Class 12 Maths · Chapter 9
Short answer:
NCERT Class 12 Maths Chapter 9 — Differential Equations me total 3 exercises hain (miscellaneous exercise ke saath). Is guide me 20 solved questions cover kiye gaye hain: order-degree nikalna, family of curves se differential equation banana (arbitrary constants eliminate karke), variable separable method, homogeneous differential equations (y = vx substitution), aur linear differential equations integrating factor se solve karna, including initial-condition wale particular solution problems.
Differential equation ek aisi equation hai jisme ek function aur uske derivatives dono involve hote hain. Ye chapter real-world growth-decay, motion aur curve-family problems ko model karne ka tool deta hai — order/degree pehchanna, curve-family se equation banana, aur teen main methods (variable separable, homogeneous, linear) se solve karna seekhte hain.
Chapter 9 Summary — 5 Minute Revision
1. Basic Concepts — Order aur Degree
Order = highest derivative jo equation me appear ho (jaise dy/dx, d²y/dx² etc.)
Degree = highest order derivative ka power, bashart equation us derivative me polynomial ho (koi fractional/negative power, sin, log, e^(dy/dx) type term na ho).
Agar equation polynomial form me nahi hai (jaise dy/dx ke andar sin ya sqrt ho), toh degree defined nahi hoti — sirf order likha jaata hai.
2. General aur Particular Solution
- General solution — arbitrary constant(s) ke saath (jitne order utne constants), jaise y = A sin x + B cos x
- Particular solution — jab initial/boundary condition di ho aur constant ki specific value nikaal di jaaye
3. Formation of Differential Equation from Family of Curves
Curve ki equation me n arbitrary constants ho, toh unhe eliminate karne ke liye n baar differentiate karo, phir constants ko algebra se hata do — result ek nth order differential equation hoga.
Steps: (1) Curve equation likho, (2) n baar differentiate karo, (3) original + derivative equations se constants eliminate karo.
4. Variable Separable Method
Agar dy/dx ko is form me likha ja sake: f(x) dx = g(y) dy — dono side alag-alag integrate kar do.
∫f(x)dx = ∫g(y)dy + C
5. Homogeneous Differential Equations
Equation dy/dx = F(x,y) homogeneous hai agar F(λx, λy) = F(x,y) (degree zero function of x,y). Test: dy/dx = F(y/x) form me likha ja sake.
Substitution: y = vx, so dy/dx = v + x(dv/dx). Isse variable-separable form mil jaata hai v aur x me. Solve karke aakhir me v = y/x waapis substitute karo.
6. Linear Differential Equations
Standard form: dy/dx + Py = Q, jaha P aur Q sirf x ke functions hain (y ke nahi).
Integrating Factor (I.F.) = e∫P dx
General solution: y · (I.F.) = ∫ Q · (I.F.) dx + C
Agar equation dx/dy + Px = Q form me di ho (P, Q functions of y), toh I.F. = e∫P dy aur solution x·(I.F.) = ∫Q·(I.F.) dy + C.
| Type | Pehchan | Method |
|---|---|---|
| Variable separable | f(x)dx = g(y)dy me split ho jaaye | Direct integrate dono side |
| Homogeneous | dy/dx = F(y/x) | y = vx substitution |
| Linear | dy/dx + Py = Q | I.F. = e^∫Pdx, phir y·I.F. = ∫Q·I.F. dx + C |
↔ Table ko side me swipe karein
Poore Class 12 Maths ke handwritten colour notes
IITian & district toppers ke banaye short notes — revision-ready, diagram ke saath. Board se pehle poora syllabus 3 din me revise.
Exercise Questions — Solutions (Q1–Q20)
Q1. Determine order and degree (if defined) of: d⁴y/dx⁴ + sin(y''') = 0
Yaha highest derivative d⁴y/dx⁴ hai, order = 4.
Lekin equation me sin(y''') term hai — y''' polynomial form me nahi hai (sin ke andar hai).
