Chapter 9 · Differential Equations

9.1 Order, Degree and Solutions of a Differential Equation

A differential equation links an unknown function with its derivatives. Its order is the highest derivative present, its degree the power of that derivative (when the equation is a polynomial in the derivatives); its solutions come as a family, from which an initial condition picks one particular solution.

A phone slips out of a hand. Ignoring air resistance, physics tells us only one thing about its fall: its acceleration is constant, $\dfrac{d^2s}{dt^2}=9.8\ \text{m/s}^2$, where $s$ is the distance fallen after $t$ seconds. That is not yet a formula for $s$ — it is a statement about the second derivative of the unknown function $s(t)$. Integrating once gives the velocity $\dfrac{ds}{dt}=9.8t+C_1$; integrating again gives $s=4.9t^2+C_1t+C_2$. Two integrations, two unknown constants — and to pin them down we need two facts, such as "the phone started at rest" ($C_1=0$) "from where we measure" ($C_2=0$). An equation about a derivative, a whole family of solutions, and starting information that selects one member of the family: that is the whole story of this chapter in miniature.

Before learning methods for solving such equations (Sections 9.2–9.4), you need a vocabulary. This section answers three questions. How do we describe a differential equation — its order and degree? What does it mean for a function to be a solution, and how do we check one? And how are the general solution, with its arbitrary constants, and a particular solution related? Along the way you will see the most useful picture in the subject, the slope field, which shows the shape of every solution before any algebra is done.

Differential Equations, Their Order and Degree

You have met equations whose unknown is a number, like $x^2-5x+6=0$. In a differential equation the unknown is a function, and the equation tells us something about its derivatives. Throughout, $x$ is the independent variable, $y$ is the dependent variable (the unknown function of $x$), and we write $y'=\dfrac{dy}{dx}$, $y''=\dfrac{d^2y}{dx^2}$, $y'''=\dfrac{d^3y}{dx^3}$.

📘 Definition 9.1 — Differential equation

A differential equation is an equation involving the derivative (or derivatives) of a dependent variable with respect to an independent variable. For example $$\dfrac{dy}{dx}=2x,\qquad \dfrac{d^2y}{dx^2}+y=0,\qquad x\dfrac{dy}{dx}-y=x^2,\qquad \left(\dfrac{dy}{dx}\right)^2+y=e^x$$ are differential equations. Because the derivatives are with respect to a single independent variable, they are called ordinary differential equations; they are the only kind in this course, so we simply say "differential equation".

An equation written with differentials, such as $x\,dy-y\,dx=0$, is also a differential equation: dividing by $dx$ gives $x\dfrac{dy}{dx}-y=0$.

📘 Definition 9.2 — Order of a differential equation

The order of a differential equation is the order of the highest-order derivative of the dependent variable that occurs in it. The order is always a positive integer.

📘 Definition 9.3 — Degree of a differential equation

Suppose a differential equation is a polynomial equation in its derivatives $y',y'',y''',\dots$ — that is, after clearing fractions and fractional powers of the derivatives, the derivatives appear only with positive integer powers (the coefficients may be any functions of $x$ and $y$). Then its degree is the highest power (positive integer index) of the highest-order derivative in it.

If the equation cannot be written as a polynomial in its derivatives — for instance if a derivative appears inside $\sin$, $\cos$, $e^{(\ )}$ or $\log$ — its degree is not defined. When defined, the degree is always a positive integer.

In words: to find the degree, first find the highest derivative (that fixes the order), then look at the power on that derivative only. A larger power on a lower derivative does not count.

✏️ Example 1 — Reading off the order and degree

Find the order and degree of each equation.
(a) $\dfrac{d^2y}{dx^2}+3\left(\dfrac{dy}{dx}\right)^4-y=0$ (b) $\left(\dfrac{d^2y}{dx^2}\right)^3+\left(\dfrac{dy}{dx}\right)^5+y=x$ (c) $\dfrac{dy}{dx}=\sin x+y^2$ (d) $\dfrac{d^3y}{dx^3}+\left(\dfrac{d^2y}{dx^2}\right)^2+\left(\dfrac{dy}{dx}\right)^7=0$ (e) $x\left(\dfrac{dy}{dx}\right)^2-2y\dfrac{dy}{dx}+x=0$

💡 Show solution▶

(a) The highest derivative is $y''$, so the order is $2$. It appears to the power $1$ (the fourth power belongs to $y'$), so the degree is $1$.

