribes. ▶ Play slides the initial point up and down so you can watch $C$ change.

Equation:

initial point $(x_0,y_0)$—

general solution—

constant $C$—

slope $y^\prime$ at the point—

⚠️ Analysis — what the family picture tells us

1. One point, one curve. For each of these equations, through every point of the plane there passes exactly one solution curve; the curves never cross. (Section 7.2 states the theorem behind this.) That is why one initial condition is enough to pin down a first-order solution.

2. The constant is not always added. For $y^\prime=2x$ the family is $x^2+C$ (shifts), but for $y^\prime=y$ it is $Ce^x$ (vertical stretches) and for $y^\prime=x+y$ it is $Ce^x-x-1$. Never assume the constant appears as “$+C$” — find it by substituting the initial condition into the correct general solution.

3. $y=0$ can be a member. For $y^\prime=y$ and $y^\prime=-2y$, the choice $C=0$ gives the constant solution $y=0$, a horizontal line. Constant solutions like this are the equilibrium solutions of Section 7.2.

Motion Problems and the First Model $y^\prime=ky$

Newton's second law turns motion into a differential equation: if $s(t)$ is position, then velocity is $v=s^\prime$ and acceleration is $a=v^\prime=s^{\prime\prime}$. Knowing the acceleration gives the second-order equation $s^{\prime\prime}=a(t)$; integrating twice and using the initial velocity and initial position solves it.

✏️ Example 5 — From acceleration to position

A ball is thrown straight up from a balcony $2$ m above the ground with initial velocity $19.6$ m/s. Ignoring air resistance its acceleration is $-9.8$ m/s². Find its height $s(t)$, its maximum height, and the time it hits the ground.

💡 Show solution▶

Set up IVP: $s^{\prime\prime}(t)=-9.8$, $s^\prime(0)=19.6$, $s(0)=2$.

Velocity $v(t)=s^\prime(t)=\int-9.8\,dt=-9.8t+C_1$; $v(0)=19.6$ gives $C_1=19.6$: $v(t)=-9.8t+19.6$.

Position $s(t)=\int(-9.8t+19.6)\,dt=-4.9t^2+19.6t+C_2$; $s(0)=2$ gives $C_2=2$: $$s(t)=-4.9t^2+19.6t+2\ \text{m}.$$

Maximum height At the top $v(t)=0$: $t=2$ s, and $s(2)=-19.6+39.2+2=21.6$ m.

Landing $s(t)=0$: $4.9t^2-19.6t-2=0$, so $t=\dfrac{19.6+\sqrt{19.6^2+4(4.9)(2)}}{9.8}\approx4.10$ s (the negative root is rejected).

Check $s^{\prime\prime}(t)=-9.8$ ✓, $s^\prime(0)=19.6$ ✓, $s(0)=2$ ✓.

The motion problem had a rate that depended only on $t$. The models in our opening paragraph are different: the rate of change of a quantity depends on the quantity itself. The simplest such law says the rate is proportional to the amount present: $$\frac{dy}{dt}=ky\qquad(k\ \text{a constant}).$$ If $k\gt0$ this is exponential growth (bacteria, compound interest); if $k\lt0$ it is exponential decay (radioactive material, a drug being cleared by the kidneys, the temperature gap in Newton's law of cooling).

📐 Theorem 7.1 — Solutions of $y^\prime=ky$

Let $k$ be a constant and let $y$ be a differentiable function on an interval $I$ containing $0$. Then $y^\prime=ky$ on $I$ if and only if $$y(t)=Ce^{kt}\quad\text{for some constant }C,$$ and the constant is the initial value: $C=y(0)$. So the IVP $y^\prime=ky$, $y(0)=y_0$ has the unique solution $y=y_0e^{kt}$.

Why. If $y=Ce^{kt}$ then $y^\prime=kCe^{kt}=ky$. Conversely, if $y^\prime=ky$, then $$\frac{d}{dt}\left[y\,e^{-kt}\right]=y^\prime e^{-kt}-kye^{-kt}=(ky-ky)e^{-kt}=0$$ on $I$, so $y\,e^{-kt}$ is a constant $C$ (a function with zero derivative on an interval is constant — a consequence of the Mean Value Theorem). Hence $y=Ce^{kt}$.