Isliye degree defined nahi hai.
Order = 4, Degree = not defined
Q2. Determine order and degree (if defined) of: (y''')² + (y'')³ + (y')⁴ + y⁵ = 0
Highest derivative y''' hai → Order = 3
Highest order derivative (y''') ka power = 2, aur puri equation polynomial form me hai.
Degree = 2
Q3. For each of the following, determine order and degree: (dy/dx)³ − 4(dy/dx)² + 7y = sin x
Highest derivative dy/dx hai → Order = 1
Highest order derivative ka power = 3, equation polynomial hai
Degree = 3
Q4. Verify that y = e^x + 1 is a solution of the differential equation y'' − y' = 0.
y = eˣ + 1
y' = eˣ
y'' = eˣ
y'' − y' = eˣ − eˣ = 0 ✓
Diya gaya y equation satisfy karta hai, isliye ye ek solution hai.
Q5. Verify that y = x sin x is a solution of the differential equation xy' = y + x√(x² − y²) (x ≠ 0 and x > y or x < −y).
y = x sin x
y' = sin x + x cos x
LHS: xy' = x sin x + x² cos x
RHS: y + x√(x²−y²) = x sin x + x√(x² − x²sin²x) = x sin x + x·x√(1−sin²x) = x sin x + x²cos x
LHS = RHS ✓ (taking cos x ≥ 0)
Hence verified.
Q6. Form the differential equation representing the family of curves y = mx, where m is an arbitrary constant.
y = mx
Differentiate: dy/dx = m
Since m = y/x, substitute: dy/dx = y/x
Required D.E.: x(dy/dx) = y
Q7. Form the differential equation representing the family of curves y = a sin(x + b), where a, b are arbitrary constants.
y = a sin(x + b) ... (1)
Differentiate: y' = a cos(x + b) ... (2)
Differentiate again: y'' = −a sin(x + b) = −y (using eq 1)
Required D.E.: y'' + y = 0
Q8. Form the differential equation of the family of circles touching the y-axis at origin.
Circles touching y-axis at origin: (x − a)² + y² = a², i.e. x² + y² = 2ax ... (1)
Differentiate: 2x + 2y y' = 2a ⟹ a = x + y y'
Substitute a in (1): x² + y² = 2x(x + y y')
x² + y² = 2x² + 2xy y'
y² − x² = 2xy y'
Required D.E.: 2xy(dy/dx) = y² − x²
Q9. Form the differential equation representing the family of ellipses having foci on x-axis and centre at origin: x²/a² + y²/b² = 1.
x²/a² + y²/b² = 1 ... (1) — two arbitrary constants a², b², so differentiate twice.
Differentiate: 2x/a² + 2y y'/b² = 0 ⟹ x/a² + y y'/b² = 0 ... (2)
Differentiate again: 1/a² + (y'² + y y'')/b² = 0 ... (3)
From (2): 1/a² = −y y'/(x b²). Substitute in (3):
−y y'/(x b²) + (y'² + y y'')/b² = 0
Multiply by b²: −y y'/x + y'² + y y'' = 0
Multiply by x: −y y' + x y'² + x y y'' = 0
Required D.E.: xy y'' + x(y')² − y y' = 0
Q10. Solve the differential equation: dy/dx = (1 + x²)/(1 + y²)
Variable separable: (1 + y²) dy = (1 + x²) dx
Integrate both sides: ∫(1+y²)dy = ∫(1+x²)dx
y + y³/3 = x + x³/3 + C
General solution: y + y³/3 = x + x³/3 + C
Q11. Solve the differential equation: dy/dx = (1 − cos x)/(1 + cos x)
1 − cos x = 2sin²(x/2), 1 + cos x = 2cos²(x/2)
dy/dx = tan²(x/2) = sec²(x/2) − 1
dy = [sec²(x/2) − 1] dx
Integrate: y = 2 tan(x/2) − x + C
General solution: y = 2 tan(x/2) − x + C
Q12. Find the particular solution of dy/dx = −4xy² given that y = 1, when x = 0.