(b) Highest derivative $y''$: order $2$. Its power is $3$: degree $3$. (The $5$ on $y'$ is irrelevant.)

(c) Only $y'$ appears, to the first power: order $1$, degree $1$. The $\sin x$ and $y^2$ are not derivatives, so they do not affect the degree.

(d) The highest derivative is $y'''$, to the power $1$: order $3$, degree $1$ — even though the equation contains a seventh power. This is the most common trap.

(e) Only $y'$ appears; its highest power is $2$: order $1$, degree $2$.

✏️ Example 2 — When the degree needs care

Find the order and degree (if defined) of: (a) $\left[1+\left(\dfrac{dy}{dx}\right)^2\right]^{3/2}=\dfrac{d^2y}{dx^2}$ (b) $y=x\dfrac{dy}{dx}+\dfrac{4}{dy/dx}$ (c) $\dfrac{d^2y}{dx^2}+\sin\left(\dfrac{dy}{dx}\right)=0$ (d) $\dfrac{d^2y}{dx^2}+\cos y=0$

💡 Show solution▶

(a) The power $\tfrac32$ is not an integer, so first make the equation a polynomial in the derivatives: squaring both sides, $$\left[1+\left(\dfrac{dy}{dx}\right)^2\right]^3=\left(\dfrac{d^2y}{dx^2}\right)^2.$$ Highest derivative $y''$, power $2$: order $2$, degree $2$. (This equation describes curves of constant curvature — circles.)

(b) The derivative appears in a denominator. Multiplying by $\dfrac{dy}{dx}$ (assumed non-zero): $y\dfrac{dy}{dx}=x\left(\dfrac{dy}{dx}\right)^2+4$. Order $1$, degree $2$.

(c) Order $2$. The derivative $y'$ sits inside a sine, and no algebra turns $\sin(y')$ into a polynomial in $y'$. Degree not defined.

(d) Here the cosine acts on $y$, not on a derivative. As a polynomial in $y''$ (with the coefficient-like term $\cos y$), the equation has order $2$, degree $1$. (It is the equation of a swinging pendulum.)

Compare (c) and (d) A non-polynomial function of $x$ or $y$ is harmless; a non-polynomial function of a derivative makes the degree undefined.

Solutions: General and Particular

To "solve" $x^2-5x+6=0$ is to find the numbers that make it true. To solve a differential equation is to find the functions that make it true — not at one value of $x$, but at every $x$ in an interval.

📘 Definition 9.4 — Solution, general solution, particular solution

A function $y=\varphi(x)$, defined on an interval $I$, is a solution of a differential equation on $I$ if, when $y$ and its derivatives are replaced by $\varphi(x)$ and its derivatives, the equation becomes true for every $x$ in $I$. A solution may also be given implicitly, by an equation linking $x$ and $y$.

A solution containing arbitrary constants, as many (independent) arbitrary constants as the order of the equation, is called a general solution (or primitive). A solution free from arbitrary constants — obtained from the general solution by giving the constants particular values — is called a particular solution.

The conditions used to find the constants, such as $y(x_0)=y_0$ and $y'(x_0)=y_1$ given at one point $x_0$, are called initial conditions. A first-order equation needs one condition; a second-order equation needs two.

🧭 Problem-solving strategy — verifying a solution

Differentiate the given function as many times as the order of the equation (for an implicit solution, differentiate implicitly, Section 5.2).
Substitute $y$, $y'$, $y''$, … into the left-hand side and simplify; separately simplify the right-hand side.
If the two sides agree for every $x$ in the interval, the function is a solution. If they differ even at one point of the interval, it is not.
Never "solve for $x$": the equation must hold identically, not at isolated values of $x$.

✏️ Example 3 — Is it a solution?

(a) Verify that $y=e^{-3x}$ is a solution of $y''+y'-6y=0$. (b) Verify that $y=xe^x$ is a solution of $y''-2y'+y=0$. (c) Show that $y=\sin x$ is not a solution of $y''-y=0$.

