NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations

NCERT Solutions for Class 12 Maths Chapter 9: The NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations help students clearly understand one of the most important and logical topics in Mathematics. This chapter explains how a function and its derivatives are related, and how they can be used to solve real-life mathematical problems. The differential equations class 12 topic is a key part of the CBSE Syllabus Class 12 Maths and also carries good weightage in CBSE Class 12 Board exams.

These differential equations NCERT solutions are designed in a step-by-step way so students can follow the concept without feeling confused. The ncert solutions class 12 differential equations PDF helps you practice different types of questions — from basic to advanced. Whether you are looking for differential equations class 12 ncert solutions chapter 9 or want to download differential equations class 12 ncert solutions pdf, these notes give you clear explanations with solved examples.

NCERT Solutions For Class 12 Maths Chapter 9 Differential Equations PDF Download

Infinity Learn gives you the access to NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations free pdf download offers clear and detailed answers to all exercise questions. These NCERT solutions cover important topics like order and degree, general and particular solutions, formation of differential equations, and methods of solving first-order equations. Perfect for CBSE Class 12 students, the PDF helps strengthen concepts, aids in board exam preparation, and supports competitive exam readiness with stepwise explanations and examples.

Questions on Class 12 Maths Chapter 9 Differential Equations with Solutions

Q1. Order & Degree

Identify order and degree of: d⁴y/dx⁴ + sin(dy/dx) = 0

Solution: Highest derivative is d⁴y/dx⁴ ⇒ order = 4. Not a polynomial in derivatives due to sin(dy/dx) ⇒ degree not defined.

Q2. Verify solution

Verify that y = e^x + 1 solves d²y/dx² − dy/dx = 0.

Solution: y' = e^x, y'' = e^x. LHS = e^x − e^x = 0. Verified.

Q3. Particular from general

General: y = C x². Find particular with y(1)=3.

Solution: C=3. Hence y = 3x².

Q4. Form DE (eliminate constants)

Family: y = a x² + b.

Solution: Differentiate: y' = 2ax, y'' = 2a. Then a = (1/2) y''. From y': y' = x y''. Required DE: y' − x y'' = 0.

Q5. Separable

dy/dx = 3x² /(2y)

Solution: 2y dy = 3x² dx ⇒ y² = x³ + C ⇒ y = ±√(x³ + C).

Q6. Linear 1st order

dy/dx + 2y = x

Solution: I.F. = e^∫2dx = e^2x. Then d(y e^2x)/dx = x e^2x. Integrate by parts ⇒ y = (1/2)x − 1/4 + C e^−2x.

Q7. Homogeneous

dy/dx = (x + y)/x

Solution: Put y = v x ⇒ y' = v + x v'. Then v + x v' = 1 + v ⇒ x v' = 1 ⇒ v = ln|x| + C. Hence y = x ln|x| + Cx.

Q8. Application (growth)

Population model: dy/dt = k y, 1999: 20000; 2004: 25000; find 2009.

Solution: y = A e^{k t}, take t=0 at 1999 ⇒ A=20000. From 2004: e^{5k}=1.25 ⇒ k=(ln1.25)/5. 2009: y(10)=20000(1.25)²=31250.

Q9. Verify family solves DE

Show y = C x² solves x y' − 2y = 0.

Solution: y' = 2Cx. Then x(2Cx) − 2(Cx²) = 0. Verified.

Q10. Substitution

dy/dx = (x − y)/(x + y)

Solution: Put y = v x ⇒ v + x v' = (1 − v)/(1 + v) ⇒ x v' = (1 − 2v − v²)/(1 + v). Separate: (1 + v)/(1 − 2v − v²) dv = dx/x. Integrate ⇒ −(1/2) ln|2 − (v + 1)²| = ln|x| + C. Replace v = y/x and simplify ⇒ x² − 2xy − y² = C.

Q11. Separable basics

dy/dx = 3y

Solution: dy/y = 3 dx ⇒ ln|y| = 3x + C ⇒ y = C e^3x.

Q12. Separable with shift

dy/dx = 2(y − 5)

Solution: dy/(y − 5) = 2 dx ⇒ ln|y − 5| = 2x + C ⇒ y = 5 + C e^2x.

Q13. Separable rational in x

dy/dx = 1/[x(1 + x)]

Solution: Partial fractions: 1/x − 1/(1 + x). Integrate ⇒ y = ln|x| − ln|1 + x| + C = ln|x/(1 + x)| + C.

Q14. Linear, constant coeff

dy/dx − 4y = 8

Solution: I.F. = e^−4x. Then d( y e^−4x)/dx = 8 e^−4x. Integrate ⇒ y = −2 + C e^4x.

Q15. Linear, variable coeff

dy/dx + (2/x) y = x, x > 0

Solution: I.F. = x². Then d(x² y)/dx = x³ ⇒ x²y = x⁴/4 + C ⇒ y = x²/4 + C/x².

Q16. Linear + IC

dy/dx + y = e^x, y(0)=2

Solution: I.F. = e^x. Then d( y e^x)/dx = e^2x ⇒ y e^x = e^2x/2 + C. Use IC ⇒ C = 3/2. Hence y = (1/2)e^x + (3/2)e^−x.

Q17. Homogeneous (y = v x)

dy/dx = (y + x)/y

Solution: y = v x ⇒ v + x v' = (v + 1)/v ⇒ x v' = (1 − v²)/v. Then v/(1 − v²) dv = dx/x ⇒ ln|1 − v²| = −2 ln|x| + C ⇒ x² − y² = C.

