NCERT Plus Two (Class 12) Mathematics • Chapter 9

Differential Equations

Q: Find the order and degree of the differential equation ( d³y dx³ )² + d²y dx² + sin( dy dx ) = 0 .

• Highest-order derivative is d³y dx³ ⇒ Order = 3 . • Because dy/dx lies inside sin(...) , it is not a polynomial equation in derivatives ⇒ Degree = Not Defined .

Q: Find the order and degree of [1 + ( dy dx )²] 3/2 = k d²y dx² .

Square both sides to clear the fractional power 3/2 : [1 + (y')²]³ = k² ( d²y dx² )² . • Highest derivative is d²y/dx² ⇒ Order = 2 . • Exponent on d²y/dx² is 2 ⇒ Degree = 2 .

Q: Verify that y = e⁻³ˣ is a solution of the differential equation d²y dx² + dy dx − 6y = 0 .

Differentiate y = e⁻³ˣ : y' = −3e⁻³ˣ , and y'' = 9e⁻³ˣ . Substitute in LHS: 9e⁻³ˣ + (−3e⁻³ˣ) − 6(e⁻³ˣ) = (9 − 3 − 6)e⁻³ˣ = 0 = RHS . Verified!

Q: Find the particular solution of dy dx = 1 + y² 1 + x² given that y = √3 when x = 1 .

Separate variables: dy 1 + y² = dx 1 + x² . Integrate: tan⁻¹ y = tan⁻¹ x + C . Plug x = 1, y = √3 : π 3 = π 4 + C ⇒ C = π 12 . Particular solution: tan⁻¹ y − tan⁻¹ x = π 12 .

Q: Solve the homogeneous differential equation (x − y) dy dx = x + 2y .

dy dx = x + 2y x − y (homogeneous of degree 0). Put y = vx ⇒ v + x dv dx = 1 + 2v 1 − v . x dv dx = 1 + 2v − v + v² 1 − v = v² + v + 1 1 − v ⇒ ∫ 1 − v v² + v + 1 dv = ∫ dx x . Write 1 − v = − 1 2 (2v + 1) + 3 2 and complete square (v + ½)² + (√3/2)² : log|x² + xy + y²| = 2√3 tan⁻¹( x + 2y √3 x ) + C .

Q: Find the general solution of (1 + x²) dy dx + 2xy = 1 1 + x² .

Divide by (1 + x²) : dy dx + 2x 1 + x² y = 1 (1 + x²)² . P = 2x 1 + x² ⇒ IF = e ∫ 2x/(1+x²) dx = e log(1+x²) = 1 + x² . Solution: y(1 + x²) = ∫ 1 (1 + x²)² · (1 + x²) dx = ∫ dx 1 + x² = tan⁻¹ x + C . Answer: y(1 + x²) = tan⁻¹ x + C .

Q: Find the particular solution of dy dx + y cot x = 2x + x² cot x given y = 0 when x = π 2 .

P = cot x ⇒ IF = e ∫ cot x dx = e log|sin x| = sin x . y sin x = ∫ (2x sin x + x² cos x) dx . Notice d dx (x² sin x) = 2x sin x + x² cos x ! So y sin x = x² sin x + C . Plug x = π/2, y = 0 ⇒ 0 = π² 4 + C ⇒ C = − π² 4 . Answer: y = x² − π² 4 sin x .

Q: In a culture, bacteria count is 1,00,000 initially and grows by 10% in 2 hours. In how many hours will it reach 2,00,000 if growth rate is proportional to count present?

dN dt = kN ⇒ N(t) = N₀ e kt with N₀ = 100000 . At t = 2, N = 1.1 N₀ ⇒ e 2k = 11 10 ⇒ k = 1 2 log( 11 10 ) . Set N = 2 N₀ : e kt = 2 ⇒ t = log 2 k = 2 log 2 log(11/10) hours .

Order & degree of ODEs, general vs particular solutions, variable separable method, homogeneous equations (y = vx), linear ODEs with integrating factor, and growth/cooling models.

📄 Official NCERT Textbook PDF

Topic 9.2: Order & Degree of Differential Equations (Polynomial-in-Derivatives Trap)

An ordinary algebra equation like x + 3 = 7 hunts for a single mystery number. A Differential Equation hunts for an entire curve y(x) by connecting y with its rates of change (y', y'', y''') — just like saying "how fast a cricket ball slows down depends on its current speed!" Before solving any differential equation, we classify it by two tags: Order and Degree.

ORDER (ALWAYS EXISTS)Highest-order derivative present in the equation (1, 2, 3...)
Note: y''' beats (y')⁴!

