Class 12 Mathematics Notes Chapter 3 (Differential Equations) – Mathematics Part-II Book

Alright class, let's get straight into Differential Equations. This chapter is crucial not just for your board exams but forms the basis for many concepts asked in various government examinations. We'll cover the key concepts from your NCERT textbook, focusing on what's essential for competitive settings.

Chapter 9: Differential Equations - Detailed Notes for Government Exam Preparation

1. Introduction & Basic Concepts

• Definition: An equation involving derivatives (or differentials) of one or more dependent variables with respect to one or more independent variables is called a differential equation.
  • Example: dy/dx = x + sin(x), d²y/dx² + y = 0, (dy/dx)² + dy/dx = cos(x)
• Ordinary Differential Equation (ODE): A differential equation involving derivatives of the dependent variable with respect to only one independent variable. Our syllabus primarily deals with ODEs.
  • Examples above are all ODEs.
• Partial Differential Equation (PDE): Involves partial derivatives with respect to more than one independent variable. (Generally outside the scope of Class 12 based exams, but good to know the distinction).
  • Example: ∂²u/∂x² + ∂²u/∂y² = 0

2. Order and Degree of a Differential Equation

• Order: The order of the highest order derivative present in the differential equation.

  • dy/dx + y = x has order 1.
  • d³y/dx³ + x(d²y/dx²) + y = 0 has order 3.
  • (dy/dx)² + y = 0 has order 1.

• Degree: The highest power (positive integer) of the highest order derivative, after the differential equation has been made free from radicals and fractions as far as the derivatives are concerned (i.e., expressed as a polynomial equation in derivatives).

  • d²y/dx² + (dy/dx)³ + y = 0: Order is 2. The highest order derivative is d²y/dx², its power is 1. So, Degree is 1.
  • (d³y/dx³)² + 2(d²y/dx²)⁵ + y = 0: Order is 3. The highest order derivative is d³y/dx³, its power is 2. So, Degree is 2.
  • dy/dx + sin(dy/dx) = 0: Order is 1. Degree is not defined because the equation cannot be expressed as a polynomial in dy/dx.
  • √(1 + (dy/dx)²) = d²y/dx²: First, square both sides: 1 + (dy/dx)² = (d²y/dx²)². Order is 2. Highest order derivative is d²y/dx², its power is 2. So, Degree is 2.

  Key Point for Exams: Questions directly asking for order and degree are common. Be careful with radicals and functions of derivatives when determining the degree.

3. Formation of a Differential Equation

• Given a family of curves (an equation with arbitrary constants), a differential equation can be formed whose solution is that family.
• Process:
  1. Write down the given equation of the family of curves.
  2. Count the number of independent arbitrary constants (say, 'n').
  3. Differentiate the equation 'n' times successively.
  4. Eliminate the 'n' arbitrary constants using the original equation and the 'n' derived equations.
  5. The resulting equation, free from arbitrary constants, is the required differential equation. Its order will be 'n'.
• Example: Form the DE for y = mx (family of lines through origin, 'm' is the constant).
  • n = 1 (only 'm').
  • Differentiate once: dy/dx = m.
  • Eliminate 'm' using the original equation (m = y/x) and the derivative: dy/dx = y/x or x(dy/dx) - y = 0. This is the required DE.

4. Solution of a Differential Equation

• General Solution: A solution which contains as many independent arbitrary constants as the order of the differential equation. It represents the family of curves satisfying the DE.
• Particular Solution: A solution obtained from the general solution by giving particular values to the arbitrary constants. These values are usually determined using given initial or boundary conditions.
  • Example: For dy/dx = 2x, the general solution is y = x² + C. If it's given that y(0) = 5, then 5 = 0² + C => C = 5. The particular solution is y = x² + 5.

5. Methods of Solving First Order, First Degree Differential Equations

These are the most important types for your exams.

• (a) Variables Separable:

  • Form: The equation can be written as f(x) dx = g(y) dy or f(x) dx + g(y) dy = 0.
  • Method: Separate the terms involving x and dx on one side and terms involving y and dy on the other side. Integrate both sides and add an arbitrary constant of integration 'C' on one side.
  • ∫ f(x) dx = ∫ g(y) dy + C
  • Example: Solve dy/dx = (1+y²)/(1+x²).
    • Separate: dy/(1+y²) = dx/(1+x²).
    • Integrate: ∫ dy/(1+y²) = ∫ dx/(1+x²) + C.
    • Solution: tan⁻¹(y) = tan⁻¹(x) + C.

