Differential Equations notes: ISC Class 12 Mathematics

1. Order, degree and solutions

A differential equation links an unknown function with its derivatives.

Order: the order of the highest derivative present.
Degree: the power of that highest derivative when the equation is a polynomial in its derivatives (clear radicals and fractions first). If a derivative appears inside $\sin$ , $\cos$ , $e^{(\ )}$ or $\log$ , the degree is not defined.
A general solution contains as many independent arbitrary constants as the order; a particular solution has the constants fixed by given conditions.
To verify a solution, differentiate it and substitute into the equation.

Order: highest derivative. Degree: its power, once the equation is a polynomial in the derivatives.
Equation	Order	Degree
$\dfrac{dy}{dx} + 4y = x$	$1$	$1$
$\left(\dfrac{d^2y}{dx^2}\right)^3 + \dfrac{dy}{dx} = 0$	$2$	$3$
$\dfrac{d^3y}{dx^3} + \left(\dfrac{dy}{dx}\right)^5 = \sin x$	$3$	$1$
$\dfrac{d^2y}{dx^2} + \cos\left(\dfrac{dy}{dx}\right) = 0$	$2$	not defined

$C = -1$
$C = 0$
$C = 2$ (through $(1, 3)$ )

The general solution is a family; a condition such as $y(1) = 3$ picks one member.

Both rows agree: $y = e^{2x} - 1$ satisfies $\dfrac{dy}{dx} = 2y + 2$.
$x$	$0$	$0.5$	$1$
$\dfrac{dy}{dx}$	$2$	$5.4366$	$14.7781$
$2y + 2$	$2$	$5.4366$	$14.7781$

Key terms

Order: The order of the highest derivative in the equation.
Degree: The power of the highest-order derivative when the equation is a polynomial in its derivatives.
Particular solution: A solution with no arbitrary constants, chosen by given conditions.

2. Forming a differential equation

Given a family with $n$ independent arbitrary constants, differentiate $n$ times and eliminate the constants. The result has order $n$ .

Examples: $y = ae^{3x}$ gives $\dfrac{dy}{dx} = 3y$ ; $y = A\sin 2x + B\cos 2x$ gives $\dfrac{d^2y}{dx^2} + 4y = 0$ ; $y = Ax^2 + Bx$ gives $x^2y'' - 2xy' + 2y = 0$ . Constants such as $A$ and $B$ in $A\cos(x + B)$ are independent; constants that can be merged (as in $Ae^{x + B}$ ) count as one.

One differentiation per constant.

Checked on two members of the family.
Step	Result
Family	$y = Ax^2 + Bx$ (two constants)
First derivative	$y' = 2Ax + B$
Second derivative	$y'' = 2A$
Eliminate $A$ and $B$	$A = \dfrac{y''}{2}$ , $B = y' - xy''$
Equation	$x^2y'' - 2xy' + 2y = 0$

Formulas

Formation rule
$n \text{ constants} \Rightarrow \text{order } n$

Key terms

Arbitrary constant: A constant that can take any value, one for each member of a family of curves.

3. Variable separable and reducible equations

If $\dfrac{dy}{dx} = f(x)g(y)$ , separate and integrate: $\displaystyle\int \dfrac{dy}{g(y)} = \int f(x)\,dx + C$ .

Reducible form: if the right side depends on $ax + by + c$ , put $v = ax + by + c$ ; then $\dfrac{dv}{dx} = a + b\dfrac{dy}{dx}$ and the new equation separates. Example: $\dfrac{dy}{dx} = (x + y)^2$ with $v = x + y$ gives $\dfrac{dv}{1 + v^2} = dx$ , so $\tan^{-1}(x + y) = x + C$ .

One constant is enough: put it on one side only.

Remember $\dfrac{dv}{dx} = 1 + \dfrac{dy}{dx}$.
Equation	Substitution	New equation
$\dfrac{dy}{dx} = (x + y)^2$	$v = x + y$	$\dfrac{dv}{dx} = 1 + v^2$
$\dfrac{dy}{dx} = \sin(x + y)$	$v = x + y$	$\dfrac{dv}{dx} = 1 + \sin v$
$\dfrac{dy}{dx} = 2(x + y) + 1$	$v = x + y$	$\dfrac{dv}{dx} = 2v + 2$

$y = \sqrt{x^2 + 4}$

Worked example 4: the solution of $y\dfrac{dy}{dx} = x$ through $(0, 2)$.

Formulas

Separation of variables
$\int \dfrac{dy}{g(y)} = \int f(x)\,dx + C$

Key terms

Variable separable: An equation that can be rearranged with all y terms on one side and all x terms on the other.

4. Homogeneous equations

$F(x, y)$ is homogeneous of degree $n$ if $F(\lambda x, \lambda y) = \lambda^nF(x, y)$ . The equation $\dfrac{dy}{dx} = F(x, y)$ is homogeneous when $F$ has degree $0$ ; then $F$ depends only on $\dfrac{y}{x}$ .

Method: put $y = vx$ , so $\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$ ; separate $v$ and $x$ ; integrate; replace $v$ by $\dfrac{y}{x}$ . (If the equation is more naturally $\dfrac{dx}{dy} = G\left(\dfrac{x}{y}\right)$ , put $x = vy$ .)