DEGREE (CHECK POLYNOMIAL!)Highest power of the highest-order derivative AFTER clearing fractional powers/radicals

NOT DEFINED TRAPDegree is NOT DEFINED if ANY derivative sits inside sin(y'), cos(y''), e^(y'), log(y')!

🧠 Memory Trick: Order = Highest Tag (y''', y'', y'); Degree = Clean Power on that Highest Tag! If any derivative is trapped inside sin, cos, e^, or log, yell "NOT DEFINED!" (Functions of x like cos x are TOTALLY FINE!)

💡 Simple Kid-Check: Think of derivatives like royal ranks: y''' is King, y'' is Duke, y' is Knight. Order just asks "Who is the highest-ranking derivative in the room?" Degree asks "What clean whole-number exponent is on that highest derivative?" — provided NO derivative is locked inside trig, log, or exponential jail!

Differential Equation	Order	Degree	Reason
`(y''')² + (y'')³ + (y')⁴ + y⁵ = 0`	3	2	Highest derivative is `y'''`, raised to power `2`
`dydx − cos x = 0`	1	1	`cos x` wraps `x` (NOT `y'`!), so polynomial in `y'`
`d⁴ydx⁴ + sin(y''') = 0`	4	Not Defined	`y'''` is trapped inside `sin(...)`!
`(1 + (y')²)^3/2 = k y''`	2	2	Square both sides: `(1+(y')²)³ = k²(y'')²`

Watch THREE classic DHSE 1-mark traps! (1) Seeing d²ydx² + cos(3x) = 0 and wrongly writing "Degree Not Defined" — cos(3x) has NO derivative inside it, so Degree = 1! (2) Seeing y''' + y² + e^(y') = 0 and looking only at y''' — even though e^(y') has a lower derivative, it STILL breaks the polynomial rule, making Degree Not Defined! (3) Forgetting to square fractional powers like √y'' or (y')^(1/3) before reading Degree!

Topic 9.3: General Solution (Arbitrary Constants = Order) vs Particular Solution

Imagine a hillside in Munnar covered in parallel ski paths! Every path obeys the exact same slope rule dydx = 2x, forming an infinite stacked family y = x² + C — that whole family is the General Solution! Plant ONE flag at (1, 3), and only ONE single track passes through it (C = 2 ⇒ y = x² + 2) — that unique curve is the Particular Solution!

GENERAL SOLUTIONContains essential arbitrary constants equal to the Order of the DE (e.g., 1st order → 1 constant C; 3rd order → 3 constants)

PARTICULAR SOLUTIONObtained by plugging initial conditions (like y(0) = 1) to eliminate ALL C's → ZERO arbitrary constants!

HOW TO VERIFYDifferentiate given y(x) up to order n, plug into LHS & verify LHS = RHS!

🧠 Memory Trick: General = Family with C's (Count of C's = Order of DE). Particular = Pinpointed single curve with ZERO C's!

💡 Simple Kid-Check: A 1st-order differential equation integrates once, producing 1 unknown constant C. A 2nd-order equation integrates twice, producing 2 constants (c₁, c₂). Give enough pinpoint coordinates to lock every constant down, and the general family collapses into ONE particular curve!

1
Count constants in General Solution: Always equal to the order of the DE (Order 4 ⇒ 4 constants).
2
Count constants in Particular Solution: Always 0 (every arbitrary constant gets solved to a specific number!).
3
Verify a given solution: Compute y', y'' and substitute directly into the differential equation — don't try to re-solve from scratch when asked to "verify"!

🔎 NCERT Ex 9.2 Verification Classic: Verify that y = a cos x + b sin x is a solution of d²ydx² + y = 0.
• Differentiate once: y' = −a sin x + b cos x.
• Differentiate twice: y'' = −a cos x − b sin x = −y.
• Add y'' + y = −y + y = 0 = RHS! Notice 2 arbitrary constants (a, b) match Order 2!

MOST MISREAD 1-MARK MCQ IN CHAPTER 9 (NCERT Ex 9.2 Q11–12):
• "Number of arbitrary constants in the general solution of a differential equation of 4th order?" → Answer is 4!
• "Number of arbitrary constants in the particular solution of a differential equation of 3rd order?" → Answer is 0 (ZERO), NOT 3! Read General vs Particular carefully!

Topic 9.4.1: Variable Separable Method with Exponential & Trig Forms

Imagine sorting two mixed laundry piles before washing: toss every white shirt (y-terms together with dy) into the left basket, and every coloured shirt (x-terms together with dx) into the right basket! Once y and x stand neatly separated on opposite sides, integrate BOTH baskets at once and drop in one + C!

