Video summary

Maths-4 Unit-1 One Shot | Partial Differential Equations | Gulshan Sir | AKTU | Gateway Classes

Main summary

Key takeaways

Educational

Main ideas / lessons conveyed

  • Purpose of the video (AKTU Unit-1): A “one-shot” focused on Engineering Mathematics – Unit 1: Partial Differential Equations (PDEs), emphasizing important exam topics rather than the full syllabus.
  • Unit-1 structure: Covers the core classification and solution methods for PDEs, especially:
    • Order and degree
    • Recognizing PDE types
    • Solving first-order PDEs using:
      • Lagrange method
      • Charpit method
    • Solving linear PDEs with constant coefficients:
      • Homogeneous vs non-homogeneous
      • Complementary Function (CF) and Particular Integral (PI)
    • Solving PDEs reducible to constant-coefficient linear PDEs via substitutions.

Methodologies & instruction lists (detailed)

A) Notation to know (used throughout Unit-1)

If (z=f(x,y)) then:

  • (p = \frac{\partial z}{\partial x} = \frac{dz}{dx})
  • (q = \frac{\partial z}{\partial y} = \frac{dz}{dy})
  • (r = \frac{\partial^2 z}{\partial x^2})
  • (s = \frac{\partial^2 z}{\partial x\partial y})
  • (t = \frac{\partial^2 z}{\partial y^2})

Key independence/dependence interpretation (as used in the video context):

  • (x) is treated as an independent variable
  • (y) as dependent (in the explained PDE examples)

B) Lagrange Method (first-order quasi-linear PDEs in standard form)

1) How to recognize when Lagrange method applies

Lagrange applies to first-order quasi-linear PDEs in standard form, typically:

  • (Pp + Qq = R) (or equivalent standard Lagrange form)

Recognition criteria emphasized:

  • The PDE must be first order (involves only first derivatives (p) and (q) in the quasi-linear sense).
  • Coefficients (P,Q,R) are functions of (x,y,z) or constants.

2) Working rule / steps (Lagrange)

  1. Convert the given PDE into standard form (Pp + Qq = R).
  2. Write the auxiliary equation using the ratios:

[ \frac{dx}{P}=\frac{dy}{Q}=\frac{dz}{R} ]

  1. Solve the auxiliary equation using either:
    • Method of grouping, or
    • Method of multipliers (when grouping doesn’t work).
  2. Obtain two independent solutions:
    • One gives (u(x,y,z)=C_1) (written as “U” in the explanation)
    • Another gives (v(x,y,z)=C_2)
  3. Final combination: combine to express the general solution in the format expected by the question.

3) Subtypes under Lagrange: total 4 types

  • Type 1: solvable by grouping (easy integrals).
  • Type 2: grouping fails partly; must use multiplier idea.
  • The speaker also highlights short vs long question patterns and emphasizes that Lagrange questions frequently appear (noted as 2 marks and 7 marks).

4) Key exam insight stressed

  • In Lagrange auxiliary-equation solving, choice of which ratios to group can change intermediate forms of (C_1, C_2).
  • If answers differ, it may be due to different grouping choices, not necessarily incorrect logic.

C) Charpit Method (first-order non-linear PDEs)

1) How to recognize Charpit applicability

Charpit solves non-linear first-order PDEs.

You must observe presence of at least one of:

  • (p^2)
  • (q^2)
  • (pq)

Distinction from Lagrange:

  • Lagrange typically deals with cases where degree in (p,q) is effectively 1.
  • Charpit involves higher degree (e.g., degree can be 2, etc.).

2) Steps to solve using Charpit

  1. Write the PDE in the Charpit standard polynomial form by bringing everything to one side:

[ f(x,y,z,p,q)=0 ]

  1. Compute required partial derivatives of (f) (as needed for the auxiliary equation):
    • Differentiate w.r.t. (x), (y), (z), (p), and (q).
  2. Write the Charpit auxiliary equation (a “big” formula emphasized to memorize/practice), following the standard pattern of five fractions, e.g.:
    • terms like (\frac{dx}{(\partial f/\partial p)\dots}),
    • (\frac{dy}{(\partial f/\partial q)\dots}),
    • (\frac{dz}{(p \partial f/\partial p + q \partial f/\partial q)\dots}) with sign patterns.
  3. Choose two of the five fractions whose integration is easiest.
  4. Integrate the chosen two fractions to get relations involving (p) and/or (q) (or a relation between them).
  5. Use these relations back in the original PDE to compute (p) and (q) as functions of (x,y,z).