💡 Show solution▶

(a) $y'=-3e^{-3x}$ and $y''=9e^{-3x}$. Then $y''+y'-6y=(9-3-6)e^{-3x}=0$ for every real $x$ ✓.

(b) Product rule: $y'=e^x+xe^x=(x+1)e^x$ and $y''=e^x+(x+1)e^x=(x+2)e^x$. So $$y''-2y'+y=\big[(x+2)-2(x+1)+x\big]e^x=0\cdot e^x=0\ ✓.$$

(c) $y''=-\sin x$, so $y''-y=-2\sin x$. This is not $0$ for all $x$ (at $x=\tfrac\pi2$ it equals $-2$), so $\sin x$ is not a solution — even though the equation happens to hold at $x=0,\pi,2\pi,\dots$

✏️ Example 4 — Two constants for a second-order equation

Show that $y=A\cos2x+B\sin2x$ is a solution of $\dfrac{d^2y}{dx^2}+4y=0$ for every choice of the constants $A$ and $B$. Then find the particular solution with $y(0)=1$ and $y'(0)=4$.

💡 Show solution▶

Verify $y'=-2A\sin2x+2B\cos2x$ and $y''=-4A\cos2x-4B\sin2x=-4y$. Hence $y''+4y=0$ for all $x$, whatever $A$ and $B$ are ✓. The order is $2$ and there are two arbitrary constants: this is the general solution.

First condition $y(0)=A\cos0+B\sin0=A$, so $A=1$.

Second condition $y'(0)=-2A\sin0+2B\cos0=2B=4$, so $B=2$.

Answer The particular solution is $y=\cos2x+2\sin2x$.

Check $y(0)=1$ ✓, $y'(0)=-2\sin0+4\cos0=4$ ✓, and at $x=0.7$ a numerical second difference gives $y''\approx-8.5635$ while $-4y\approx-8.5635$ ✓.

✏️ Example 5 — An implicit general solution

Show that the curves $x^2+y^2=C$ ($C\gt0$) give solutions of $\dfrac{dy}{dx}=-\dfrac xy$ (where $y\ne0$). Find the particular solution through the point $(3,-4)$, and state it as an explicit function on an interval.

💡 Show solution▶

Verify Differentiating $x^2+y^2=C$ implicitly: $2x+2y\dfrac{dy}{dx}=0$, so $\dfrac{dy}{dx}=-\dfrac xy$ wherever $y\ne0$ ✓. One arbitrary constant, first-order equation: a general solution (a family of circles centred at the origin).

Particular Through $(3,-4)$: $C=9+16=25$, so $x^2+y^2=25$.

Explicit form The circle is not the graph of one function. The point has $y\lt0$, so the solution through it is the lower semicircle $y=-\sqrt{25-x^2}$, valid on $-5\lt x\lt5$ (at $x=\pm5$, $y=0$ and the slope $-\tfrac xy$ is undefined).

Check At $x=3$: $y=-4$ ✓, and $y'=\dfrac{x}{\sqrt{25-x^2}}=\dfrac34$, while $-\dfrac xy=-\dfrac{3}{-4}=\dfrac34$ ✓.

✏️ Example 6 — The falling phone

A phone is dropped from rest from the top of a $20$ m building. Ignoring air resistance, the distance $s$ (in m) it has fallen after $t$ seconds satisfies $\dfrac{d^2s}{dt^2}=9.8$. (a) Find the general solution. (b) Find the particular solution for this drop. (c) How long does the phone take to reach the ground, and how fast is it moving then?

💡 Show solution▶

(a) Integrate twice: $\dfrac{ds}{dt}=9.8t+C_1$, then $s=4.9t^2+C_1t+C_2$. Two arbitrary constants for a second-order equation: the general solution.

(b) Initial conditions: $s(0)=0$ gives $C_2=0$; starting from rest, $s'(0)=0$ gives $C_1=0$. So $s=4.9t^2$.

(c) speed $\dfrac{ds}{dt}=9.8t\approx9.8\times2.020\approx19.8\ \text{m/s}$ (about $71$ km/h — which is why phones crack).

The faint curves are members of the general solution for several values of the constant $C$; Play sweeps a blue member through the family. Drag the orange point $(x_0,y_0)$ anywhere: the one member through it — the particular solution — is drawn in red, with its value of $C$. The last readout substitutes that curve back into the equation at the point.

