Differential Equations – JEE Main Maths Formula Sheet & Class 12 Notes | All Methods, Formulas & Solution Types

Definitions — Order, Degree, Linear, Ordinary and Partial Differential Equations

Why Precise Definitions Are the First JEE Main Differential Equations Question Type

Differential Equation (from PDF — Differential Equations): An equation involving dependent and independent variables together with derivatives of the dependent variable is called a Differential Equation. All of the following are examples of Differential Equations: d²y/dx² + dy/dx + y = 0; f(x, y, dy/dx, d²y/dx², …) = 0; (dy/dx)² – y·d²y/dx² = y/f(x); x·(d⁴x/dt⁴) + y·(d²x/dt²) = te.

Types of Differential Equations (from PDF):

(1) Ordinary Differential Equation: A differential equation involving only one independent variable and derivatives with respect to this single independent variable. Examples: d²x/dt² + 4·(dx/dt) = te; d⁴x/dt⁴ + y·(d²x/dt²) = te; dy/dx = x + y.

(2) Partial Differential Equation: A differential equation involving partial derivatives with respect to more than one independent variable. Example: ∂²u/∂x² + ∂²u/∂y² + ∂²u/∂z² = 0 (Laplace equation).

(3) Linear Differential Equation: A differential equation is called linear if: (i) every dependent variable and every derivative involved occurs in the first degree (no y², no (dy/dx)² etc.), and (ii) no product of dependent variable and/or derivatives occurs. General form of nth order linear Differential Equation:

aₙ(x)·dⁿy/dxⁿ + aₙ₋₁(x)·dⁿ⁻¹y/dxⁿ⁻¹ + … + a₁(x)·dy/dx + a₀(x)·y = f(x)

A Differential Equation that is not linear is called non-linear. Examples of non-linear: (dy/dx)² + y = x (squared derivative); y·(d²y/dx²) = 0 (product of y and derivative).

Order of a Differential Equation (from PDF — JEE Main): The order of the highest order derivative involved in the Differential Equation. The order is always a positive integer. Examples: dy/dx = x+y → order 1; d²y/dx² + y = 0 → order 2; (d³y/dx³)² + (dy/dx)³ = 0 → order 3.

Degree of a Differential Equation (from PDF — JEE Main): The degree of the highest order derivative which occurs in the Differential Equation, after the equation has been made free of radicals and fractions as far as derivatives are concerned. The degree is defined only when the Differential Equation is a polynomial in derivatives. If the Differential Equation cannot be expressed as a polynomial in derivatives (due to sin(dy/dx), eˢⁱⁿ⁽ᵈʸ/ᵈˣ⁾, log(dy/dx) etc.), the degree is not defined. Examples: (d²y/dx²)³ + dy/dx = x → degree 3 (highest derivative d²y/dx² has power 3); √(1+(dy/dx)²) = d²y/dx² → square both: 1+(dy/dx)² = (d²y/dx²)² → degree 2; eˢⁱⁿ⁽ᵈʸ/ᵈˣ⁾ = x → degree not defined. Download the Free PDF for Differential Equations for all order-degree determination examples for JEE Main.

Formation of Differential Equations and Types of Solutions

Why Formation and Solution Types Are Essential Differential Equations Concepts for JEE Main

Formation of a Differential Equation (from PDF — Differential Equations): A Differential Equation can be formed from a given relation involving dependent and independent variables (with arbitrary constants) by the process of differentiation and elimination of the arbitrary constants. The number of times we differentiate equals the number of arbitrary constants to be eliminated. The resulting Differential Equation has order equal to the number of arbitrary constants in the original relation.

Working method for formation (from PDF — Differential Equations JEE Main):

Step 1: Count the number n of arbitrary constants in the given relation — this will be the order of the Differential Equation formed.

Step 2: Differentiate the relation n times to get n additional equations.

Step 3: Eliminate all n arbitrary constants from the (n+1) equations (original + n derivatives). The resulting equation is the required Differential Equation.

Example: y = Asinx + Bcosx (two constants A, B → will give 2nd order DE). y' = Acosx – Bsinx. y'' = –Asinx – Bcosx = –y → y'' + y = 0. This is the Differential Equation for the given family of curves.

Types of solutions (from PDF — Differential Equations):

General Solution (Complete Primitive): A solution containing a number of independent arbitrary constants equal to the order of the Differential Equation. It represents a family of curves in the xy-plane.

Particular Solution: A solution obtained from the general solution by giving specific values to one or more arbitrary constants (using initial conditions or boundary conditions). It represents a single specific curve.

