Differential Equations Formula Sheet

A scannable, exam-day revision sheet covering every key definition, standard form, solution method, and application from the Differential Equations chapter.

Basic Definitions

A differential equation (DE) contains derivatives of a dependent variable with respect to an independent variable.

Term	Definition
Order	Highest order derivative present in the DE
Degree	Power of the highest order derivative, when the DE is a polynomial in derivatives
General solution	Solution containing arbitrary constants equal to the order
Particular solution	Obtained by giving specific values to the constants (using initial conditions)

$$\boxed{\text{Order} = \text{Highest order derivative in the DE}}$$$$\boxed{\text{Degree} = \text{Power of highest order derivative (if polynomial in derivatives)}}$$

Standard Forms

$$\boxed{\text{First order, first degree: } \frac{dy}{dx} = f(x, y)}$$$$\boxed{\text{Second order, first degree: } \frac{d^2y}{dx^2} + P(x)\frac{dy}{dx} + Q(x)y = R(x)}$$

A DE is linear if there are no products or powers of $y$ and its derivatives (e.g. no $y\,y'$, no $(y')^2$).

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Formation of Differential Equations

Formation = creating a DE from a relation by eliminating arbitrary constants (reverse of solving).

$$\boxed{\text{Number of arbitrary constants} = \text{Order of the DE}}$$

Procedure (D-E-R): Differentiate the relation as many times as there are constants → Eliminate the constants → Rearrange to standard form. The final DE must contain only $x, y$ and derivatives.

Standard Results (eliminate the constant → DE)

Given Relation	Differential Equation
$y = ae^{kx}$	$\dfrac{dy}{dx} = ky$
$y = ae^{x} + be^{-x}$	$\dfrac{d^2y}{dx^2} - y = 0$
$y = a\cos kx + b\sin kx$	$\dfrac{d^2y}{dx^2} + k^2 y = 0$
$x^2 + y^2 = a^2$	$x + y\dfrac{dy}{dx} = 0$
$y^2 = 4ax$	$y\dfrac{dy}{dx} = 2a$ (or $y = 2x\,y'$ if $a$ arbitrary)
$(y - a)^2 = 4a(x - b)$	$(y')^2 = 4a\,y''$ (after elimination)

Special Exponential–Trigonometric Forms

$$y = e^{ax}(A\cos bx + B\sin bx) \;\Rightarrow\; \boxed{\frac{d^2y}{dx^2} - 2a\frac{dy}{dx} + (a^2 + b^2)y = 0}$$$$y = (c_1 + c_2 x)e^{ax} \;\Rightarrow\; \boxed{\frac{d^2y}{dx^2} - 2a\frac{dy}{dx} + a^2 y = 0}$$

Conic Section Results

Curve	DE after one differentiation
Circle $x^2 + y^2 = r^2$	$x + y\,y' = 0$
Ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$	$\dfrac{x}{a^2} + \dfrac{y}{b^2}y' = 0$
Parabola $y^2 = 4ax$	$y\,y' = 2a$

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Method 1: Variable Separable

A DE is variable separable if it can be written as a product of an $x$-function and a $y$-function.

$$\boxed{\frac{dy}{dx} = h(x)\cdot k(y) \quad\Longrightarrow\quad \frac{dy}{k(y)} = h(x)\,dx}$$

General solution:

$$\boxed{\int \frac{dy}{k(y)} = \int h(x)\,dx + C}$$

Form	Separation	Solution
$\dfrac{dy}{dx} = f(x)g(y)$	$\dfrac{dy}{g(y)} = f(x)dx$	$\displaystyle\int \frac{dy}{g(y)} = \int f(x)dx + C$
$\dfrac{dy}{dx} = \dfrac{f(x)}{g(y)}$	$g(y)dy = f(x)dx$	$\displaystyle\int g(y)dy = \int f(x)dx + C$
$f_1(x)g_1(y)dx + f_2(x)g_2(y)dy = 0$	$\dfrac{f_1(x)}{f_2(x)}dx = -\dfrac{g_2(y)}{g_1(y)}dy$	integrate both sides