Equation:

general solution—

point $(x_0,y_0)$—

constant for this point—

sweeping member—

check $y'-f(x,y)$ at the point—

Slope Fields: Seeing Solutions Before Solving

A first-order equation of the form $\dfrac{dy}{dx}=f(x,y)$ is a set of instructions: "at the point $(x,y)$, the solution curve through it must have slope $f(x,y)$". Draw a short dash with that slope at many points of the plane and you get the equation's slope field (also called a direction field). A solution curve is a curve that is tangent to the dashes at every one of its points — like a leaf carried along by the current in a river. For $\dfrac{dy}{dx}=x-y$, for example, the slope is $0$ along the line $y=x$, positive below it and negative above it.

📐 Theorem 9.1 — One solution curve through each point

Let $f(x,y)$ be continuous on a rectangle $a\lt x\lt b$, $c\lt y\lt d$ containing the point $(x_0,y_0)$, and suppose the derivative of $f$ with respect to $y$ (treating $x$ as a constant) is also continuous there. Then the problem $$\dfrac{dy}{dx}=f(x,y),\qquad y(x_0)=y_0$$ has exactly one solution on some interval containing $x_0$.

We accept this theorem without proof; it is not examined. It explains two facts you will use all the time: one initial condition is exactly what is needed to fix the single constant of a first-order general solution, and two solution curves of such an equation never cross inside the rectangle.

When we cannot (or do not want to) find a formula, a computer follows the dashes numerically. The simplest scheme, Euler's method, steps a distance $h$ along the current dash: $y_{n+1}=y_n+h\,f(x_n,y_n)$. Better schemes (such as the classical Runge–Kutta method used in the lab below) sample several slopes in each step and are far more accurate. You will not be examined on these schemes; they are how the curves below are drawn.

Grey dashes show the slope $f(x,y)$ at each point. Tap or click anywhere to launch the solution curve through that point; it is traced forwards and backwards in $x$ by following the dashes numerically. Play launches curves from preset points. Switch the method to Euler's (step $h=0.25$) to see a crude scheme drift away from the true curve; the readouts compare both with the exact solution two units to the right of the launch point.

Equation:method:

launch point—

slope there $f(x_0,y_0)$—

numerical $y(x_0+2)$—

exact $y(x_0+2)$—

error—

⚠️ Common mistakes when studying alone

1. Taking the highest power instead of the power of the highest derivative. In $y'''+(y')^7=0$ the degree is $1$, not $7$. Find the order first; then look only at that derivative.

2. Reading the degree before making the equation a polynomial. Clear fractional powers and denominators containing derivatives first: $\sqrt{1+y'}=y''$ becomes $1+y'=(y'')^2$, of degree $2$.

3. Calling the degree undefined because of $\sin y$ or $e^x$. Only a non-polynomial function of a derivative ($\sin y'$, $e^{y''}$, $\log y'$) makes the degree undefined.

4. Checking a solution at one point. A solution must satisfy the equation for every $x$ in an interval. $\sin x$ satisfies $y''-y=0$ at $x=0$, but it is not a solution.

5. Counting constants that are not independent. $y=Ae^{x+B}=(Ae^B)e^x$ really has only one constant, $Ae^B$. The general solution of an order-$n$ equation needs $n$ constants that cannot be merged.

6. Forgetting which branch an implicit solution takes. Through $(3,-4)$ the solution of $y'=-\tfrac xy$ is $y=-\sqrt{25-x^2}$, not the whole circle and not the upper half.

🚀 Beyond the board syllabus (JEE Advanced)

In JEE Advanced 2026; not in the CBSE Class 12 syllabus or JEE Main. JEE Advanced also runs the process backwards: given a family of curves with $n$ arbitrary constants, find the differential equation they all satisfy — called forming the differential equation. Differentiate the family $n$ times, giving $n+1$ equations, and eliminate the $n$ constants. The result has order $n$ (and no constants).

Worked example. Form the differential equation of the family $y=Ae^x+Be^{-2x}$.