Q18. Homogeneous with IC

dy/dx = (x − 2y)/(x + y), y(1)=1

Solution: Put y = v x ⇒ v + x v' = (1 − 2v)/(1 + v) ⇒ x v' = (1 − 3v − v²)/(1 + v). Separate and integrate (partial fractions) to obtain implicit relation (y + x)² = C x (x + 3y). Apply (x,y)=(1,1) ⇒ C=1. Hence (y + x)² = x(x + 3y).

Q19. Exactness test

(2xy + y²) dx + (x² + 2xy) dy = 0

Solution: M = 2xy + y², N = x² + 2xy. M_y = 2x + 2y = N_x. Exact.

Q20. Solve exact

(2xy + y²) dx + (x² + 2xy) dy = 0

Solution: Potential φ: integrate M in x ⇒ φ = x²y + x y² + g(y). Match φ_y=N ⇒ g'(y)=0. Solution: x²y + x y² = C.

Q21. Exact directly

(y + 2x) dx + (x + 2y) dy = 0

Solution: Exact. φ = x y + x² + y² = C.

Q22. Bernoulli

y' + (2/x) y = x³ y², x>0

Solution: Let v = y⁻¹. Then v' − (2/x) v = −x³. I.F. = x⁻². So d(v x⁻²)/dx = −x ⇒ v = −(1/2) x⁴ + C x². Hence 1/y = C x² − x⁴/2.

Q23. Clairaut

y = x (dy/dx) + 3

Solution: Put p = dy/dx. Then y = xp + 3. Differentiate: y' = p = p + x dp/dx ⇒ x dp/dx = 0 ⇒ p constant. General solution: y = p x + 3. Envelope (singular): eliminate p from y = px + 3 and x = 0 gives y = 3.

Q24. Orthogonal trajectories (lines)

Family: y = m x

Solution: Slope = y/x. Orthogonal slope = −x/y. DE: y dy = −x dx ⇒ x² + y² = C.

Q25. Orthogonal trajectories (circles)

Family: x² + y² = 2 a x

Solution: Differentiate: 2x + 2y y' = 2a ⇒ y' = (a − x)/y. Replace a = (x² + y²)/(2x). Then y' = (y² − x²)/(2xy). Orthogonal slope = −2xy/(y² − x²). Integrate ⇒ y/x = C ⇒ lines through origin.

Q26. Logistic-type

dy/dx = k y (M − y)

Solution: dy/[y(M − y)] = k dx. Partial fractions ⇒ (1/M)(ln|y| − ln|M − y|) = kx + C ⇒ y = M /(1 + C e^{−k M x}).

Q27. Decay + half-life

dy/dt = −λ y

Solution: y = y₀ e^{−λ t}. Half-life T_1/2 = (ln 2)/λ.

Q28. Mixing tank

Volume constant V, rate r, inflow conc c

Solution: dy/dt = r c − r(y/V). Linear ⇒ y = c V + C e^−(r/V)t.

Q29. Newton cooling

dT/dt = −k (T − T_a)

Solution: T(t) − T_a = (T₀ − T_a) e^{−k t}.

Q30. Form DE from solution

y = A e^2x + B e^−2x

Solution: y'' = 4y. Required DE: y'' − 4y = 0.

Q31. Form DE for circle

x² + y² = a²

Solution: Differentiate: 2x + 2y y' = 0 ⇒ x + y y' = 0.

Q32. Verify by substitution

Check y = x ln x solves x y' − y = x.

Solution: y' = ln x + 1. LHS = x(ln x + 1) − x ln x = x. Verified.

Q33. Linear with trig P(x)

y' + (tan x) y = sin x

Solution: I.F. = e^{∫ tan x dx} = sec x. Then d(y sec x)/dx = tan x. Integrate ⇒ y = cos x (−ln|cos x| + C).

Q34. Exact quadratic form

(2x − y) dx + (−x + 2y) dy = 0

Solution: Exact. Potential φ = x² − x y + y² = C.

Q35. Bernoulli simple

y' + y = y²

Solution: y' = y² − y. Separate: dy/[y(y − 1)] = dx. PF ⇒ ln|y − 1| − ln|y| = x + C ⇒ y = 1/(1 − C e^x).

Q36. Separable with trig

y' = y cot x

Solution: dy/y = cot x dx ⇒ ln|y| = ln|sin x| + C ⇒ y = C sin x.

NCERT Solutions for Class 12 Maths Chapter 9 – Differential Equations

The NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations is an essential part of the Calculus unit, which carry around 35 marks in final CBSE Class 12 Board exams. This chapter has six exercises and also one miscellaneous exercise to help students get deep understanding of the basic concepts. These differential equations class 12 ncert solutions chapter 9 are given in step-by-step format to make learning easier.

Adifferential equation is an equation that contain derivative of a dependent variable with respect to one or more independent variables. The order of a differential equation tells the highest order derivative present, and its degree shows the power of that highest derivative (only if it is a polynomial). The differential equations ncert solutions explain how to find both general and particular solutions depending on the number of arbitrary constants involved.

Students also learn how to form a differential equation from a given function by differentiating and removing the constants. One important method, known as variable separable method, is used for equations where the variables x and y can be separated — terms with y go with dy, and terms with x go with dx.

Key Features of NCERT Solutions for Differential Equations

• Clear explanation of definition, order, and degree of differential equations.
• Diagrammatic and solved examples to support conceptual clarity.
• Step-wise solutions for differential equations of first order and first degree.
• Practice sets for the homogeneous differential equations and variable separable method.
• Easy references for theory and formula from CBSE Syllabus Class 12 Maths and differential equations class 12 ncert solutions pdf.

These ncert solutions class 12 differential equations help students in board exam as well as in competitive exams like JEE Exam.