Singular Solution: A solution which cannot be obtained from the general solution by any particular choice of the arbitrary constants but which satisfies the Differential Equation. (Exists for some non-linear Differential Equations — generally not tested in JEE Main.)

Note from PDF: A solution of a Differential Equation does not involve the derivatives of the dependent variables — it is purely a relation between the variables. Download the Free PDF for Differential Equations for formation examples for JEE Main.

Solution by Inspection — All 16 Exact Differential Forms from the PDF

Why These 16 Inspection Formulas Are the Fastest Differential Equations Method for JEE Main

The inspection method (also called the exact differential method) in Differential Equations recognises that many expressions in the integrand are already the total differential d(something). Identifying the exact form converts the Differential Equation into a direct integration problem. All 16 forms below are listed exactly from the Aakash PDF.

All 16 Inspection Forms (from PDF — Differential Equations JEE Main):

(i) d(xy) = x dy + y dx → ∫d(xy) = xy + c

(ii) d(eˣ/y) = (y·eˣdx – eˣdy)/y² → y·eˣdx – eˣdy = y²·d(eˣ/y)

(iii) d(eʸ/x) = (x·eʸdy – eʸdx)/x² → x·eʸdy – eʸdx = x²·d(eʸ/x)

(iv) d(log(x²+y²)) = 2(x dx + y dy)/(x²+y²) → ∫d(log(x²+y²)) = log(x²+y²) + c

(v) d(log(y/x)) = (x dy – y dx)/(xy) → x dy – y dx = xy·d(log(y/x))

(vi) d(log(x/y)) = (y dx – x dy)/(xy) → y dx – x dy = xy·d(log(x/y))

(vii) d(tan⁻¹(y/x)) = (x dy – y dx)/(x²+y²) → x dy – y dx = (x²+y²)·d(tan⁻¹(y/x))

(viii) d(tan⁻¹(x/y)) = (y dx – x dy)/(x²+y²) → y dx – x dy = (x²+y²)·d(tan⁻¹(x/y))

(ix) d(xy)/(xy) = (x dy + y dx)/(xy) → d(log(xy)) = (x dy + y dx)/(xy)

(x) d(xy·log(xy)) = (x dy + y dx)(1 + log(xy))

(xi) d(x²/y²) = 2(y²·x dx – x²·y dy)/y⁴ = (2xy²dx – 2x²y dy)/y⁴ → (xy dx – x² dy)/y² = (y/2)d(x²/y²) ... from PDF exactly: d(x²/y²) = 2(x dy·y²dx – x²·y dy)/y⁴... cleaner notation: d(x/y) = (y dx – x dy)/y² and d(y/x) = (x dy – y dx)/x²

(xii) d(y/x) = (x dy – y dx)/x²

(xiii) d(x/y) = (y dx – x dy)/y²

(xiv) d(x²/y) = (2xy dx – x²dy)/y²

(xv) d(y²/x) = (2xy dy – y²dx)/x²

(xvi) From PDF exactly — the two forms for x²/y and y²/x: d(x²/y²) = 2[xy²dx – x²y dy]/y⁴ = (2x/y²)dx – (2x²/y³)dy

Compact list of the most important inspection forms (from PDF — Differential Equations JEE Main):

d(xy) = x dy + y dx

d(x/y) = (y dx – x dy)/y²

d(y/x) = (x dy – y dx)/x²

d(log(x/y)) = (y dx – x dy)/(xy)

d(log(y/x)) = (x dy – y dx)/(xy)

d(log(xy)) = (y dx + x dy)/(xy) = (x dy + y dx)/(xy)

d(tan⁻¹(y/x)) = (x dy – y dx)/(x²+y²)

d(tan⁻¹(x/y)) = (y dx – x dy)/(x²+y²)

d(x²+y²) = 2x dx + 2y dy → d(log(x²+y²)) = 2(x dx + y dy)/(x²+y²)

d(x/y²) = (y²dx – 2xy dy)/y⁴

d(y/x²) = (x²dy – 2xy dx)/x⁴

d(x²/y) = (2xy dx – x²dy)/y²

d(y²/x) = (2xy dy – y²dx)/x²

d(eˣ/y) = (y eˣdx – eˣdy)/y²

d(eʸ/x) = (x eʸdy – eʸdx)/x²

d(xy·log(xy)) = (x dy + y dx)(1 + log(xy)) = d(xy)·(1+log(xy))

How to use inspection method in Differential Equations (JEE Main): Scan the Differential Equation for pairs (x dy – y dx), (y dx – x dy), (x dy + y dx), (x dx + y dy) and the functions multiplying them (x², y², xy, x²+y²). Match to the inspection table above. The Differential Equation splits directly into d(something) terms, which are integrated in one step. Download the Free PDF for Differential Equations for all inspection method examples for JEE Main.