✏️ Example 6 — Drug elimination

A patient receives a $200$ mg injection of a drug. The body removes the drug at a rate proportional to the amount present: $\dfrac{dA}{dt}=-0.25A$, with $t$ in hours. (a) Find $A(t)$. (b) How much remains after $4$ hours? (c) When does the amount fall to $50$ mg?

💡 Show solution▶

a By Theorem 7.1 with $k=-0.25$ and $A(0)=200$: $$A(t)=200e^{-0.25t}\ \text{mg}.$$

b $A(4)=200e^{-1}\approx73.6$ mg.

c $200e^{-0.25t}=50\iff e^{-0.25t}=\tfrac14\iff -0.25t=\ln\tfrac14=-\ln4\iff t=4\ln4\approx5.55$ hours.

Check $A^\prime(t)=200(-0.25)e^{-0.25t}=-0.25A(t)$ ✓, $A(0)=200$ ✓. Notice the amount halves every $4\ln2\approx2.77$ h — a fixed half-life, the signature of exponential decay.

⚠️ Common mistakes

Writing the solution of $y^\prime=ky$ as $y=e^{kt}+C$. Check it: $y^\prime=ke^{kt}$, but $ky=ke^{kt}+kC$ — these differ unless $C=0$. The constant multiplies: $y=Ce^{kt}$.
Forgetting units. $k$ has units of “per unit time” (here h$^{-1}$), so $kt$ is dimensionless, as an exponent must be.
Confusing “solution” with “initial condition”: $y(0)=5$ is not a solution; it is a condition the solution must satisfy.

Exercises

Order and verification. For parts 1–4 state the order of the equation. For parts 5–10 decide, by substitution, whether the given function is a solution.

1 $y^{\prime\prime\prime}-4y^\prime+y=x$
2 $\left(y^\prime\right)^2+y=\sin x$
3 $x^2y^{\prime\prime}+xy^\prime+y=0$
4 $\dfrac{d^4y}{dx^4}=y$
5 $y=3e^{-2x}$ for $y^\prime+2y=0$
6 $y=x^2+2x+2$ for $y^\prime=y-x^2$
7 $y=\sin3x$ for $y^{\prime\prime}+9y=0$
8 $y=\dfrac{1}{1+x}$ for $y^\prime=-y^2$ on $(-1,\infty)$
9 $y=e^x-x-1$ for $y^\prime=x+y$
10 Which of $y=e^x$, $y=e^{2x}$, $y=e^{3x}$ are solutions of $y^{\prime\prime}-3y^\prime+2y=0$?

💡 Show solutions▶

1 Highest derivative $y^{\prime\prime\prime}$: third-order.

2 Only $y^\prime$ appears (squared): first-order.

3 Highest derivative $y^{\prime\prime}$: second-order.

4 Fourth-order.

5 $y^\prime=-6e^{-2x}$; $y^\prime+2y=-6e^{-2x}+6e^{-2x}=0$. Yes, a solution.

6 $y^\prime=2x+2$ and $y-x^2=2x+2$. Yes.

7 $y^\prime=3\cos3x$, $y^{\prime\prime}=-9\sin3x$; $y^{\prime\prime}+9y=-9\sin3x+9\sin3x=0$. Yes.

8 $y=(1+x)^{-1}$, $y^\prime=-(1+x)^{-2}$, and $-y^2=-(1+x)^{-2}$. Yes, on $(-1,\infty)$ (where $y$ is defined and differentiable).

9 $y^\prime=e^x-1$ and $x+y=x+e^x-x-1=e^x-1$. Yes.

10 For $y=e^{rx}$: $y^{\prime\prime}-3y^\prime+2y=\left(r^2-3r+2\right)e^{rx}=(r-1)(r-2)e^{rx}$. This is $0$ for $r=1$ and $r=2$, so $e^x$ and $e^{2x}$ are solutions; for $r=3$ the residual is $2e^{3x}\ne0$, so $e^{3x}$ is not.

General solutions and initial-value problems. Find the general solution (parts 1–3) or solve the IVP (parts 4–10).