Differentiate: $y'=Ae^x-2Be^{-2x}$. Adding twice the original removes $B$: $y'+2y=3Ae^x$. Differentiate this once more: $y''+2y'=3Ae^x$. The right-hand sides are equal, so $y''+2y'=y'+2y$, that is $$\dfrac{d^2y}{dx^2}+\dfrac{dy}{dx}-2y=0,$$ an equation of order $2$ — matching the two constants. Check with $y=e^{-2x}$ ($A=0$, $B=1$): $4-2-2=0$ ✓.

Exercises

Find the order and the degree (if defined) of each differential equation.

1 $\dfrac{d^4y}{dx^4}+\sin\left(\dfrac{d^3y}{dx^3}\right)=0$
2 $\dfrac{dy}{dx}+5y=0$
3 $\left(\dfrac{d^2y}{dx^2}\right)^2+\cos\left(\dfrac{dy}{dx}\right)=0$
4 $\left(\dfrac{ds}{dt}\right)^4+3s\dfrac{d^2s}{dt^2}=0$
5 $y'''+2y''+y'=0$
6 $\left(\dfrac{d^2y}{dx^2}\right)^2+\left(\dfrac{dy}{dx}\right)^3=\sqrt x$
7 $\dfrac{dy}{dx}=\sqrt{1+\left(\dfrac{d^2y}{dx^2}\right)^3}$
8 $\dfrac{d^2y}{dx^2}=\left[y+\left(\dfrac{dy}{dx}\right)^2\right]^{1/4}$
9 $(y''')^2+(y'')^3+(y')^4+y^5=0$

💡 Show solutions▶

1 Order $4$; the derivative $y'''$ is inside a sine, so the degree is not defined.

2 Order $1$, degree $1$.

3 Order $2$; $\cos(y')$ is not a polynomial in $y'$, so the degree is not defined.

4 The highest derivative is $\dfrac{d^2s}{dt^2}$, to the power $1$: order $2$, degree $1$.

5 Order $3$, degree $1$.

6 Order $2$, degree $2$ ($\sqrt x$ involves no derivative).

7 Squaring: $\left(\dfrac{dy}{dx}\right)^2=1+\left(\dfrac{d^2y}{dx^2}\right)^3$. Order $2$, degree $3$.

8 Raising to the fourth power: $\left(\dfrac{d^2y}{dx^2}\right)^4=y+\left(\dfrac{dy}{dx}\right)^2$. Order $2$, degree $4$.

9 Highest derivative $y'''$, power $2$: order $3$, degree $2$.

Verify that the given function (explicit or implicit) is a solution of the differential equation, or show that it is not. $A$, $C$, $a$, $b$ are constants.

1 $y=e^x+1$; $y''-y'=0$
2 $y=x^2+2x+C$; $y'-2x-2=0$
3 $y=\cos x+C$; $y'+\sin x=0$
4 $y=Ax$; $xy'=y$ ($x\ne0$)
5 $y=e^{2x}(a+bx)$; $y''-4y'+4y=0$
6 $y=\tan^{-1}x$; $(1+x^2)y''+2xy'=0$
7 $y=x^2+1$; $xy'=2y$
8 $x^2-xy+y^2=C$; $(x-2y)\dfrac{dy}{dx}=2x-y$ (where $x\ne2y$)

💡 Show solutions▶

1 $y'=e^x$, $y''=e^x$, so $y''-y'=0$ ✓.

2 $y'=2x+2$, so $y'-2x-2=0$ ✓.

3 $y'=-\sin x$, so $y'+\sin x=0$ ✓.

4 $y'=A$, so $xy'=Ax=y$ ✓.

5 $y'=e^{2x}(2a+b+2bx)$ and $y''=e^{2x}(4a+4b+4bx)$. Then $y''-4y'+4y$ equals $e^{2x}$ times $(4a+4b+4bx)$ $-\,(8a+4b+8bx)$ $+\,(4a+4bx)$, and this bracket is $0$ ✓.

6 $y'=\dfrac{1}{1+x^2}$, $y''=-\dfrac{2x}{(1+x^2)^2}$. Then $(1+x^2)y''+2xy'=-\dfrac{2x}{1+x^2}+\dfrac{2x}{1+x^2}=0$ ✓.

7 $xy'=x\cdot2x=2x^2$, but $2y=2x^2+2$. They differ for every $x$, so $y=x^2+1$ is not a solution. (Its relative $y=x^2$ is.)