  6. Use total differentiation:

    • Substitute (p=\frac{\partial z}{\partial x}) and (q=\frac{\partial z}{\partial y}) into:

    [ dz = p\,dx + q\,dy ]

    • Integrate to obtain the final solution.

D) Linear PDEs with Constant Coefficients: CF & PI framework

1) Recognition: linear with constant coefficients

  • Linear PDE: degree = 1 (power of derivative terms is 1)
  • Constant coefficient: coefficients multiplying derivatives are constants
  • Homogeneous vs non-homogeneous:
    • Homogeneous: all derivative terms are of the same order
    • Non-homogeneous: derivative terms are of different orders (or include mismatched terms)

2) Core output for linear PDEs

  • General solution:

[ \text{CF} + \text{PI} ]

E) Finding CF (homogeneous linear PDE with constant coefficients)

Speaker’s procedure:

  1. Convert the PDE into operator form using substitutions:
    • (\frac{\partial}{\partial x} \to d)
    • (\frac{\partial}{\partial y} \to d’)
    • Higher derivatives:
      • (\frac{\partial^2}{\partial x^2} \to d^2)
      • (\frac{\partial^2}{\partial x\partial y} \to dd’)
      • etc.
  2. Substitute:
    • (d \to m), (d’ \to 1) (in the described auxiliary construction)
  3. Form the auxiliary equation by setting the coefficient of (z) to zero.
  4. Find roots of the auxiliary equation:
    • Case 1: distinct roots
    • Case 2: repeated roots
    • Case 3: common factor structure (repeated/common factors)
  5. Write CF using the standard root-to-expression rule:
    • Factors correspond to expressions like (f(y+mx)) or (x)-shifted / multiplier variants depending on multiplicity.

F) Finding PI (homogeneous & non-homogeneous) — case strategy

  • The video emphasizes multiple cases based on the form on the right-hand side.
  • Highlighted general principle:
    • For homogeneous PDEs where RHS = 0:
      • PI = 0
    • For non-homogeneous PDEs:
      • PI depends on the RHS pattern

G) PI cases for homogeneous linear PDEs (as taught)

  • Case 1: RHS of the form (f(ax+by)), or explicitly:
    • (e^{ax+by})
    • trigonometric patterns like (\sin(ax+by)), (\cos(ax+by)), etc.

Denominator-failure handling (important):

  • If after substitution the denominator becomes 0, it’s a failure case.
  • Then apply the rule:
    • differentiate the denominator w.r.t. the variable corresponding to the dominant power,
    • multiply by the corresponding variable factor to remove the zero denominator.

Many examples focus on checking whether substitution makes denominator = 0 and applying the corrected differentiation/multiplication rule.

H) Non-homogeneous linear PDEs: CF + PI with more PI cases

  • CF extraction is similar (from the homogeneous part).
  • PI extraction has four cases (as described), based mainly on RHS structure, including:
    • (e^{ax+by})
    • (\sin(ax+by))/(\cos(ax+by)) with variations in phase/structure
    • Rational power forms like ((x^m y^n))
    • Products like (e^{ax+by}\cdot V) (reduced to earlier cases)

I) “Reducible to linear PDE with constant coefficient”

Meaning

  • PDE coefficients may be variable initially, but using substitutions it becomes:
    • linear with constant coefficients

Standard substitutions taught (core instruction set)

  • Substitute:
    • (x = e^u) (so (u=\log x))
    • (y = e^v) (so (v=\log y))
  • Replace derivative operators using chain-rule equivalents, e.g.:
    • (x\,\frac{\partial}{\partial x} \to) an operator in terms of (u)
    • similar mappings for higher derivatives (as required).

Final instruction

  • After solving in the substituted variables, convert back:
    • (e^u \to x), (e^v \to y), and (u \to \log x) etc.

Speakers / sources featured

  • Gulshan Sir (main instructor)
  • Gateway Classes (channel/organization referenced)

Original video