The last step, back to $y$, is often forgotten.

Replace $x, y$ by $\lambda x, \lambda y$ and look for a power of $\lambda$.
$F(x, y)$	Degree of homogeneity
$x^2 + 3xy$	$2$
$\dfrac{x^2 + y^2}{xy}$	$0$
$\sqrt{x^2 + y^2}$	$1$
$x + y + 1$	not homogeneous

Checked by substituting back at three points.
Step	Result
Substitute $y = vx$	$x\dfrac{dv}{dx} = \cos^2 v$
Separate	$\sec^2 v\,dv = \dfrac{dx}{x}$
Integrate	$\tan v = \log x + C$
Through $\left(1, \dfrac{\pi}{4}\right)$	$C = 1$
Solution	$y = x\tan^{-1}(1 + \log x)$

Formulas

Homogeneous substitution
$y = vx \Rightarrow \dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$

Key terms

Homogeneous function: A function satisfying F(λx, λy) = λⁿ F(x, y) for all λ > 0.

5. Linear equations

Form 1: $\dfrac{dy}{dx} + Py = Q$ , with $P$ and $Q$ functions of $x$ . Integrating factor $\text{IF} = e^{\int P\,dx}$ , and $y \times \text{IF} = \displaystyle\int Q \times \text{IF}\,dx + C$ .

Form 2: $\dfrac{dx}{dy} + Px = Q$ , with $P$ and $Q$ functions of $y$ . Integrating factor $e^{\int P\,dy}$ , and $x \times \text{IF} = \displaystyle\int Q \times \text{IF}\,dy + C$ .

Always divide by the coefficient of the derivative first, and keep the sign of $P$ .

For $\dfrac{dx}{dy} + Px = Q$, swap the roles of $x$ and $y$.

Use $e^{\log f} = f$ to simplify.
Equation	$P$	Integrating factor
$\dfrac{dy}{dx} + 2y = x$	$2$	$e^{2x}$
$\dfrac{dy}{dx} + \dfrac{y}{x} = x^2$	$\dfrac{1}{x}$	$x$
$\dfrac{dy}{dx} - y\tan x = 1$	$-\tan x$	$\cos x$

$y = 2x - 1 + e^{-2x}$

Worked example 8: the exponential part dies away, leaving the line $y = 2x - 1$.

Formulas

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

Key terms

Linear differential equation: A first-order equation in which the dependent variable and its derivative appear only to the first power.
Integrating factor: The multiplier e^(∫P) that turns the left side into an exact derivative.

6. Growth and decay models

If a quantity changes at a rate proportional to its size, $\dfrac{dN}{dt} = kN$ and $N = N_0e^{kt}$ ( $k > 0$ growth, $k < 0$ decay). Equal time steps multiply $N$ by the same factor, so the doubling time (or half-life) is $\dfrac{\log 2}{|k|}$ .

Newton's law of cooling: $\dfrac{dT}{dt} = -k(T - T_s)$ gives $T = T_s + (T_0 - T_s)e^{-kt}$ , where $T_s$ is the surrounding temperature.

$N = 500 \cdot 3^{t/3}$

Growth at a rate proportional to size: $\dfrac{dN}{dt} = kN$.

Two equal time steps multiply the excess temperature by the same fraction.
Step	Result
Model	$\dfrac{dT}{dt} = -k(T - 35)$ , so $T = 35 + 185e^{-kt}$
Use $T(4) = 160$	$e^{-4k} = \dfrac{125}{185}$
At $t = 8$	$T = 35 + 185\left(\dfrac{125}{185}\right)^2 = 35 + \dfrac{125^2}{185}$
Value	$119.46$ °C

Formulas

Exponential growth or decay
$N = N_0e^{kt}$

Key terms

Half-life: The time a decaying quantity takes to fall to half its value.
Newton's law of cooling: The rate of cooling is proportional to the excess temperature over the surroundings.

Exam tips

How ISC marks are awarded

Order and degree (1–2 marks): one mark each; the degree only after clearing radicals.
Formation (2–3 marks): each differentiation, the elimination, the final equation.
Separable or reducible (2–3 marks): separation, integration with the constant, the particular value.
Homogeneous (3–5 marks): proof of homogeneity, substitution, separation, integration, return to $y$ .
Linear (3–5 marks): standard form, integrating factor, integration, constant from the condition.

Common mistakes

Giving a degree when a derivative sits inside a function.
Adding a constant on both sides, or forgetting it.
Missing the $1$ in $\dfrac{dv}{dx} = 1 + \dfrac{dy}{dx}$ .
Reading $P$ before dividing by the coefficient of the derivative, or losing its sign.
Leaving the answer in $v$ instead of $\dfrac{y}{x}$ .

1. Order, degree and solutions

Key terms

2. Forming a differential equation

Formulas

Key terms

3. Variable separable and reducible equations

Formulas

Key terms

4. Homogeneous equations

Formulas

Key terms

5. Linear equations

Formulas

Key terms

6. Growth and decay models

Formulas

Key terms

Exam tips

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