Unit 5: Higher Order DEs & PDEs
Table of Contents
- 1. Introduction: General Solution
- 2. Homogeneous Linear DEs with Constant Coefficients
- 3. Non-Homogeneous Linear DEs: The Particular Integral (PI)
- 4. Different Forms of Particular Integrals
- 5. Formation of Partial Differential Equations (PDEs)
- 6. Solution of Partial Differential Equations (Lagrange's Method)
1. Introduction: General Solution
This unit focuses on linear DEs of order 'n' with constant coefficients:
General Solution of a Non-Homogeneous DE
The complete solution (General Solution) is the sum of two parts:
y = yc + yp
- yc (Complementary Function):
The general solution of the corresponding homogeneous equation (i.e., setting R(x) = 0). It will contain 'n' arbitrary constants. - yp (Particular Integral):
A *specific* solution (with no arbitrary constants) of the full non-homogeneous equation.
2. Homogeneous Linear DEs with Constant Coefficients (Finding yc)
To find the Complementary Function (yc), we solve the homogeneous equation:
Where D is the differential operator (D = d/dx, D² = d²/dx², etc.)
Step-by-Step Procedure:
- Form the Auxiliary Equation (AE):
Replace 'D' with a variable 'm' and set the polynomial to zero.anmn + ... + a1m + a0 = 0 - Find the roots (m1, m2, ..., mn) of the AE.
- Write yc based on the roots:
3. Non-Homogeneous Linear DEs: The Particular Integral (PI)
To find the Particular Integral (yp), we "solve" for y from the equation f(D)y = R(x).
Here, 1/f(D) is an "inverse operator". The method depends on the form of R(x).
4. Different Forms of Particular Integrals
This is a set of "short-cut" rules for finding yp.
Form 1: R(x) = eax
Rule: Replace every 'D' with 'a'.
Case of Failure: If f(a) = 0, this means (D-a) is a factor of f(D).
Rule: If f(D) = (D-a)kg(D), where g(a)≠0, then:
Form 2: R(x) = sin(ax) or cos(ax)
Rule: Replace every 'D²' with '(-a²)'. (Do *not* replace D with -a).
Case of Failure: If f(-a²) = 0.
Rule: The method is similar to the "Case of Failure" for eax, but more complex. A common trick is to use yp = x · [solution] (for a single failure).
Form 3: R(x) = xm (a polynomial)
Rule: 1. Rewrite [ 1 / f(D) ] using algebraic manipulation (e.g., long division or binomial theorem) to get a series in *ascending* powers of D.
(e.g., 1 / (1-D) = 1 + D + D² + ... ) 2. