Differential Equations

2.1 Core concepts

• An equation involving the derivative(s) of a dependent variable with respect to an independent variable is a differential equation; if only one independent variable is involved, it is an ordinary differential equation (ODE), otherwise a partial differential equation — this chapter studies ODEs only (NCERT §9.2, p. 300–301). Examples introduction include dy/dx = eˣ, d²y/dx² + y = 0, and xy(d²y/dx²) + x(dy/dx)² − y(dy/dx) = 0.
• Standard derivative notation used throughout: dy/dx = y′, d²y/dx² = y″, d³y/dx³ = y‴, and yₙ for the nth-order derivative (NCERT §9.2 Note, p. 301). The prime notation is preferred in compact statements and is the form CUET uses in MCQ stems.
• Order of a differential equation is the order of the highest-order derivative appearing in it; e.g. dy/dx = eˣ has order 1, d²y/dx² + y = 0 has order 2, d³y/dx³ + x²(d²y/dx²)³ = 0 has order 3 (NCERT §9.2.1, p. 301–302). Order is always a positive integer and is determined unambiguously by inspection.
• Degree is defined only if the equation is a polynomial in the derivatives y′, y″, y‴, …; if defined, it is the highest (positive integral) power of the highest-order derivative (NCERT §9.2.2, p. 302). The qualifier "polynomial in derivatives" is the single most common source of conceptual errors.
• Equations such as (dy/dx) + sin(dy/dx) = 0 are not polynomial in y′, so their degree is not defined; order and degree (when defined) are always positive integers (NCERT §9.2.2 and Note, p. 302). Other examples of non-polynomial DEs include those involving e^(y′), log(y″), √(y′), or fractional powers of derivatives — none of these admit a degree.
• A solution is a function y = φ(x) which, when substituted into the equation, reduces L.H.S. = R.H.S.; the curve y = φ(x) is the solution / integral curve (NCERT §9.3, p. 304). Verification of a candidate solution is done by direct substitution.
• General solution contains as many arbitrary constants as the order of the equation; a particular solution is obtained by giving definite values to those constants (NCERT §9.3, p. 305). A first-order DE has a one-parameter family of solutions; a second-order DE has a two-parameter family; etc.
• Formation of a DE whose general solution is given: differentiate the family enough times to eliminate every arbitrary constant; an n-parameter family produces an n-th-order DE (concept used in Exercise 9.2 Q11–Q12 and miscellaneous examples).
• Variables-separable form dy/dx = h(y)·g(x): rewrite as dy/h(y) = g(x) dx and integrate both sides to get H(y) = G(x) + C (NCERT §9.4.1, p. 306–307). The technique works whenever the right-hand side factors into a function of y times a function of x.
• Worked example: dy/dx = (x + 1)/(2 − y) separates to (2 − y) dy = (x + 1) dx and integrates to x² + y² + 2x − 4y + C = 0 (NCERT Example 4, p. 307). The implicit form is acceptable as the "general solution".
• Application — bank principal growing continuously at 5% per year obeys dP/dt = P/20, a first-order separable DE giving P = 1000 e^(t/20); doubling time t = 20 logₑ 2 ≈ 13.86 years (NCERT Example 9, p. 310).
• Homogeneous function: F(λx, λy) = λⁿ F(x, y) for some constant n called the degree of homogeneity; the corresponding DE dy/dx = F(x, y) is homogeneous when F is homogeneous of degree zero, i.e. F can be written purely in terms of the ratio y/x as g(y/x) (NCERT §9.4.2, p. 312–313).
• Standard substitution y = v·x (so dy/dx = v + x·dv/dx) reduces the equation to a variables-separable form ∫ dv/[g(v) − v] = ∫ dx/x + C (NCERT §9.4.2, p. 313–314). After integration, back-substitute v = y/x to obtain the general solution in (x, y).
• If the equation is written as dx/dy = h(x/y), use x = v·y instead, so dx/dy = v + y·dv/dy (NCERT §9.4.2 Note, p. 314). The "direction" of the substitution depends on whether dy/dx or dx/dy is more naturally expressed.
• Linear differential equation in y: dy/dx + Py = Q where P, Q are constants or functions of x only; multiplying both sides by the integrating factor I.F. = e^(∫P dx) makes the L.H.S. an exact derivative d/dx[y · I.F.] (NCERT §9.4.3, p. 322–323).
• Solution of a linear DE: y · (I.F.) = ∫ Q · (I.F.) dx + C (NCERT §9.4.3, p. 323). This three-step recipe — rewrite in standard form, compute I.F., integrate Q × I.F. — is the workhorse of CUET DE questions.
• Mirror form: dx/dy + P₁x = Q₁ (P₁, Q₁ functions of y only) has I.F. = e^(∫P₁ dy) and solution x · (I.F.) = ∫ Q₁ · (I.F.) dy + C (NCERT §9.4.3, p. 324). Use this when the equation is more naturally linear in x.
• Worked examples: dy/dx − y = cos x → P = −1, I.F. = e^(−x), and y = [(sin x − cos x)/2] + C eˣ (Example 14, p. 324); x dy/dx + 2y = x² rewrites as dy/dx + (2/x)y = x with I.F. = x², solution y = x²/4 + C x⁻² (Example 15, p. 325).
• Many physical, biological, and economic laws are most naturally expressed as differential equations — Newton's second law, radioactive decay, Newton's law of cooling, RC-circuit voltage decay, logistic growth — so these techniques generalise far beyond mathematics (NCERT §9.4 closing remarks, p. 326).