1 $y^\prime=2x+3$
2 $y^\prime=e^{2x}$
3 $y^\prime=\dfrac1x$, $x\gt0$
4 $y^\prime=6x^2-4$, $y(0)=1$
5 $y^\prime=\cos x$, $y(\pi)=2$
6 $\dfrac{dy}{dt}=te^{t}$, $y(0)=0$
7 $y^\prime=\dfrac{1}{1+x^2}$, $y(1)=0$
8 $y^{\prime\prime}=12x$, $y(0)=1$, $y^\prime(0)=-2$
9 Verify that $y=Ce^{-x}+x-1$ solves $y^\prime+y=x$ for every $C$, then find the solution with $y(0)=3$.
10 The general solution of $y^{\prime\prime}+y=0$ is $y=C_1\cos x+C_2\sin x$. Find the solution with $y(0)=2$, $y^\prime(0)=-1$.

Check: the derivative of your solution (right panel) must retrace the right-hand side of the equation

Part:

solution $y$ with its tangent

$y^\prime$ being traced

$x$—

$y(x)$—

$y^\prime(x)$—

$y$ is…—

💡 Show solutions▶

1 $y=\displaystyle\int(2x+3)\,dx=x^2+3x+C$.

2 $y=\displaystyle\int e^{2x}dx=\tfrac12e^{2x}+C$.

3 $y=\displaystyle\int\frac{dx}{x}=\ln x+C$ (on $x\gt0$, $|x|=x$).

4 $y=2x^3-4x+C$; $y(0)=C=1$. So $y=2x^3-4x+1$.

5 $y=\sin x+C$; $y(\pi)=0+C=2$. So $y=\sin x+2$.

6 By parts ($u=t$, $dv=e^tdt$): $y=te^t-e^t+C=(t-1)e^t+C$. $y(0)=-1+C=0$, so $C=1$: $y=(t-1)e^t+1$.

7 $y=\tan^{-1}x+C$; $y(1)=\tfrac{\pi}{4}+C=0$, so $y=\tan^{-1}x-\dfrac{\pi}{4}$.

8 $y^\prime=6x^2+C_1$, $y^\prime(0)=C_1=-2$; $y=2x^3-2x+C_2$, $y(0)=C_2=1$. So $y=2x^3-2x+1$.

9 $y^\prime=-Ce^{-x}+1$, so $y^\prime+y=-Ce^{-x}+1+Ce^{-x}+x-1=x$ ✓ for every $C$. Then $y(0)=C-1=3$ gives $C=4$: $y=4e^{-x}+x-1$.

10 $y(0)=C_1=2$. $y^\prime=-C_1\sin x+C_2\cos x$, so $y^\prime(0)=C_2=-1$. Hence $y=2\cos x-\sin x$.

III

Motion and exponential models. Give units in every answer. Use $g=9.8$ m/s² where needed.

1 A car accelerates at a constant $3$ m/s² from an initial speed of $5$ m/s. Find its position $s(t)$ (with $s(0)=0$) and the distance covered in the first $10$ s.
2 A stone is dropped (from rest) from a cliff $80$ m high. When does it hit the water below, and how fast is it going then?
3 A car travelling at $30$ m/s brakes with constant deceleration $6$ m/s². How long does it take to stop, and how far does it travel while braking?
4 A bacterial culture satisfies $\dfrac{dP}{dt}=0.4P$ ($t$ in hours) with $P(0)=500$. Find $P(5)$ and the time at which $P=10\,000$.
5 Carbon-14 decays according to $y^\prime=-ky$ with a half-life of $5730$ years. Find $k$ and the fraction of the original carbon-14 left after $2000$ years.
6 A drug has a half-life of $6$ hours in the body and is eliminated at a rate proportional to the amount present. If $400$ mg are given at once, how much remains after $24$ hours?
7 A quantity satisfies $y^\prime=ky$ with $y(0)=10$ and $y(3)=40$. Find $k$.
8 Modelling — cooling coffee. Newton's law of cooling says the temperature $T$ of a cup in a $20^\circ$C room changes at a rate proportional to the temperature difference: $\dfrac{dT}{dt}=-k(T-20)$, $k\gt0$. (a) Verify that $T=20+70e^{-kt}$ satisfies the equation and $T(0)=90$. (b) If $T(5)=60$ ($t$ in minutes), find $k$. (c) When does the coffee reach $30^\circ$C?