Variable Separable Method — Differential Equations Type I

Why Variable Separable Is the First and Most Common Differential Equations Method in JEE Main

Variable Separable Differential Equation (from PDF — Differential Equations): A Differential Equation in which it is possible to separate all terms involving x and dx to one side and all terms involving y and dy to the other side is called a variable separable Differential Equation. The equation is of the form: f(x) dx = g(y) dy, OR f₁(x) dx + f₂(y) dy = 0.

Method for Variable Separable Differential Equations (JEE Main):

Step 1: Write the Differential Equation as dy/dx = f(x)/g(y), which gives g(y) dy = f(x) dx.

Step 2: Integrate both sides: ∫g(y) dy = ∫f(x) dx + c.

Step 3: The result is the general solution. Apply initial conditions if given to find particular solution.

Reducible to Variable Separable (from PDF — Differential Equations JEE Main): Sometimes a Differential Equation is not directly in variable separable form, but a suitable substitution reduces it to that form.

Standard reducible types:

(a) dy/dx = f(ax + by + c): Substitute v = ax + by + c → dv/dx = a + b·(dy/dx) = a + b·f(v). Now dv/[a + b·f(v)] = dx → variable separable in v and x.

(b) dy/dx = f(ax + by): Same substitution v = ax + by → dv/dx = a + bf(v) → variable separable.

(c) (a₁x + b₁y + c₁)dx + (a₂x + b₂y + c₂)dy = 0 with a₁/a₂ = b₁/b₂ ≠ c₁/c₂: substitute ax + by = v to reduce to variable separable. (When ratios of all three are equal, it's linear; when only first two ratios are equal, use this substitution.)

(d) For dy/dx = f(ax+by): if a₁b₂ – a₂b₁ ≠ 0, substitute X = x – h, Y = y – k where (h, k) is the intersection of a₁x+b₁y+c₁=0 and a₂x+b₂y+c₂=0 → reduces to homogeneous Differential Equation.

Standard variable separable examples in Differential Equations (JEE Main):

dy/dx = (1+y²)/(1+x²) → dy/(1+y²) = dx/(1+x²) → tan⁻¹y = tan⁻¹x + c → tan⁻¹y – tan⁻¹x = c.

y' = eˣ⁺ʸ = eˣ·eʸ → e⁻ʸ dy = eˣ dx → –e⁻ʸ = eˣ + c.

dy/dx = (x²+1)/(y²+1) → (y²+1)dy = (x²+1)dx → y³/3 + y = x³/3 + x + c. Download the Free PDF for Differential Equations for all variable separable examples and reductions for JEE Main.

Homogeneous Differential Equations — Recognition and y = vx Substitution

Why Homogeneous Differential Equations and the y = vx Method Are Direct JEE Main Questions

Homogeneous Differential Equation (from PDF — Differential Equations): A Differential Equation of the form dy/dx = f(x,y)/g(x,y), where f(x,y) and g(x,y) are homogeneous expressions in x and y of the same degree, is called a Homogeneous Differential Equation. It can be written in the form: dy/dx = F(y/x).

Recognition test for Homogeneous Differential Equations (JEE Main): A function f(x,y) is homogeneous of degree n if f(λx, λy) = λⁿ f(x,y). For the Differential Equation dy/dx = f(x,y)/g(x,y): if both f and g are homogeneous of the same degree → the Differential Equation is homogeneous. Alternatively, if dy/dx can be written purely in terms of the ratio y/x (after dividing numerator and denominator by the appropriate power of x), the Differential Equation is homogeneous.

Working Method for Homogeneous Differential Equations (from PDF — 3 steps):

Step 1: Put y = vx (so y/x = v). Differentiate: dy/dx = v + x·(dv/dx) [by product rule: dy/dx = d(vx)/dx = v + x·dv/dx].

Step 2: Substitute in the given Differential Equation. The equation becomes: v + x·(dv/dx) = F(v) → x·(dv/dx) = F(v) – v. This is now a variable separable Differential Equation in v and x: dv/[F(v) – v] = dx/x.