2.2 Definitions to memorise

2.3 Diagrams / processes to remember

• Three-step recipe for a linear DE (NCERT §9.4.3, p. 324): (i) write the equation in the form dy/dx + Py = Q, (ii) compute I.F. = e^(∫P dx), (iii) write y · (I.F.) = ∫ Q · (I.F.) dx + C. Skipping step (i) — failing to divide through to make the y′ coefficient 1 — is the single biggest source of wrong I.F.s.
• Substitution flow for a homogeneous DE (NCERT §9.4.2, p. 313–314): put y = vx → dy/dx = v + x·dv/dx → original equation becomes x·dv/dx = g(v) − v → separate variables → integrate → replace v by y/x. Forgetting the chain rule on dy/dx (writing dy/dx = v instead of v + x·dv/dx) is a recurring error.
• Recognition test for homogeneity (NCERT p. 312): in the candidate function F(x, y), replace x by λx and y by λy and simplify; if every λ cancels (i.e. λ⁰ remains), the DE is homogeneous of degree 0. If λⁿ remains with n ≠ 0, F is homogeneous of degree n but the DE itself is not "a homogeneous DE"'s strict sense.
• Recognition test for linearity: rewrite the DE so that y′ has coefficient 1; if what remains is y′ + (function of x only) · y = (function of x only), the equation is linear in y. If the equation contains y² or sin y or eʸ, it is non-linear.
• Growth/decay template (NCERT Example 9, p. 309–310): "rate proportional to amount" ⇒ dP/dt = kP ⇒ P = P₀ e^(kt); used for bank principal, bacterial culture (Exercise 9.3 Q20–Q22), and balloon volume (Q19). Always identify P₀ from the initial condition before applying.
• Formation-of-DE flow: given y = f(x, c₁, c₂, …, cₙ), differentiate n times to obtain n + 1 equations; eliminate the n constants among them; the surviving equation is the required DE of order n.
• Verification flow: to check whether y = φ(x) solves a given DE, compute φ′(x), φ″(x), … as needed, substitute into LHS, and confirm it equals RHS identically in x.
• General-solution check: count the arbitrary constants; they must equal the order of the equation. A "first-order" solution with two constants signals algebraic error.