• (b) Homogeneous Differential Equations:

  • Definition: A function F(x, y) is homogeneous of degree 'n' if F(λx, λy) = λⁿ F(x, y). A first-order DE of the form dy/dx = F(x, y) (or dx/dy = G(x, y)) is homogeneous if F(x, y) is a homogeneous function of degree zero (i.e., can be expressed as g(y/x) or h(x/y)).
    • Often recognizable as equations where each term has the same total degree in x and y. Example: dy/dx = (x² + y²)/(xy).
  • Method:
    1. Substitute y = vx. Then dy/dx = v + x(dv/dx).
    2. Substitute these into the original equation. The equation will transform into a variables separable form in terms of 'v' and 'x'.
    3. Separate variables (v and dv on one side, x and dx on the other).
    4. Integrate both sides.
    5. Replace v with y/x to get the solution in terms of x and y.
    • (Alternatively, if the form is dx/dy = H(x, y) where H is homogeneous of degree zero, use substitution x = vy).
  • Example: Solve dy/dx = (x+y)/x.
    • dy/dx = 1 + y/x. This is homogeneous.
    • Put y = vx => dy/dx = v + x(dv/dx).
    • v + x(dv/dx) = 1 + v => x(dv/dx) = 1.
    • Separate: dv = dx/x.
    • Integrate: ∫ dv = ∫ dx/x + C => v = ln|x| + C.
    • Replace v = y/x: y/x = ln|x| + C or y = x(ln|x| + C).

• (c) Linear Differential Equations:

  • Form: dy/dx + P(x) y = Q(x), where P(x) and Q(x) are functions of x only (or constants).
  • Method:
    1. Identify P(x) and Q(x).
    2. Calculate the Integrating Factor (I.F.): I.F. = e^(∫ P(x) dx).
    3. The general solution is given by: y * (I.F.) = ∫ [Q(x) * (I.F.)] dx + C.
  • Alternative Form: dx/dy + P₁(y) x = Q₁(y), where P₁(y) and Q₁(y) are functions of y only (or constants).
    • I.F. = e^(∫ P₁(y) dy).
    • Solution: x * (I.F.) = ∫ [Q₁(y) * (I.F.)] dy + C.
  • Example: Solve dy/dx + (1/x)y = x².
    • This is linear with P(x) = 1/x and Q(x) = x².
    • ∫ P(x) dx = ∫ (1/x) dx = ln|x|.
    • I.F. = e^(ln|x|) = |x|. Assuming x > 0, I.F. = x.
    • Solution: y * x = ∫ (x² * x) dx + C = ∫ x³ dx + C.
    • xy = x⁴/4 + C.

Key Strategy for Exams:

1. Check if the equation is Variables Separable.
2. If not, check if it's Homogeneous (can it be written as dy/dx = g(y/x)?).
3. If not, check if it's Linear (dy/dx + Py = Q or dx/dy + P₁x = Q₁).

Mastering these three forms and their solution methods is essential. Remember to practice integration thoroughly, as it's integral (pun intended!) to solving differential equations.

Multiple Choice Questions (MCQs)

1. The order and degree of the differential equation (d²y/dx²)³ + (dy/dx)⁴ + sin(x) + 1 = 0 are respectively:
   (a) 2, 4
   (b) 2, 3
   (c) 3, 2
   (d) 4, 2

2. The order and degree of the differential equation [1 + (dy/dx)²]^(3/2) = 5 (d²y/dx²) are respectively:
   (a) 2, 2
   (b) 2, 3
   (c) 1, 3
   (d) 2, degree not defined

3. The differential equation representing the family of curves y = A cos(x) + B sin(x), where A and B are arbitrary constants, is:
   (a) d²y/dx² + y = 0
   (b) d²y/dx² - y = 0
   (c) dy/dx + y = 0
   (d) d²y/dx² + (dy/dx)² = 0

4. The general solution of the differential equation dy/dx = e^(x-y) is:
   (a) e^y = e^x + C
   (b) e^x + e^y = C
   (c) e^(-y) = e^x + C
   (d) e^x + e^(-y) = C