8 Implicitly: $2x-\left(y+x\dfrac{dy}{dx}\right)+2y\dfrac{dy}{dx}=0$, so $(2y-x)\dfrac{dy}{dx}=y-2x$; multiplying by $-1$, $(x-2y)\dfrac{dy}{dx}=2x-y$ ✓.

III

General and particular solutions.

1 How many arbitrary constants are there in the general solution of a differential equation of order $4$? In a particular solution?
2 $y=Ce^{-x}$ is the general solution of $y'+y=0$. Find the particular solution with $y(0)=5$.
3 $y=A\cos x+B\sin x$ is the general solution of $y''+y=0$. Find the particular solution with $y(0)=2$, $y'(0)=-1$.
4 Verify that $y=C_1e^x+C_2e^{-x}$ satisfies $y''=y$, and find the particular solution with $y(0)=0$, $y'(0)=2$.
5 Verify that $y=x^3+Cx$ is a solution of $xy'-y=2x^3$, and find the solution curve passing through $(1,3)$.
6 Which of these are particular solutions of $y''+9y=0$? (a) $\cos3x$ (b) $A\sin3x$ (c) $\sin3x+2\cos3x$ (d) $e^{3x}$
7 Explain why $y=Ae^{x+B}$ is not the general solution of $y''-y=0$, even though it satisfies the equation and contains two constants.

💡 Show solutions▶

1 Four (as many as the order); a particular solution has none.

2 $5=Ce^0=C$, so $y=5e^{-x}$.

3 $y(0)=A=2$; $y'=-A\sin x+B\cos x$ gives $y'(0)=B=-1$. So $y=2\cos x-\sin x$.

4 $y''=C_1e^x+C_2e^{-x}=y$ ✓. Conditions: $C_1+C_2=0$ and $C_1-C_2=2$, so $C_1=1$, $C_2=-1$: $y=e^x-e^{-x}$.

5 $y'=3x^2+C$, so $xy'-y=3x^3+Cx-x^3-Cx=2x^3$ ✓. Through $(1,3)$: $1+C=3$, $C=2$, so $y=x^3+2x$.

6 (a) $y''=-9\cos3x$, so $y''+9y=0$: a particular solution. (b) It satisfies the equation but contains an arbitrary constant, so it is not a particular solution (nor the general one — it has only one constant). (c) A particular solution ($A=2$, $B=1$ in $A\cos3x+B\sin3x$). (d) $y''+9y=18e^{3x}\ne0$: not a solution. Answer: (a) and (c).

7 $Ae^{x+B}=(Ae^B)e^x=Ke^x$: the two constants merge into one, so there is really only one arbitrary constant. It also misses solutions such as $e^{-x}$. The general solution is $y=C_1e^x+C_2e^{-x}$.

Applications and slope fields.

1 A cricket ball thrown upwards. A ball is thrown vertically upwards from ground level at $14.7\ \text{m/s}$. Its height $h$ (m) after $t$ s satisfies $\dfrac{d^2h}{dt^2}=-9.8$. Find $h(t)$, the greatest height and the time when it returns to the ground.
2 Braking. A car travelling at $20\ \text{m/s}$ brakes with constant deceleration, so its position $x$ (m) satisfies $\dfrac{d^2x}{dt^2}=-5$, with $x(0)=0$, $x'(0)=20$. Find $x(t)$, the time to stop and the stopping distance.
3 Bacteria (illustrative data). Show that $N=500e^{0.2t}$ is a solution of $\dfrac{dN}{dt}=0.2N$ with $N(0)=500$, where $t$ is in hours. Estimate $N$ after $5$ hours.
4 For $\dfrac{dy}{dx}=x-y$: find the slope of the solution curve through each of $(2,1)$, $(0,0)$ and $(1,1)$; describe the set of points where the slope is $0$; and verify that the straight line $y=x-1$ is a solution.
5 Cooling (illustrative data). Verify that $T=25+65e^{-0.08t}$ solves $\dfrac{dT}{dt}=-0.08(T-25)$ with $T(0)=90$, and find the temperature after $10$ minutes.