Step 3: Integrate both sides: ∫dv/[F(v)–v] = ∫dx/x = ln|x| + c. After integration, replace v by y/x to get the solution in terms of x and y.

Standard Homogeneous Differential Equation examples (from PDF — Differential Equations JEE Main):

x·(dy/dx) = y + x·sin(y/x): here F(y/x) = y/x + sin(y/x). Substitute y = vx: v + x·dv/dx = v + sin(v). x·dv/dx = sinv → dv/sinv = dx/x → –cosecv·cotv…integrate → –loge|cosecv+cotv| = log|x| + c → log|x(cosecv+cotv)| = c. Substitute v = y/x for final answer.

dy/dx = (y²–x²)/(2xy): substitute y = vx → v + xv' = (v²x²–x²)/(2x·vx) = (v²–1)/(2v). So xv' = (v²–1)/(2v) – v = (v²–1–2v²)/(2v) = –(v²+1)/(2v) → 2v dv/(v²+1) = –dx/x → log(v²+1) = –logx + c → log[(y/x)²+1] + logx = c → log(x²+y²)/x = c. Download the Free PDF for Differential Equations for all homogeneous examples and standard results for JEE Main.

Linear Differential Equations — Integrating Factor Method

Why the Linear Differential Equation IF = e^∫Pdx Formula Is the Most-Used Differential Equations Method in JEE Main

Linear Differential Equation (from PDF — Differential Equations): A Differential Equation of the form: dy/dx + P·y = Q, where P and Q are either constants or functions of x only (not y), is called a linear Differential Equation (also known as Lagrange's Differential Equation).

Complete solution method for linear Differential Equations (from PDF — 3 steps):

Step 1: Write the Differential Equation in the standard form dy/dx + P·y = Q (with coefficient of dy/dx = 1). Identify P(x) and Q(x).

Step 2: Find the Integrating Factor: IF = e^{∫P dx}

Step 3: Multiply both sides by IF. The left side becomes d/dx[y·IF]. Integrate: y·(IF) = ∫Q·(IF) dx + c. The general solution is y·e^{∫Pdx} = ∫Q·e^{∫Pdx} dx + c.

Key properties of the Integrating Factor (from PDF — Differential Equations):

(i) IF = e^{∫P dx} — find ∫P dx first, then exponentiate. Do not add c to ∫P dx here (the constant of IF is absorbed into c).

(ii) After multiplying by IF, the left side always becomes: d/dx[y·e^{∫Pdx}] = (dy/dx + P·y)·e^{∫Pdx} = Q·e^{∫Pdx}. This is why the method works.

(iii) The solution y·IF = ∫Q·IF dx + c can also be written as: y = e^{–∫Pdx}[∫Q·e^{∫Pdx} dx + c].

Linear Differential Equation in x (from PDF — Differential Equations JEE Main): If the Differential Equation is of the form dx/dy + P₁·x = Q₁, where P₁ and Q₁ are constants or functions of y only: IF = e^{∫P₁ dy}. Solution: x·(IF) = ∫Q₁·(IF) dy + c (treat y as the independent variable throughout).

Working rule for identifying and solving linear Differential Equations (JEE Main):

Recognise linear Differential Equation: no y², no y·(dy/dx), no (dy/dx)² — only first power of both y and dy/dx. Rearrange to dy/dx + P(x)·y = Q(x) form. Find ∫P dx (use standard integral formulas). Write IF. Multiply through. Integrate right side. Write solution y·IF = ∫Q·IF dx + c. Substitute initial conditions if given to find particular solution. Download the Free PDF for Differential Equations for all linear DE examples and IF computations for JEE Main.

Bernoulli's Equation and Equations Reducible to Linear

Why Bernoulli's Equation and Functional Substitution Are Advanced Differential Equations Methods for JEE Main

Bernoulli's Differential Equation (from PDF — Differential Equations): An equation of the form: dy/dx + P·y = Q·yⁿ, where P and Q are constants or functions of x, and n ≠ 0 and n ≠ 1 (if n=0, linear; if n=1, variable separable), is called Bernoulli's Differential Equation.

Method for Bernoulli's Differential Equation (from PDF — Differential Equations JEE Main):

Step 1: Divide both sides of the Bernoulli Differential Equation by yⁿ: y⁻ⁿ·(dy/dx) + P·y¹⁻ⁿ = Q.

Step 2: Substitute v = y^(1–n). Differentiate: dv/dx = (1–n)·y⁻ⁿ·(dy/dx), so y⁻ⁿ·(dy/dx) = (1/(1–n))·(dv/dx).

