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MathematicsCalculusHigh Yield (2-3 Qs in JEE)

Differential Equations Formulas

First-order linear differential equations, integrating factor (IF), homogeneous forms, and variable separable methods.

8 Formulas•828+ Questions in Bank

Core Formulas & Mathematical Identities

8
01

Linear Differential Equation (First Order)

dydx+P(x)y=Q(x)  ⟹  y⋅e∫Pdx=∫(Q⋅e∫Pdx)dx+C\frac{dy}{dx} + P(x) y = Q(x) \implies y \cdot e^{\int P dx} = \int \left( Q \cdot e^{\int P dx} \right) dx + C

Multiplying by Integrating Factor IF = e^(∫ P dx) makes LHS a total derivative.

Examiner Tip: If dy/dx looks complicated, check if dx/dy + P(y)x = Q(y) is simpler.
02

Bernoulli's Differential Equation

dydx+P(x)y=Q(x)yn  ⟹  11−ndzdx+P(x)z=Q(x)(z=y1−n)\frac{dy}{dx} + P(x) y = Q(x) y^n \implies \frac{1}{1-n} \frac{dz}{dx} + P(x) z = Q(x) \quad (z = y^{1-n})

Reduces non-linear Bernoulli equations into standard first-order linear form.

Examiner Tip: Divide entire equation by y^n before substituting z = y^(1-n).
03

Homogeneous Differential Equation

dydx=f(yx)  ⟹  y=vx,dydx=v+xdvdx\frac{dy}{dx} = f\left(\frac{y}{x}\right) \implies y = vx, \quad \frac{dy}{dx} = v + x\frac{dv}{dx}

Substitute y = vx to convert homogeneous equations into variable separable form.

Examiner Tip: Always substitute back v = y/x into the final expression.
04

Exact Differential Form Substitutions

x dy+y dx=d(xy),x dy−y dxx2=d(yx),x dx+y dyx2+y2=d(12ln⁡(x2+y2))x \, dy + y \, dx = d(xy), \quad \frac{x \, dy - y \, dx}{x^2} = d\left(\frac{y}{x}\right), \quad \frac{x \, dx + y \, dy}{x^2 + y^2} = d\left(\frac{1}{2}\ln(x^2 + y^2)\right)

Recognizing exact differentials allows one-line integration without lengthy substitutions.

Examiner Tip: Group x dy and y dx terms together first.
05

Orthogonal Trajectories

Replace dydx⟶−dxdy in the differential equation\text{Replace } \frac{dy}{dx} \longrightarrow -\frac{dx}{dy} \text{ in the differential equation}

Finds the family of curves that intersect the given family at right angles everywhere.

Examiner Tip: Eliminate any arbitrary constants before replacing the derivative.
06

Clairaut's Equation & Singular Solution

y=px+f(p)  (p=dydx)  ⟹  General: y=cx+f(c),Singular: Eliminate p via x+f′(p)=0y = px + f(p) \; \left(p = \frac{dy}{dx}\right) \implies \text{General: } y = cx + f(c), \quad \text{Singular: Eliminate } p \text{ via } x + f'(p) = 0

Standard non-linear differential equation whose general solution is a family of straight lines and singular solution is the envelope.

Examiner Tip: Substitute dy/dx = c directly into the differential equation to get the general solution in 2 seconds.
07

Linear Fractional Form Reductions

dydx=a1x+b1y+c1a2x+b2y+c2  ⟹  {If a2=−b1: Cross-multiply and integrate directlyIf a1a2=b1b2: Substitute t=a1x+b1y\frac{dy}{dx} = \frac{a_1 x + b_1 y + c_1}{a_2 x + b_2 y + c_2} \implies \begin{cases} \text{If } a_2 = -b_1: \text{ Cross-multiply and integrate directly} \\ \text{If } \frac{a_1}{a_2} = \frac{b_1}{b_2}: \text{ Substitute } t = a_1 x + b_1 y \end{cases}

Shortcuts for non-homogeneous rational fractional linear differential equations.

Examiner Tip: When a₂ = -b₁, (b₁ y dx + a₂ x dy) becomes exact: d(b₁ x y).
08

Order, Degree & Newton's Law of Cooling

Degree is undefined if derivatives are within transcendental functions;dTdt=−k(T−T0)  ⟹  T(t)=T0+ΔT0 e−kt\text{Degree is undefined if derivatives are within transcendental functions}; \quad \frac{dT}{dt} = -k(T - T_0) \implies T(t) = T_0 + \Delta T_0 \, e^{-kt}

Criteria for polynomial degree in derivatives, and standard differential model for thermal cooling.

Examiner Tip: dy/dx = sin(y) has degree 1, but sin(dy/dx) = y has undefined degree.

Top Examiner Traps & Pitfalls

Frequently tested misconceptions in Differential Equations

  • •Forgetting the sign in integrating factor: IF = e^(∫ P dx). If equation has -P(x)y, IF has a negative exponent.
  • •Dividing by zero without checking singular solutions when separating variables.