2.4 Common confusions / NTA trap points

• Degree-not-defined trap. If the DE contains sin(y′), e^(y′), log(y′), etc., the equation is not polynomial in derivatives — degree is "not defined," even though order is perfectly defined (NCERT §9.2.2, p. 302; Q11 of Exercise 9.1, p. 303).
• Highest-power vs highest-order. Degree is the power of the highest-order derivative, not the largest power appearing anywhere. In (y‴)² + (y″)³ + (y′)⁴ + y⁵ = 0, order = 3 and degree = 2, not 4 or 5 (NCERT Exercise 9.1 Q6, p. 303).
• Number of arbitrary constants. A general solution of an n-th order DE has exactly n arbitrary constants; a particular solution has zero constants (NCERT Exercise 9.2 Q11–Q12, p. 306).
• Homogeneity check. F(x, y) = sin x + cos y is not homogeneous (cannot be written as λⁿ F(x, y)); whereas y² + 2xy, 2x − 3y and cos(y/x) are homogeneous of degrees 2, 1, 0 respectively (NCERT §9.4.2, p. 312). Only degree-0 functions give "homogeneous DEs" in the textbook's sense.
• Wrong substitution direction. If the DE comes naturally as dx/dy = h(x/y), substitute x = vy, not y = vx (NCERT §9.4.2 Note, p. 314; Exercise 9.4 Q16, p. 321).
• Integrating factor sign. For dy/dx − y = cos x, P = −1, so I.F. = e^(−x), not eˣ (NCERT Example 14, p. 324). Students often forget the sign of P after rewriting in standard form.
• Failure to normalise y′ coefficient. In x(dy/dx) − y = 2x², one must first divide by x to read off P = −1/x; reading off P = −1 from the un-divided form gives a wrong I.F.
• Confusing linear with first-degree. "Linear" requires the dependent variable and its derivatives to appear linearly. dy/dx = y² is first-degree (degree 1 in y′) but not linear.
• Mistaking dy/dx = e^(x+y) for non-separable. It separates: e^(x+y) = eˣ·eʸ, so the equation is variables-separable (Exercise 9.3 Q23).
• Integration constant placement. Place +C on the right side only, after both integrations; placing it before back-substitution often produces an algebraically equivalent but exam-marker-unfriendly form.
• Failure to back-substitute v = y/x. After solving the separated form in (v, x), some students forget to write the answer in (x, y).
• Treating "particular solution" as having one arbitrary constant. Once initial conditions are applied, all constants are determined; the particular solution is constant-free.

2.5 Key formulas & theorems

2.6 Solved examples (NCERT-grounded)

Example A (NCERT Example 4, p. 307). Solve dy/dx = (x + 1)/(2 − y).

Step 1 — separate variables: (2 − y) dy = (x + 1) dx. Step 2 — integrate both sides: 2y − y²/2 = x²/2 + x + C₁. Step 3 — rearrange: x² + y² + 2x − 4y + C = 0 where C = −2C₁. Answer: general solution in implicit form.

Example B (NCERT Example 9, p. 309–310). A bank pays continuously compounded interest at 5%; initial deposit ₹1000. Find amount after t years and doubling time.

Step 1 — write DE: dP/dt = (5/100) P = P/20. Step 2 — separate and integrate: dP/P = dt/20 ⇒ log P = t/20 + C; using P(0) = 1000 gives C = log 1000, so P = 1000 e^(t/20). Step 3 — doubling time: 2 = e^(t/20) ⇒ t = 20 log 2 ≈ 13.86 years. Answer: P(t) = 1000 e^(t/20); doubling in 20 ln 2 years.

Example C (NCERT Example 14, p. 324). Solve dy/dx − y = cos x.

Step 1 — identify P, Q: P = −1, Q = cos x; I.F. = e^(∫ −1 dx) = e^(−x). Step 2 — apply linear-solution formula: y · e^(−x) = ∫ cos x · e^(−x) dx + C. Step 3 — evaluate integral: ∫ e^(−x) cos x dx = e^(−x)(sin x − cos x)/2, so y · e^(−x) = e^(−x)(sin x − cos x)/2 + C ⇒ y = (sin x − cos x)/2 + C eˣ. Answer: y = (sin x − cos x)/2 + C eˣ.

Example D (NCERT Example 15, p. 325). Solve x dy/dx + 2y = x².

Step 1 — divide by x: dy/dx + (2/x) y = x; here P = 2/x, Q = x. Step 2 — compute I.F.: I.F. = e^(∫ 2/x dx) = e^(2 log x) = x². Step 3 — integrate Q · I.F.: y · x² = ∫ x · x² dx + C = x⁴/4 + C ⇒ y = x²/4 + C/x². Answer: y = x²/4 + C x⁻².

Example E (Exercise 9.3 Q23, p. 312). Solve dy/dx = e^(x + y).

Step 1 — factor: e^(x + y) = eˣ · eʸ, so dy/dx = eˣ · eʸ — separable. Step 2 — separate and integrate: e^(−y) dy = eˣ dx ⇒ −e^(−y) = eˣ − C. Step 3 — rearrange: eˣ + e^(−y) = C. Answer: eˣ + e^(−y) = C.