💡 Show solutions▶

1 $v=3t+5$, $s=1.5t^2+5t$ m. $s(10)=150+50=200$ m.

2 Taking up as positive: $s^{\prime\prime}=-9.8$, $s^\prime(0)=0$, $s(0)=80$, so $s=80-4.9t^2$. $s=0$ when $t=\sqrt{80/4.9}\approx4.04$ s; then $|v|=9.8t\approx39.6$ m/s.

3 $v=30-6t=0$ at $t=5$ s. Distance $s(5)=30(5)-3(5)^2=75$ m.

4 $P=500e^{0.4t}$. $P(5)=500e^{2}\approx3695$ bacteria. $500e^{0.4t}=10\,000\iff t=\dfrac{\ln20}{0.4}\approx7.49$ h.

5 $y=y_0e^{-kt}$ and $e^{-5730k}=\tfrac12$, so $k=\dfrac{\ln2}{5730}\approx1.21\times10^{-4}$ per year. Fraction left: $e^{-2000k}=2^{-2000/5730}\approx0.785$ (about $78.5\%$).

6 $24$ h is $4$ half-lives: $400\cdot\left(\tfrac12\right)^4=25$ mg. (Equivalently $A=400e^{-(\ln2/6)t}$, $A(24)=25$ mg.)

7 $10e^{3k}=40\iff e^{3k}=4\iff k=\dfrac{\ln4}{3}\approx0.462$ (per unit of time).

8 (a) $T^\prime=-70ke^{-kt}$ and $-k(T-20)=-k\cdot70e^{-kt}$ ✓; $T(0)=20+70=90$ ✓. (b) $20+70e^{-5k}=60\iff e^{-5k}=\tfrac47\iff k=\tfrac15\ln\tfrac74\approx0.112$ per minute. (c) $20+70e^{-kt}=30\iff e^{-kt}=\tfrac17\iff t=\dfrac{\ln7}{k}\approx17.4$ minutes.

🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. How much of the caffeine from a morning coffee is still in your body at bedtime?

Why it matters. Pharmacologists model how the body clears caffeine, medicines and alcohol with exactly the equation $A^\prime=-kA$. Dosing schedules for antibiotics, painkillers and anaesthetics around the world are designed from half-lives measured this way, and sleep researchers use the same model to advise on late-afternoon coffee.

What to do.

Assume a cup of coffee contains about $95$ mg of caffeine, drunk at 08:00, and that caffeine is cleared at a rate proportional to the amount present with a half-life of $5$ hours (for healthy adults it typically lies somewhere between about $3$ and $7$ hours).
Find $k$ and write $A(t)$ for the amount (in mg) $t$ hours after 08:00.
Compute how much is left at 22:00. Then add a second $95$ mg cup at 14:00 (the two doses simply add) and recompute.
Repeat with half-lives of $3$ h and $7$ h and tabulate the results. Plot $A(t)$ for each case.

Tools: Calculator, spreadsheet or Desmos

Hand in: A half-page note with your table and graph, a recommendation for the latest sensible time for a coffee, and one limitation of the model.

Going further. Look up the half-life of a common medicine (for example paracetamol or ibuprofen) and explain why its recommended dosing interval is a few hours long.

Instructor note: with a 5 h half-life, $k=\ln2/5\approx0.139$ h$^{-1}$; one cup leaves $95\cdot2^{-14/5}\approx13.6$ mg at 22:00, and the second cup adds $95\cdot2^{-8/5}\approx31.3$ mg, about $45$ mg in total. Good answers note that absorption is not instantaneous and that half-lives vary with genetics, pregnancy, smoking and medication.

🧪 Self-check quiz

0/5 correct

Q1. What is the order of $\left(y^\prime\right)^3+x^2y^{\prime\prime}=\sin x$?

Q2. Which function is a solution of $y^{\prime\prime}+4y=0$?

Q3. The solution of $y^\prime=2x$, $y(1)=4$ is

Q4. The general solution of $\dfrac{dy}{dt}=-3y$ is

Q5. How many initial conditions are normally needed to single out one solution of a second-order equation?

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