MATDSM-251/252: Differential Equations FYUG Even Semester Exam, 2025

Programme: FYUG (4th Semester)

Course Code: MATDSM-251/252

Full Marks: 70

Time Duration: 3 Hours

UNIT I: First Order Differential Equations

Question 1 (a)

[2 Marks]

Write one solution for dy/dx = 3x².

Integrating both sides with respect to x:

y = ∫ 3x² dx = x³ + C

One specific solution (letting C = 0) is y = x³.

Question 1 (c)

[2 Marks]

Determine whether (2xy + y²)dx + (x² + 2xy)dy = 0 is exact.

Here, M = 2xy + y² and N = x² + 2xy.

  • ∂M/∂y = 2x + 2y
  • ∂N/∂x = 2x + 2y

Since ∂M/∂y = ∂N/∂x, the differential equation is exact.

Question 2 (b)

[5 Marks]

Solve dy/dx + 2y/x = x³.

This is a linear differential equation of the form dy/dx + Py = Q, where P = 2/x and Q = x³.

  1. Integrating Factor (IF): IF = e^(∫ P dx) = e^(∫ 2/x dx) = e^(2 ln x) = x².
  2. General Solution: y * (IF) = ∫ Q * (IF) dx + C
  3. y * x² = ∫ x³ * x² dx + C = ∫ x⁵ dx + C
  4. y * x² = x⁶/6 + C

Final Answer: y = x⁴/6 + C/x².

UNIT II: Higher Order & Clairaut's Equations

Question 3 (c)

[2 Marks]

Write the general form of Clairaut's equation.

y = xp + f(p), where p = dy/dx.

Question 4 (a)

[5 Marks]

Find the general solution of y'' + 3y' + 2y = e⁻²ˣ.

1. Complementary Function (CF): Characteristic equation is m² + 3m + 2 = 0.

  • (m + 1)(m + 2) = 0 ⇒ m = -1, -2.
  • CF = C₁e⁻ˣ + C₂e⁻²ˣ.

2. Particular Integral (PI): PI = [1 / (D² + 3D + 2)] e⁻²ˣ.

Since D = -2 makes the denominator zero, we use the formula for a repeated root: PI = x * [1 / (2D + 3)] e⁻²ˣ = x * [1 / (2(-2) + 3)] e⁻²ˣ = -xe⁻²ˣ.

General Solution: y = C₁e⁻ˣ + C₂e⁻²ˣ - xe⁻²ˣ.

UNIT III: Cauchy-Euler & Variation of Parameters

Question 5 (b)

[2 Marks]

What substitution reduces x²y'' - 4xy' + 6y = 0 into a constant coefficient equation?

The standard substitution is x = eᶻ or z = ln x.

The transformed equation becomes: [D(D-1) - 4D + 6]y = 0 ⇒ (D² - 5D + 6)y = 0, where D = d/dz.

Question 6 (b)

[5 Marks]

Solve x²y'' + 5xy' + 6y = 0 using substitution.

Substitute x = eᶻ, then x(dy/dx) = Dy and x²(d²y/dx²) = D(D-1)y.

Equation becomes: [D(D-1) + 5D + 6]y = 0 ⇒ (D² + 4D + 6)y = 0.

Characteristic roots: m = [-4 ± √(16 - 24)] / 2 = -2 ± i√2.

y = e⁻²ᶻ [C₁ cos(√2 z) + C₂ sin(√2 z)].

Final Answer: y = x⁻² [C₁ cos(√2 ln x) + C₂ sin(√2 ln x)].

UNIT IV: Simultaneous & Total Differential Equations

Question 7 (c)

[2 Marks]

Check if (2x + 3y)dx + (3x + 4y)dy = 0 is a total differential equation.

Here M = 2x + 3y and N = 3x + 4y.

  • ∂M/∂y = 3
  • ∂N/∂x = 3

Since ∂M/∂y = ∂N/∂x, it is a total (exact) differential equation.

UNIT V: Partial Differential Equations (PDE)

Question 9 (b)

[2 Marks]

Find the order and degree of the PDE: ∂²z/∂x² + (∂z/∂x)² = 0.

  • Order: 2 (highest derivative is second order).
  • Degree: 1 (the highest order derivative is raised to the power 1).

Question 10 (a)

[5 Marks]

Solve (x-y)∂z/∂x + (x+y)∂z/∂y = z using Lagrange's method.

Lagrange's auxiliary equations are: dx / (x-y) = dy / (x+y) = dz / z.

  1. From dy/dx = (x+y)/(x-y), this is homogeneous. Let y=vx.
  2. Solving leads to ln(x² + y²) - 2 arctan(y/x) = C₁.
  3. Using multipliers (dx-dy)/( (x-y)-(x+y) ) = dz/z leads to second characteristic C₂.

General Solution: Φ(C₁, C₂) = 0.

Exam Focus Enhancements

Exam Tips

Important Formulas List

Answer Presentation Strategy

Clearly separate the Complementary Function (CF) and Particular Integral (PI) when solving second-order linear equations to ensure partial marks.