💡 Show solutions▶

1 $h'=-9.8t+C_1$, $h=-4.9t^2+C_1t+C_2$. $h(0)=0$ gives $C_2=0$ and $h'(0)=14.7$ gives $C_1=14.7$: $h=14.7t-4.9t^2$. Highest when $h'=0$: $t=1.5$ s, $h=22.05-11.025\approx11.0$ m. Back on the ground when $t(14.7-4.9t)=0$: $t=3$ s.

2 $x'=-5t+20$, $x=-2.5t^2+20t$. It stops when $x'=0$: $t=4$ s, after $x(4)=80-40=40$ m.

3 $\dfrac{dN}{dt}=500(0.2)e^{0.2t}=0.2N$ ✓ and $N(0)=500$ ✓. $N(5)=500e^1\approx1359$ bacteria.

4 Slopes: $2-1=1$; $0$; $0$. The slope is $0$ exactly on the line $y=x$. For $y=x-1$: $y'=1$ and $x-y=x-(x-1)=1$ ✓.

5 $\dfrac{dT}{dt}=65(-0.08)e^{-0.08t}=-0.08(T-25)$ ✓, $T(0)=90$ ✓. $T(10)=25+65e^{-0.8}$ $\approx25+29.2=54.2\ ^\circ\text{C}$.

🎯 Exam corner Board · 1 markBoard · 3 marksJEE MainJEE Advanced

How this is examined: almost every Board paper has a 1-mark question on the order and degree (often with a trap — a fractional power, or a derivative inside a sine), and a 2- or 3-mark question asking you to verify a solution or find a particular solution. JEE combines order and degree into numerical answers and asks which functions satisfy a given equation.

E1. Board · 1 mark The order and degree of $\left(\dfrac{d^2y}{dx^2}\right)^3+\left(\dfrac{dy}{dx}\right)^4+\sin y=0$ are respectively (A) $2,4$ (B) $2,3$ (C) $1,4$ (D) $2$, not defined

💡 Show solution▶

Idea Highest derivative $y''$ (order $2$), raised to the power $3$ (degree $3$). The term $\sin y$ is a function of $y$, not of a derivative, so the equation is still a polynomial in the derivatives.

Answer (B). Trap (A) takes the highest power anywhere; trap (D) thinks any sine makes the degree undefined.

E2. Board · 1 mark Assertion–Reason. Assertion (A): the degree of $\dfrac{d^2y}{dx^2}+\sin\left(\dfrac{dy}{dx}\right)=0$ is not defined. Reason (R): the degree of a differential equation is defined only when it is a polynomial equation in its derivatives. (A) Both A and R are true and R is the correct explanation of A (B) Both A and R are true but R is not the correct explanation of A (C) A is true but R is false (D) A is false but R is true

💡 Show solution▶

Idea R is exactly Definition 9.3, and A follows from it: $\sin(y')$ cannot be written as a polynomial in $y'$.

Answer (A).

E3. Board · 3 marks Show that $y=ae^x+be^{-x}+x^2$ is a solution of $\dfrac{d^2y}{dx^2}-y+x^2=2$ for all constants $a,b$. Hence find the particular solution with $y(0)=1$ and $y'(0)=0$.

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Verify $y'=ae^x-be^{-x}+2x$, $y''=ae^x+be^{-x}+2$. Then $$y''-y+x^2=(ae^x+be^{-x}+2)$$ $$\quad-(ae^x+be^{-x}+x^2)+x^2=2\ ✓.$$ (1 mark)

Conditions $y(0)=a+b=1$ and $y'(0)=a-b=0$. (1 mark)

Answer $a=b=\tfrac12$: $y=\dfrac{e^x+e^{-x}}{2}+x^2$. (1 mark)

E4. JEE Main If $y=e^{mx}$ is a solution of $\dfrac{d^2y}{dx^2}-5\dfrac{dy}{dx}+6y=0$, then the sum of all possible values of $m$ is (A) $5$ (B) $6$ (C) $-5$ (D) $1$

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Method Substituting, $(m^2-5m+6)e^{mx}=0$ for all $x$. Since $e^{mx}\ne0$, $m^2-5m+6=0$, so $m=2$ or $m=3$.

Answer (A) $5$. Trap (B) is the product of the roots. ⏱ Fast route: the sum of the roots of $m^2-5m+6=0$ is $5$ directly.