Step 3: Substitute: (1/(1–n))·(dv/dx) + P·v = Q → dv/dx + (1–n)P·v = (1–n)Q. This is now a standard linear Differential Equation in v and x with P_new = (1–n)P and Q_new = (1–n)Q.

Step 4: Solve the linear Differential Equation using IF = e^{∫(1–n)P dx}. Write solution: v·IF = ∫(1–n)Q·IF dx + c.

Step 5: Substitute back v = y^(1–n) to get the solution in terms of x and y.

Bernoulli's Differential Equation examples (JEE Main):

dy/dx + y/x = y³·x² (n=3, not 0 or 1 → Bernoulli). Divide by y³: y⁻³dy/dx + y⁻²/x = x². Let v = y⁻² → dv/dx = –2y⁻³dy/dx. So –(1/2)dv/dx + v/x = x² → dv/dx – 2v/x = –2x² (linear in v). IF = e^{∫–2/x dx} = e^{–2lnx} = x⁻². Solution: v·x⁻² = ∫–2x²·x⁻² dx = ∫–2 dx = –2x + c → v = –2x³ + cx² → y⁻² = –2x³ + cx².

Differential Equations Reducible to Linear (from PDF — Differential Equations JEE Main): If the Differential Equation is of the form f(y)·(dy/dx) + P·f(y) = Q, where P and Q are constants or functions of x: Substitute f(y) = v → f'(y)·(dy/dx) = dv/dx → the equation becomes: dv/dx + P·v = Q, which is a linear Differential Equation in v and x. Solve by IF method, then replace v = f(y) at the end.

Common Reducible Forms (Differential Equations — JEE Main):

If f(y) = sin y: (sin y)·dy/dx + P·sin y = Q (multiply sinx·dy/dx type).

If f(y) = eʸ: eʸ·dy/dx + P·eʸ = Q → let v = eʸ → dv/dx = eʸ·dy/dx → dv/dx + Pv = Q (linear). Download the Free PDF for Differential Equations for all Bernoulli and reducible-to-linear examples for JEE Main.

Solution of d²y/dx² = f(x) and Second Order Differential Equations

Why Second Order Differential Equations by Double Integration Are JEE Main Direct Questions

Differential Equation of type d²y/dx² = f(x) (from PDF — Differential Equations): These Differential Equations are solved by integrating twice successively. The method:

Step 1: Integrate both sides of d²y/dx² = f(x) once: d/dx[dy/dx] = f(x) → integrate → dy/dx = ∫f(x)dx + C₁ = φ(x) + C₁ where φ(x) = ∫f(x)dx.

Step 2: Integrate once more: y = ∫[φ(x) + C₁]dx + C₂ = ∫φ(x)dx + C₁x + C₂.

The general solution contains exactly two arbitrary constants C₁ and C₂, consistent with the order of the Differential Equation (order = 2).

Notation from PDF: Let Φ(x) = ∫φ(x)dx = ∫∫f(x)dx·dx (double integral of f(x)). Then: y = Φ(x) + C₁x + C₂.

Second order Differential Equation examples (Differential Equations — JEE Main):

d²y/dx² = 6x: Integrate once → dy/dx = 3x² + C₁. Integrate again → y = x³ + C₁x + C₂.

d²y/dx² = sin x: Integrate → dy/dx = –cos x + C₁. Integrate again → y = –sin x + C₁x + C₂.

d²y/dx² = e^(2x): Integrate → dy/dx = e^(2x)/2 + C₁. Integrate again → y = e^(2x)/4 + C₁x + C₂.

With initial conditions y(0) = 1 and y'(0) = 0: C₁ = y'(0) = 0 (from dy/dx expression), C₂ = y(0) – (value at 0) = 1. These pinpoint the particular solution among the two-parameter family.

Other directly integrable second-order Differential Equations (Differential Equations — JEE Main):

d²y/dx² = g(dy/dx): Let p = dy/dx → dp/dx = g(p) → variable separable in p and x. After solving for p(x) = dy/dx, integrate again to get y.

d²y/dx² = h(y): Let p = dy/dx → p·dp/dy = h(y) (using d²y/dx² = p·dp/dy). Now separate: p dp = h(y) dy → integrate → p² = 2∫h(y)dy + C₁ → p = dy/dx = ±√(2H(y)+C₁). Then separate again to integrate for y in terms of x. Download the Free PDF for Differential Equations for all second-order double-integration examples for JEE Main.