Differential Equations — Class 12 IMO

Order, Degree and Formation

A differential equation relates a function to its derivatives. For example $\dfrac{dy}{dx}=2x$ or $\dfrac{d^2y}{dx^2}+y=0$.

Order and degree

The order is the order of the highest derivative present.
The degree is the power of the highest-order derivative, after the equation is made free of radicals and fractions in the derivatives. Degree is defined only when the equation is polynomial in its derivatives.

For instance, $\left(\dfrac{d^2y}{dx^2}\right)^3+\dfrac{dy}{dx}=0$ has order $2$ and degree $3$.

General and particular solutions

A solution containing arbitrary constants is the general solution; fixing the constants using given conditions gives a particular solution. An $n$th-order equation has $n$ arbitrary constants in its general solution.

Forming a differential equation

Given a family of curves with $n$ arbitrary constants, differentiate $n$ times and eliminate the constants. The result is a differential equation of order $n$ that the whole family satisfies.

Example 1: Find the order and degree of $\dfrac{d^2y}{dx^2}+3\left(\dfrac{dy}{dx}\right)^2+y=0$.

Highest derivative is $\dfrac{d^2y}{dx^2}$ (order $2$); its power is $1$, so degree $1$.

Example 2: Find the order and degree of $\left(\dfrac{dy}{dx}\right)^3+2y=x$.

Highest derivative is first order; its highest power is $3$. Order $1$, degree $3$.

Example 3: Form the differential equation of the family $y=mx$ ($m$ arbitrary).

Differentiate: $\dfrac{dy}{dx}=m$. Eliminate $m$ using $m=\dfrac{y}{x}$: $\dfrac{dy}{dx}=\dfrac{y}{x}$, i.e. $x\dfrac{dy}{dx}=y$.

Example 4: Verify that $y=e^{2x}$ is a solution of $\dfrac{dy}{dx}=2y$.

$\dfrac{dy}{dx}=2e^{2x}=2y$. The equation is satisfied, so $y=e^{2x}$ is a solution.

Quick recap

Order = highest derivative present; degree = power of that derivative (polynomial form).
General solution has $n$ arbitrary constants for order $n$; particular solution fixes them.
To form a DE from a family with $n$ constants: differentiate $n$ times, eliminate constants.
Verify a solution by substituting it back into the equation.

✓ Quick check

The integrating factor of the differential equation dy/dx − y = cos x is:

P = −1. IF = e^(∫P dx) = e^(∫−1 dx) = e⁻ˣ.

The family y²=x²+C leads to the differential equation:

Differentiating gives 2y(dy/dx)=2x.

Variable-Separable Equations

A first-order equation is variable-separable if it can be written so that all the $y$-terms (with $dy$) are on one side and all the $x$-terms (with $dx$) on the other:

$$\frac{dy}{dx}=g(x)\,h(y)\ \Longrightarrow\ \frac{dy}{h(y)}=g(x)\,dx.$$

Then integrate both sides. A single constant of integration $C$ captures the family.

Method

Separate the variables.
Integrate each side.
Combine constants into one $C$.
Apply any initial condition to find $C$ for a particular solution.

Example 1: Solve $\dfrac{dy}{dx}=\dfrac{x}{y}$.

Separate: $y\,dy=x\,dx$. Integrate: $\dfrac{y^2}{2}=\dfrac{x^2}{2}+C_1$, i.e. $y^2-x^2=C$.

Example 2: Solve $\dfrac{dy}{dx}=ky$.

$\dfrac{dy}{y}=k\,dx \Rightarrow \ln|y|=kx+C_1 \Rightarrow y=Ce^{kx}$ (the exponential growth/decay law).

Example 3: Solve $\dfrac{dy}{dx}=(1+x)(1+y)$.

$\dfrac{dy}{1+y}=(1+x)\,dx \Rightarrow \ln|1+y|=x+\dfrac{x^2}{2}+C.$

Example 4: Solve $\dfrac{dy}{dx}=e^{x-y}$.

$e^{x-y}=e^{x}e^{-y}$, so $e^{y}\,dy=e^{x}\,dx \Rightarrow e^{y}=e^{x}+C.$

Quick recap

Separable form: $\dfrac{dy}{h(y)}=g(x)\,dx$; integrate both sides.
Combine all constants into a single $C$.
$\dfrac{dy}{dx}=ky \Rightarrow y=Ce^{kx}$ (growth/decay).
Use an initial condition to pin down $C$.

✓ Quick check

The particular solution of dy/dx = 2x, given that y(1) = 2, is:

dy = 2x dx. Integrating gives y = x² + C. At x=1, y=2 → 2 = (1)² + C → C = 1. So, y = x² + 1.

If y=x²+C, then the differential equation is:

Differentiating gives dy/dx=2x.

Linear Differential Equations

A first-order linear differential equation has the standard form

$$\frac{dy}{dx}+P(x)\,y=Q(x),$$

where $P$ and $Q$ are functions of $x$ only. It is solved using an integrating factor.

The integrating factor (IF)

$$\text{IF}=e^{\int P\,dx}.$$

Multiplying through by the IF turns the left side into the derivative of (IF $\times y$). The solution is then

$$y\cdot(\text{IF})=\int Q\cdot(\text{IF})\,dx + C.$$

Method

Put the equation in standard form and read off $P$ and $Q$.
Compute IF $=e^{\int P\,dx}$.
Write $y\cdot\text{IF}=\int Q\cdot\text{IF}\,dx+C$ and integrate.

Example 1: Find the integrating factor of $\dfrac{dy}{dx}+\dfrac{y}{x}=x$.

$P=\dfrac1x$, so IF $=e^{\int \frac1x dx}=e^{\ln x}=x.$

Example 2: Solve $\dfrac{dy}{dx}+\dfrac{y}{x}=x$.

IF $=x$. Then $y\cdot x=\int x\cdot x\,dx=\int x^2 dx=\dfrac{x^3}{3}+C$. So $y=\dfrac{x^2}{3}+\dfrac{C}{x}.$

Example 3: Find the IF of $\dfrac{dy}{dx}+2y=e^{x}$.

$P=2$, IF $=e^{\int 2\,dx}=e^{2x}.$

Example 4: Solve $\dfrac{dy}{dx}+2y=e^{x}$.

IF $=e^{2x}$. $y e^{2x}=\int e^{x}e^{2x}dx=\int e^{3x}dx=\dfrac{e^{3x}}{3}+C$. So $y=\dfrac{e^{x}}{3}+Ce^{-2x}.$

Quick recap

Standard linear form: $\dfrac{dy}{dx}+P(x)y=Q(x)$.
Integrating factor IF $=e^{\int P\,dx}$.
Solution: $y\cdot\text{IF}=\int Q\cdot\text{IF}\,dx+C$.
Always put the equation in standard form before reading $P$ and $Q$.

✓ Quick check

An object cools according to dT/dt=−0.5(T−20). If T=20°C, then dT/dt equals:

Substituting T=20 gives dT/dt=0.

A water tank leaks water at a rate proportional to the square root of the volume V of water remaining. The differential equation is:

Leaking means the volume is decreasing over time. Proportional to square root of V means dV/dt = −k√V (where k is a positive constant).