Integration Formulas Needed

Integral	Result
$\displaystyle\int \frac{dx}{x}$	$\ln\lvert x\rvert + C$
$\displaystyle\int \frac{dx}{x^2 + a^2}$	$\dfrac{1}{a}\tan^{-1}\!\left(\dfrac{x}{a}\right) + C$
$\displaystyle\int \frac{dx}{\sqrt{a^2 - x^2}}$	$\sin^{-1}\!\left(\dfrac{x}{a}\right) + C$
$\displaystyle\int e^{ax}\,dx$	$\dfrac{1}{a}e^{ax} + C$
$\displaystyle\int \tan x\,dx$	$\ln\lvert \sec x\rvert + C$
$\displaystyle\int \cot x\,dx$	$\ln\lvert \sin x\rvert + C$
$\displaystyle\int \sec x\,dx$	$\ln\lvert \sec x + \tan x\rvert + C$
$\displaystyle\int \csc x\,dx$	$\ln\lvert \csc x - \cot x\rvert + C$

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Method 2: Homogeneous Differential Equations

A function $f(x,y)$ is homogeneous of degree $n$ if:

$$\boxed{f(\lambda x, \lambda y) = \lambda^n f(x, y)}$$

A DE is homogeneous if $\dfrac{dy}{dx} = \dfrac{f(x,y)}{g(x,y)}$ with $f, g$ homogeneous of the same degree, equivalently:

$$\boxed{\frac{dy}{dx} = F\!\left(\frac{y}{x}\right)}$$

Substitution:

$$\boxed{y = vx \quad\Rightarrow\quad \frac{dy}{dx} = v + x\frac{dv}{dx}}$$

After substitution the DE becomes variable separable in $v$ and $x$; solve, then replace $v = \dfrac{y}{x}$.

Standard form	Notes
$\dfrac{dy}{dx} = f\!\left(\dfrac{y}{x}\right)$	explicit ratio
$\dfrac{dy}{dx} = \dfrac{ax + by}{cx + dy}$	degree 1 / degree 1
$(ax + by)dx + (cx + dy)dy = 0$	degrees match

graph TD
    A[Is DE homogeneous?] -->|Yes| B[Substitute y = vx]
    B --> C["dy/dx = v + x dv/dx"]
    C --> D[Substitute into original DE]
    D --> E[Simplify to variable separable in v and x]
    E --> F[Solve by separation]
    F --> G[Replace v = y/x]
    G --> H[Final solution]

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Method 3: Linear Differential Equations

Standard form (linear in $y$):

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

Linear in $x$ (when $x$ is the dependent variable):

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

Integrating Factor:

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

General solution:

$$\boxed{y\cdot \text{IF} = \int Q\cdot \text{IF}\,dx + C}$$

The left side is recognised as $\dfrac{d}{dx}\left(y\cdot \text{IF}\right)$ after multiplying through by IF.

Common Integrating Factors

$P(x)$	$\text{IF} = e^{\int P\,dx}$
constant $a$	$e^{ax}$
$\dfrac{1}{x}$	$x$
$\dfrac{n}{x}$	$x^{n}$
$\tan x$	$\sec x$
$\cot x$	$\sin x$
$\sec^2 x$	$e^{\tan x}$
$\dfrac{2x}{1+x^2}$	$1 + x^2$
$\dfrac{1}{x\log x}$	$\log x$

Choosing the Method

Method	Use when	Form
Variable Separable	$x$ and $y$ separate cleanly	$\dfrac{dy}{dx} = f(x)g(y)$
Homogeneous	RHS is a function of $\dfrac{y}{x}$	$\dfrac{dy}{dx} = F(y/x)$
Linear	first power of $y$ and $y'$ only	$\dfrac{dy}{dx} + Py = Q$

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Applications of Differential Equations

Growth and Decay

$$\boxed{\frac{dN}{dt} = kN \quad\Longrightarrow\quad N(t) = N_0 e^{kt}}$$

• $k > 0$: exponential growth; $k < 0$: exponential decay ($\frac{dN}{dt} = -kN$).
• $N_0$ = initial quantity at $t = 0$.