5. The differential equation dy/dx = (x²+y²)/(2xy) is:
   (a) Linear
   (b) Homogeneous
   (c) Variables Separable
   (d) None of these

6. The Integrating Factor (I.F.) for the differential equation x(dy/dx) - y = 2x² is:
   (a) x
   (b) -1/x
   (c) 1/x
   (d) e^(-1/x)

7. The solution of the differential equation dy/dx + y tan(x) = sec(x) is:
   (a) y sec(x) = tan(x) + C
   (b) y tan(x) = sec(x) + C
   (c) y cos(x) = x + C
   (d) y sin(x) = x + C

8. The particular solution of the differential equation dy/dx = y/x, given that y=1 when x=1, is:
   (a) y = x
   (b) y = 1/x
   (c) y = ln|x|
   (d) y = e^x

9. The differential equation formed by eliminating the arbitrary constant 'c' from y² = 4ax is: (Assuming 'a' is a fixed constant)
   (a) y = 2x (dy/dx)
   (b) y² = x (dy/dx)
   (c) 2y = x (dy/dx)
   (d) y = x² (dy/dx)
   (Self-correction: The question should likely eliminate 'a' treating it as the arbitrary constant. Let's assume 'a' is the arbitrary constant)
   The differential equation formed by eliminating the arbitrary constant 'a' from y² = 4ax is:
   (a) y = 2x (dy/dx)
   (b) y² = x (dy/dx)
   (c) 2y = x (dy/dx)
   (d) y = x² (dy/dx)

10. The general solution of dy/dx = (y+x)/x is:
    (a) y = x log|x| + Cx
    (b) y = x log|x| + C
    (c) log|y| = x + C
    (d) y = log|x| + Cx

Answer Key for MCQs:

1. (b) Order is 2 (highest derivative d²y/dx²), Degree is 3 (power of highest derivative).
2. (a) Square both sides: [1 + (dy/dx)²]³ = 25 (d²y/dx²)². Highest derivative is d²y/dx², its highest power is 2. Order = 2, Degree = 2.
3. (a) y = A cos(x) + B sin(x). dy/dx = -A sin(x) + B cos(x). d²y/dx² = -A cos(x) - B sin(x) = -(A cos(x) + B sin(x)) = -y. So, d²y/dx² + y = 0.
4. (a) dy/dx = e^x / e^y => e^y dy = e^x dx. Integrate: ∫ e^y dy = ∫ e^x dx + C => e^y = e^x + C.
5. (b) dy/dx = (1 + (y/x)²)/(2(y/x)). It's a function of y/x, hence homogeneous.
6. (c) First write in standard form: dy/dx - (1/x)y = 2x. Here P(x) = -1/x. ∫ P(x) dx = ∫ (-1/x) dx = -ln|x| = ln|1/x|. I.F. = e^(ln|1/x|) = 1/x (assuming x>0).
7. (a) P(x) = tan(x), Q(x) = sec(x). ∫ P(x) dx = ∫ tan(x) dx = ln|sec(x)|. I.F. = e^(ln|sec(x)|) = sec(x). Solution: y * sec(x) = ∫ sec(x) * sec(x) dx + C = ∫ sec²(x) dx + C = tan(x) + C. So, y sec(x) = tan(x) + C.
8. (a) dy/y = dx/x. Integrate: ln|y| = ln|x| + C₁. ln|y/x| = C₁ => y/x = e^(C₁) = C. So, y = Cx. Given y=1 when x=1, 1 = C*1 => C=1. Particular solution is y = x.
9. (a) y² = 4ax. Differentiate wrt x: 2y (dy/dx) = 4a. So, 4a = 2y (dy/dx). Substitute back into original equation: y² = (2y dy/dx) * x. Since y is generally not 0, divide by y: y = 2x (dy/dx).
10. (a) dy/dx = y/x + 1. Put y = vx => dy/dx = v + x(dv/dx). v + x(dv/dx) = v + 1 => x(dv/dx) = 1. dv = dx/x. Integrate: v = ln|x| + C. Replace v = y/x: y/x = ln|x| + C => y = x ln|x| + Cx.

Study these concepts thoroughly, practice identifying the type of first-order DE, and be very comfortable with integration techniques. Good luck with your preparation!