Separate variables: dy/y² = −4x dx
Integrate: ∫y⁻² dy = ∫−4x dx
−1/y = −2x² + C
Apply condition x = 0, y = 1: −1 = 0 + C ⟹ C = −1
−1/y = −2x² − 1 ⟹ 1/y = 2x² + 1
Particular solution: y = 1/(1 + 2x²)
Q13. Show that the differential equation (x − y) dy/dx = x + 2y is homogeneous and solve it.
dy/dx = (x + 2y)/(x − y) — dividing num and denom by x: dy/dx = (1 + 2(y/x))/(1 − (y/x)) = F(y/x)
Hence homogeneous.
Put y = vx ⟹ dy/dx = v + x(dv/dx)
v + x(dv/dx) = (1 + 2v)/(1 − v)
x(dv/dx) = (1 + 2v)/(1 − v) − v = (1 + 2v − v + v²)/(1 − v) = (1 + v + v²)/(1 − v)
(1 − v)/(1 + v + v²) dv = dx/x
Split: ∫[1/(1+v+v²)] dv − ∫[v/(1+v+v²)] dv = ln|x| + C
Standard integration gives: (1/2)ln|x²+xy+y²| − √3 tan⁻¹[(2y+x)/(√3 x)] = ln|x| + C (final compact form after simplification)
General solution: log|x² + xy + y²| − 2√3 tan⁻¹[(2y+x)/(√3 x)] = C (constants absorbed)
Q14. Show that the given differential equation x dy − y dx = √(x² + y²) dx is homogeneous and solve it.
dy/dx = [y + √(x²+y²)]/x = y/x + √(1 + (y/x)²) = F(y/x)
Hence homogeneous.
Put y = vx ⟹ dy/dx = v + x(dv/dx)
v + x(dv/dx) = v + √(1+v²)
x(dv/dx) = √(1+v²)
dv/√(1+v²) = dx/x
Integrate: ln|v + √(1+v²)| = ln|x| + ln C = ln|Cx|
v + √(1+v²) = Cx, substitute v = y/x:
y/x + √(1 + y²/x²) = Cx ⟹ y + √(x²+y²) = Cx²
General solution: y + √(x² + y²) = Cx²
Q15. Show that the given differential equation (x² + xy) dy = (x² + y²) dx is homogeneous and solve it.
dy/dx = (x² + y²)/(x² + xy) — divide by x²: dy/dx = (1 + (y/x)²)/(1 + y/x) = F(y/x)
Hence homogeneous.
Put y = vx ⟹ dy/dx = v + x(dv/dx)
v + x(dv/dx) = (1 + v²)/(1 + v)
x(dv/dx) = (1+v²)/(1+v) − v = (1 + v² − v − v²)/(1+v) = (1 − v)/(1 + v)
(1+v)/(1−v) dv = dx/x
Write (1+v)/(1−v) = −1 + 2/(1−v):
∫[−1 + 2/(1−v)] dv = ∫dx/x
−v − 2 ln|1 − v| = ln|x| + C
Substitute v = y/x: −(y/x) − 2 ln|1 − y/x| = ln|x| + C
General solution: −y/x − 2 ln|(x−y)/x| = ln|x| + C, jo simplify hoke: x²(x−y)² e^(y/x)... form me bhi likha ja sakta hai (log form sabse safe hai).
Q16. Solve the differential equation: dy/dx + 2y = sin x (linear differential equation).
Standard linear form dy/dx + Py = Q, P = 2, Q = sin x
I.F. = e^∫2 dx = e^(2x)
Solution: y · e^(2x) = ∫sin x · e^(2x) dx + C
∫e^(2x) sin x dx = e^(2x)(2 sin x − cos x)/5 (standard reduction formula ∫e^(ax)sin bx dx = e^(ax)(a sin bx − b cos bx)/(a²+b²))
So y e^(2x) = e^(2x)(2 sin x − cos x)/5 + C
General solution: y = (2 sin x − cos x)/5 + C e^(−2x)
Q17. Solve the differential equation: x dy/dx + 2y = x², (x ≠ 0).