SEPARABLE FORMdydx = f(x) · g(y)
⇒ ∫ dyg(y) = ∫ f(x) dx + C

LOG CONSTANT TRICKIf EVERY integrated term is a log or tan⁻¹, write constant as log C (or tan⁻¹ C) for clean combining!

EXPONENTIAL SPLITSpot e^(x − y) = eˣ / eʸ or e^(x + y) = eˣ · eʸ to separate powers!

🧠 Memory Trick: dy and dx MUST sit upstairs (numerators), NEVER in the denominator! Put all y's with dy on Left, all x's with dx on Right!

💡 Simple Kid-Check: You can ONLY separate variables when dy/dx factors cleanly as (x-stuff) × (y-stuff) or (x-stuff) / (y-stuff)! Cross-multiply so dy stays on top of its y-fraction and dx stays on top of its x-fraction, then slap ∫ on both sides!

1
Factor into x and y blocks: E.g., dydx = e^(x−y) + x²e^(−y) = e^(−y)(eˣ + x²)!
2
Separate baskets: Move y-factor beside dy and dx beside x-factor: eʸ dy = (eˣ + x²) dx.
3
Integrate both sides & plug initial point (if a particular solution is requested) to solve C.

⚡ Two Board Exam Classics:
1. NCERT Ex 9.3 Q5 (sec² x tan y dx + sec² y tan x dy = 0): Divide by tan x tan y:
sec² xtan x dx + sec² ytan y dy = 0 ⇒ log|tan x| + log|tan y| = log C ⇒ tan x · tan y = C!
2. NCERT Example 6 (dydx = −4xy², y(0) = 1): ∫ y⁻² dy = ∫ −4x dx ⇒ −1y = −2x² + C. Put (0, 1) ⇒ C = −1 ⇒ y = 12x² + 1!

Watch THREE common errors! (1) Writing ∫ dyy² = log(y²) — WRONG! Only 1/y integrates to log|y|; 1/y² is y⁻² whose integral is −1/y! (2) Trying to separate dydx = x + y directly — addition CANNOT be separated by dividing! (3) Forgetting the sign when rewriting e^(x−y)!

Topic 9.4.2: Homogeneous Differential Equations (y = vx Substitution)

Look at dydx = x² + y²xy — notice how x and y are welded together so you can't pull them apart directly, YET every single term (x², y², xy) has exact total degree 2! That balanced-degree clue unlocks a magic zipper: substitute y = vx (the ratio v = y/x), and the equation unzips instantly into Variable Separable form!

HOMOGENEOUS CHECKReplace x → λx, y → λy in dydx = F(x, y):
If all λ's cancel (F(λx, λy) = λ⁰ F(x, y)), it's Homogeneous of degree 0!

MAGIC SUBSTITUTIONPut y = vx ⇒
dydx = v + x dvdx

ALTERNATE (x/y) FORMIf e^(x/y) or cos(x/y) appears, write dxdy = G(x/y) & put x = vy ⇒ dxdy = v + y dvdy!

🧠 Memory Trick: Equal powers in every term? Unzip with y = vx! Never forget Product Rule on vx: dy/dx = v + x(dv/dx) — and swap v = y/x back at the finish!

💡 Simple Kid-Check: Count the combined power of x and y in every single term (ignoring constants): in (x − y) dy = (x + 2y) dx, every term has power 1; in (x² + xy) dx + y² dy = 0, every term has power 2! Equal powers = Homogeneous!

1
Write dydx = F(x, y) and prove F(λx, λy) = F(x, y) (needed for full 4-mark board credit!).
2
Substitute y = vx and dydx = v + xdvdx, then cancel x to get v + xdvdx = g(v).
3
Separate v and x: ∫ dvg(v) − v = ∫ dxx → integrate and restore v = yx!

🔓 Classic Homogeneous Example (NCERT Ex 9.4 Q1): Solve (x² + xy) dy = (x² + y²) dx.
• dydx = x² + y²x² + xy (every term degree 2). Put y = vx ⇒ v + xdvdx = 1 + v²1 + v.
• Subtract v: xdvdx = 1 + v² − v − v²1 + v = 1 − v1 + v ⇒ ∫ 1 + v1 − v dv = ∫ dxx.
• Rewrite 1 + v1 − v = −1 + 21 − v ⇒ −v − 2 log|1 − v| = log|x| + C ⇒ (x − y)² = k x e^(y/x)!

Watch THREE deadly board traps! (1) Putting y = vx and writing dydx = xdvdx (forgetting the leading v + from Product Rule!). (2) Leaving your final answer in v instead of replacing v = y/x! (3) When you spot e^(x/y) (note x/y on top of e!), DO NOT use y = vx — use x = vy!