E5. JEE Main (Numerical value.) If $p$ and $q$ are the order and degree of $\left[1+\left(\dfrac{dy}{dx}\right)^2\right]^{3/2}=5\,\dfrac{d^2y}{dx^2}$, find $p+q$.

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Method Square to clear the fractional power: $\left[1+(y')^2\right]^3=25(y'')^2$. Order $p=2$, degree $q=2$.

Answer $p+q=\mathbf{4}$. (Reading the degree as $\tfrac32$ or $1$ before squaring is the trap.)

E6. JEE Advanced (Numerical answer; uses the 🚀 box.) The differential equation of the family of parabolas $y^2=4a(x-b)$, where $a$ and $b$ are arbitrary constants, has order $p$ and degree $q$. Find $p+q$.

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Eliminate $b$ Differentiate: $2yy'=4a$, i.e. $yy'=2a$.

Eliminate $a$ Differentiate again: $(y')^2+yy''=0$. This equation contains no constants; it has order $2$ and degree $1$.

Answer $p+q=\mathbf{3}$. ⏱ Fast route: two essential constants give order $2$; the elimination is by differentiation only, so $y''$ appears to the first power.

🔬 Research–Teaching NexusMini project · about 60 minutes

Research question. A plastic bottle with a small hole near the bottom drains through the hole. Which differential equation does the water height obey: is the rate of fall $\dfrac{dh}{dt}$ proportional to $h$, or to $\sqrt h$?

Why it matters. Engineers size the outlets of water tanks, dams, fuel tanks and drip-irrigation reservoirs with exactly this kind of model, and hospitals use it to predict how long an IV bag lasts. Choosing the right differential equation from data — before solving anything — is the first step of every model.

What to do.

Make a hole about $4$ mm across near the bottom of a straight-sided $1.5$ L bottle (ask an adult to help). Stick a paper strip marked in centimetres up the side, with $0$ at the hole.
Fill to $h=20$ cm, film the bottle with a phone while the water drains, and read off $h$ every $10$ s from the video.
For each interval estimate the rate $\dfrac{\Delta h}{\Delta t}$ and the average height. Plot the rate against $h$, and separately against $\sqrt h$.
Whichever plot is closer to a straight line through the origin tells you the model: $\dfrac{dh}{dt}=-kh$ or $\dfrac{dh}{dt}=-k\sqrt h$. Estimate $k$ from the slope, and state the order and degree of your equation.

Tools: Bottle, ruler or paper scale, phone camera, spreadsheet.

Hand in: Your table of $t$, $h$ and rates, both plots, your choice of model with a reason, the value of $k$ with units, and one paragraph on the sources of error.

Going further. Torricelli's law predicts $\dfrac{dh}{dt}=-\dfrac aA\sqrt{2gh}$ for a hole of area $a$ in a bottle of cross-section $A$. Compare its $k$ with yours; the ratio is called the discharge coefficient. After Section 9.2 you will be able to solve your equation and predict the time to empty.

Instructor note: the data almost always favour $\sqrt h$. For a bottle of radius $4.5$ cm ($A\approx63.6\ \text{cm}^2$) and a $4$ mm hole ($a\approx0.126\ \text{cm}^2$), the ideal $k=\tfrac aA\sqrt{2g}\approx0.087\\text{cm}^{1/2}\text{/s}$ (with $g=981\ \text{cm/s}^2$) and the ideal time to drain from $20$ cm is $\tfrac{2\sqrt{20}}{k}\approx102$ s; measured times are typically $1.5$ to $1.7$ times longer (discharge coefficient about $0.6$–$0.65$).

🧪 Self-check quiz

0/5 correct

Q1. The order of $\dfrac{d^3y}{dx^3}+\left(\dfrac{dy}{dx}\right)^4=x$ is

Q2. The degree of $\left(\dfrac{d^2y}{dx^2}\right)^2+\left(\dfrac{dy}{dx}\right)^3=0$ is

Q3. The general solution of a differential equation of order $2$ contains

Q4. Which function is a solution of $\dfrac{dy}{dx}=2y$?

Q5. The particular solution of $\dfrac{dy}{dx}=\cos x$ with $y(0)=1$ is

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