Divide by x: dy/dx + (2/x)y = x
Here P = 2/x, Q = x
I.F. = e^∫(2/x)dx = e^(2 ln x) = x²
Solution: y·x² = ∫x · x² dx + C = ∫x³ dx + C = x⁴/4 + C
General solution: y x² = x⁴/4 + C, i.e. y = x²/4 + C/x²
Q18. Find the particular solution of the differential equation dy/dx + y cot x = 2x + x² cot x (x ≠ 0), given that y = 0 when x = π/2.
Linear form: P = cot x, Q = 2x + x² cot x
I.F. = e^∫cot x dx = e^(ln|sin x|) = sin x
Solution: y sin x = ∫(2x + x² cot x) sin x dx + C
= ∫2x sin x dx + ∫x² cos x dx + C
∫2x sin x dx = −2x cos x + 2 sin x (by parts)
∫x² cos x dx = x² sin x − ∫2x sin x dx = x² sin x − (−2x cos x + 2 sin x) = x² sin x + 2x cos x − 2 sin x
Sum: (−2x cos x + 2 sin x) + (x² sin x + 2x cos x − 2 sin x) = x² sin x
So y sin x = x² sin x + C
Apply x = π/2, y = 0: 0 = (π/2)²(1) + C ⟹ C = −π²/4
Particular solution: y sin x = x² sin x − π²/4
Q19. Find the equation of a curve passing through the point (0, 0) and whose differential equation is y' = eˣ sin x.
dy/dx = eˣ sin x
dy = eˣ sin x dx
Integrate: ∫eˣ sin x dx = eˣ(sin x − cos x)/2 (standard formula, a=1,b=1)
y = eˣ(sin x − cos x)/2 + C
Apply (0,0): 0 = 1(0 − 1)/2 + C = −1/2 + C ⟹ C = 1/2
Curve: y = eˣ(sin x − cos x)/2 + 1/2
Q20. In a bank, principal increases continuously at the rate of 5% per year. In how many years will Rs 1000 double itself, given log_e 2 = 0.6931?
Let principal = P at time t. Given dP/dt = (5/100) P = P/20
Separate: dP/P = dt/20
Integrate: ln P = t/20 + C
At t=0, P = 1000: ln 1000 = C
So ln(P/1000) = t/20
When P = 2000: ln 2 = t/20
t = 20 × ln 2 = 20 × 0.6931
t ≈ 13.86 years
Important Equations — Ek Nazar Me
| Concept | Formula / Rule |
|---|---|
| Order | Highest derivative present in the equation |
| Degree | Power of highest order derivative, jab equation derivatives me polynomial ho |
| Variable separable form | f(x) dx = g(y) dy ⟹ ∫f(x)dx = ∫g(y)dy + C |
| Homogeneous test | dy/dx = F(y/x) (degree-zero homogeneous function) |
| Homogeneous substitution | y = vx ⟹ dy/dx = v + x(dv/dx); solve karke v = y/x waapis daalo |
| Linear D.E. (in y) | dy/dx + Py = Q, where P, Q are functions of x only |
| Integrating Factor | I.F. = e^∫P dx |
| Linear D.E. solution | y · (I.F.) = ∫ Q · (I.F.) dx + C |
| Linear D.E. (in x) | dx/dy + Px = Q, I.F. = e^∫P dy, solution: x·(I.F.) = ∫Q·(I.F.) dy + C |
| ∫e^(ax) sin bx dx | e^(ax)(a sin bx − b cos bx)/(a² + b²) + C |
| ∫e^(ax) cos bx dx | e^(ax)(a cos bx + b sin bx)/(a² + b²) + C |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- Order aur degree confuse karna. Order = highest derivative kaunsa hai (1st, 2nd...); degree = uska power. Ye do alag cheezein hain, students inhe mix kar dete hain.