Topic 9.4.3: First-Order Linear Differential Equations (dy/dx + Py = Q) & IF

What if an equation like dydx + y tan x = sec x is NEITHER separable NOR homogeneous? Meet the secret rescue key: multiply BOTH sides by one tailored function called the Integrating Factor (IF), and the entire left side magically snaps shut into a single Product Rule derivative ddx(y · IF) — letting one clean integral finish the job!

TYPE 1 STANDARD FORMdydx + Py = Q (P, Q of x only)
IF = e^{∫ P dx}
y · (IF) = ∫ Q · (IF) dx + C

TYPE 2 (dx/dy) FORMdxdy + P₁x = Q₁ (P₁, Q₁ of y only)
IF = e^{∫ P₁ dy}
x · (IF) = ∫ Q₁ · (IF) dy + C

EXPONENTIAL-LOG RULEe^{k log x} = e^{log(x^k)} = x^k
(NEVER kx! Move k to power first!)

🧠 Memory Trick: Step 0: Make coefficient of dy/dx equal to 1! Grab P WITH its +/− sign, compute IF = e^{∫ P dx}, and write y × IF = ∫ Q × IF dx + C!

💡 Simple Kid-Check: Notice why it's called Linear: both y and dy/dx appear ONLY to power 1 and never multiply each other! Once you multiply by IF = e^{∫ P dx}, the left side ALWAYS collapses to ddx(y · IF) — no guesswork!

1
Divide out anything multiplying dy/dx: Turn x dydx + 2y = x² into dydx + 2x y = x FIRST!
2
Find Integrating Factor IF = e^{∫ P dx}: Remember e^{−log x} = e^log(x⁻¹) = 1x!
3
Write final integral formula: y · (IF) = ∫ Q · (IF) dx + C and integrate RHS cleanly!

🏆 Top 4-Mark Kerala Board Classic (NCERT Ex 9.5 Q5 / Ex 15): Solve x dydx + 2y = x² log x.
• Divide by x: dydx + (2x) y = x log x ⇒ P = 2x, Q = x log x.
• IF = e^{∫ (2/x) dx} = e^{2 log x} = e^log(x²) = x².
• Solution: y · x² = ∫ (x log x) · x² dx = ∫ x³ log x dx (By Parts: u = log x, v = x³):
y x² = x⁴4 log x − x⁴16 + C!

Watch THREE HUGE DHSE TRAPS! (1) Reading P = 2 from x dy/dx + 2y = x² WITHOUT dividing by x first (true P = 2/x!). (2) Dropping the minus sign on P in dy/dx − y = cos x (true P = −1 ⇒ IF = e^−x!). (3) Simplifying e^{−log x} as −x instead of 1/x (and e^{log(sin x)} = sin x)!

Topic 9.4: Real-Life Models: Population Growth, Cooling & Compound Interest

Why does money in a continuously compounded bank account grow faster each year, while a hot cup of Kozhikode Sulaimani cools fastest right when poured? Both follow Nature's favourite differential law: "How fast a quantity changes is proportional to how much is currently there!" Solving that single 1st-order DE reveals exponential curves everywhere in science and finance!

CONTINUOUS GROWTH / BANKdPdt = r100 P
⇒ P(t) = P₀ e^(r/100)t

DECAY / COOLINGdydt = −k y
⇒ y(t) = y₀ e^−kt (k > 0)

CURVE SLOPE WORDING"Slope of tangent at (x, y) is ..." ⇒ immediately write dydx = ...!

🧠 Memory Trick: "Rate proportional to current amount" = dy/dt = k·y ⇒ Instant Exponential y = y₀ e^kt! Set t = 0 FIRST to find C = log y₀!

💡 Simple Kid-Check: "Continuous growth at r% per year" simply translates sentence-by-sentence into algebra: "Rate of change of Principal" = dP/dt, "equals r% of Principal" = (r/100) P! Separate dP/P = (r/100) dt, integrate to log P, and exponentiate!

1
Translate words into a DE: "Slope of tangent" = dy/dx; "Rate of growth of bacteria N" = dN/dt = kN.
2
Solve by Variable Separable: ∫ dPP = ∫ k dt ⇒ log P = kt + C ⇒ P = A e^kt.
3
Plug t = 0 first to lock A = P₀, then plug the second condition (e.g. "doubles in 10 years") to solve k or target time!