Radioactive decay / half-life:

$$\boxed{N(t) = N_0\left(\frac{1}{2}\right)^{t/T_{1/2}}}, \qquad k = \frac{\ln 2}{T_{1/2}}$$

Carbon dating: $\;t = \dfrac{1}{k}\ln\!\left(\dfrac{N_0}{N}\right)$, with $k = \dfrac{\ln 2}{5730}$ (C-14 half-life 5730 yr).

Newton’s Law of Cooling

$$\boxed{\frac{dT}{dt} = -k(T - T_s) \quad\Longrightarrow\quad T(t) = T_s + (T_0 - T_s)e^{-kt}}$$

Time to cool from $T_1$ to $T_2$ in surroundings $T_0$:

$$t = \frac{1}{k}\ln\!\left(\frac{T_1 - T_0}{T_2 - T_0}\right)$$

As $t \to \infty$, $T \to T_s$ (body temperature approaches the surroundings).

Motion Problems

$$\boxed{m\frac{dv}{dt} = mg - kv}, \qquad v_{\text{terminal}} = \frac{mg}{k}$$

Solution: $v = \dfrac{mg}{k}\left(1 - e^{-kt/m}\right)$.

Simple Harmonic Motion:

$$\boxed{\frac{d^2x}{dt^2} + \omega^2 x = 0}, \quad \omega = \sqrt{\frac{k}{m}}$$

General solution: $x(t) = A\cos(\omega t) + B\sin(\omega t) = C\cos(\omega t + \phi)$.

Electric Circuits

Circuit	Equation	Solution	Time constant
RC (charging)	$R\dfrac{dQ}{dt} + \dfrac{Q}{C} = V_0$	$Q = CV_0\left(1 - e^{-t/RC}\right)$, $\;I = \dfrac{V_0}{R}e^{-t/RC}$	$\tau = RC$
RL	$L\dfrac{dI}{dt} + RI = V_0$	$I = \dfrac{V_0}{R}\left(1 - e^{-Rt/L}\right)$	$\tau = \dfrac{L}{R}$

Orthogonal Trajectories

Curves cutting a given family at right angles.

Method: find the DE of the family (eliminate the constant), then replace $\dfrac{dy}{dx}$ with $-\dfrac{dx}{dy}$ and solve.

Given family	Orthogonal trajectories
$x^2 + y^2 = c^2$ (circles)	$y = Ax$ (lines through origin)
$xy = c$ (rectangular hyperbolas)	$y^2 - x^2 = K$ (hyperbolas)
$y^2 = 4ax$ (parabolas)	$y^2 + 2x^2 = C$ (ellipses)

Mixing (Tank) Problems

$$\boxed{\frac{dQ}{dt} = \text{Rate in} - \text{Rate out}}$$

• Rate in $=$ (concentration in) $\times$ (inflow rate)
• Rate out $=$ (tank concentration $\tfrac{Q}{V}$) $\times$ (outflow rate)

This yields a linear DE solved by the integrating factor method.

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Master Summary Table

Application	Differential Equation	Solution
Growth	$\dfrac{dN}{dt} = kN$	$N = N_0 e^{kt}$
Decay	$\dfrac{dN}{dt} = -kN$	$N = N_0 e^{-kt}$
Cooling	$\dfrac{dT}{dt} = -k(T - T_s)$	$T = T_s + (T_0 - T_s)e^{-kt}$
RC Circuit	$R\dfrac{dQ}{dt} + \dfrac{Q}{C} = V$	$Q = CV(1 - e^{-t/RC})$
Free Fall	$m\dfrac{dv}{dt} = mg - kv$	$v = \dfrac{mg}{k}(1 - e^{-kt/m})$

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Related Topics

• Order and Degree of Differential Equations
• Formation of Differential Equations
• Variable Separable Method
• Homogeneous Differential Equations
• Linear Differential Equations
• Applications of Differential Equations