- Degree har jagah define hoti hai maan lena. Agar derivative sin, log, sqrt ya exponent ke andar ho (jaise sin(dy/dx) ya e^(d²y/dx²)), toh equation polynomial form me nahi hai — degree defined hi nahi hoti. Sirf order likho.
- Homogeneous test galat karna. Direct dy/dx dekh ke bol dena 'homogeneous hai' galat hai — pehle poora expression F(y/x) form me convert karke check karo dono terms same degree ke hain.
- y = vx substitution ke baad v ko wapas y/x se replace karna bhool jaana. Final answer x aur y me hona chahiye, v me nahi — ye sabse common exam mistake hai.
- Integrating factor ke exponent me sign ki galti. I.F. = e^∫P dx hai, na ki e^(−∫P dx). P ka sign equation se sahi copy karo, especially jab equation ko standard form dy/dx + Py = Q me convert karte waqt divide karna padta hai.
- General solution me arbitrary constant +C likhna bhool jaana. Integration ke baad +C zaroor likho — bina C ke answer 'particular case' ban jaata hai, general solution nahi. Particular solution wale question me hi C ki value nikaal ke substitute karo.
Board-Style Important Questions
- 1 mark: Write the order and degree (if defined) of the differential equation given.
- 1 mark: State whether a given first-order differential equation is homogeneous or not.
- 2 marks: Form the differential equation representing a given family of curves with one arbitrary constant.
- 4 marks: Solve a variable separable differential equation.
- 6 marks: Show a given differential equation is homogeneous and solve it using y = vx substitution.
- 6 marks: Find the particular solution of a linear differential equation using integrating factor, given an initial condition.
Aksar Poochhe Jaane Wale Sawaal
Differential equation ka order aur degree kaise nikale?
Order = equation me jo sabse highest derivative present hai (dy/dx ho toh 1, d2y/dx2 ho toh 2, waise hi). Degree = us highest order derivative ka power, lekin sirf tab jab poori equation derivatives me polynomial ho — agar derivative sin, log, ya exponent ke andar hai toh degree define nahi hoti.
Homogeneous differential equation kaise pehchane?
Equation ko dy/dx = F(x,y) form me likho. Agar F(x,y) ko sirf y/x (ya x/y) ke function ke roop me likha ja sake — matlab x aur y dono ka same degree ho aur unhe divide karke ek variable v = y/x ban jaaye — toh equation homogeneous hai.
y = vx substitution kyu use karte hain?
Homogeneous equation dy/dx = F(y/x) me directly variables separate nahi hote. y = vx substitute karne se dy/dx = v + x(dv/dx) ban jaata hai, jisse equation v aur x ke terms me variable-separable form me aa jaati hai — solve karne ke baad v = y/x wapas daal do.
Integrating factor (I.F.) kya hota hai aur kyu zaroori hai?
Linear differential equation dy/dx + Py = Q ko directly integrate nahi kar sakte kyunki LHS ek perfect derivative nahi hai. I.F. = e^∫Pdx se multiply karne par LHS d/dx(y · I.F.) ban jaata hai, jo directly integrate ho jaata hai — isliye I.F. zaroori hai.
General solution aur particular solution me kya farak hai?
General solution me arbitrary constant(s) hote hain (jaise +C), jo curves ki poori family represent karta hai. Jab question me koi initial ya boundary condition di ho (jaise x=0 pe y=1), toh us condition se C ki specific value nikaal ke particular solution milta hai.
Family of curves se differential equation kaise banate hain?
Curve ki equation me jitne arbitrary constants hon, utni baar equation ko differentiate karo. Fir original equation aur derivative wali equations ko combine karke algebra se constants eliminate kar do — jo equation bachegi wahi required differential equation hai.
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