🏦 NCERT Example 9 & Ex 9.3 Q19 (Continuous Bank Interest): Principal increases continuously at 5% per year. In how many years does ₹1000 double itself?
• Model: dPdt = 5100 P = P20 ⇒ ∫ dPP = ∫ 120 dt ⇒ log P = t20 + C.
• At t = 0, P = 1000 ⇒ C = log 1000. Set P = 2000: log(20001000) = t20 ⇒ t = 20 log 2 years (≈ 13.86 yrs)!

Watch TWO classic word-problem traps! (1) Writing dPdt = 5 P instead of 5100 P for 5% interest — forgetting the percent /100 makes ₹1000 explode in seconds! (2) In curve geometry problems like "passes through (0, −2) with slope equal to 2y/x", always evaluate C with the given point at the end!

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (Kerala HSE March • 1 Marks): Find the order and degree of the differential equation (d³ydx³)² + d²ydx² + sin(dydx) = 0.

• Highest-order derivative is d³ydx³ ⇒ Order = 3.
• Because dy/dx lies inside sin(...), it is not a polynomial equation in derivatives ⇒ Degree = Not Defined.

Q2 (Kerala HSE Say • 2 Marks): Find the order and degree of [1 + (dydx)²]^3/2 = k d²ydx².

Square both sides to clear the fractional power 3/2:
[1 + (y')²]³ = k² (d²ydx²)².
• Highest derivative is d²y/dx² ⇒ Order = 2.
• Exponent on d²y/dx² is 2 ⇒ Degree = 2.

Q3 (Kerala HSE March • 2 Marks): Verify that y = e⁻³ˣ is a solution of the differential equation d²ydx² + dydx − 6y = 0.

Differentiate y = e⁻³ˣ:
y' = −3e⁻³ˣ, and y'' = 9e⁻³ˣ.
Substitute in LHS: 9e⁻³ˣ + (−3e⁻³ˣ) − 6(e⁻³ˣ) = (9 − 3 − 6)e⁻³ˣ = 0 = RHS. Verified!

Q4 (Kerala HSE March • 3 Marks): Find the particular solution of dydx = 1 + y²1 + x² given that y = √3 when x = 1.

Separate variables: dy1 + y² = dx1 + x².
Integrate: tan⁻¹ y = tan⁻¹ x + C.
Plug x = 1, y = √3: π3 = π4 + C ⇒ C = π12.
Particular solution: tan⁻¹ y − tan⁻¹ x = π12.

Q5 (Kerala HSE Model • 4 Marks): Solve the homogeneous differential equation (x − y) dydx = x + 2y.

dydx = x + 2yx − y (homogeneous of degree 0). Put y = vx ⇒ v + x dvdx = 1 + 2v1 − v.
x dvdx = 1 + 2v − v + v²1 − v = v² + v + 11 − v ⇒ ∫ 1 − vv² + v + 1 dv = ∫ dxx.
Write 1 − v = −12(2v + 1) + 32 and complete square (v + ½)² + (√3/2)²:
log|x² + xy + y²| = 2√3 tan⁻¹(x + 2y√3 x) + C.

Q6 (Kerala HSE March • 4 Marks): Find the general solution of (1 + x²) dydx + 2xy = 11 + x².

Divide by (1 + x²): dydx + 2x1 + x² y = 1(1 + x²)².
P = 2x1 + x² ⇒ IF = e^{∫ 2x/(1+x²) dx} = e^log(1+x²) = 1 + x².
Solution: y(1 + x²) = ∫ 1(1 + x²)² · (1 + x²) dx = ∫ dx1 + x² = tan⁻¹ x + C.
Answer: y(1 + x²) = tan⁻¹ x + C.

Q7 (NCERT / DHSE Say • 4 Marks): Find the particular solution of dydx + y cot x = 2x + x² cot x given y = 0 when x = π2.

P = cot x ⇒ IF = e^{∫ cot x dx} = e^{log|sin x|} = sin x.
y sin x = ∫ (2x sin x + x² cos x) dx. Notice ddx(x² sin x) = 2x sin x + x² cos x!
So y sin x = x² sin x + C. Plug x = π/2, y = 0 ⇒ 0 = π²4 + C ⇒ C = −π²4.
Answer: y = x² − π²4 sin x.

Q8 (Kerala HSE Model • 3 Marks): In a culture, bacteria count is 1,00,000 initially and grows by 10% in 2 hours. In how many hours will it reach 2,00,000 if growth rate is proportional to count present?

dNdt = kN ⇒ N(t) = N₀ e^kt with N₀ = 100000.
At t = 2, N = 1.1 N₀ ⇒ e^2k = 1110 ⇒ k = 12 log(1110).
Set N = 2 N₀: e^kt = 2 ⇒ t = log 2k = 2 log 2log(11/10) hours.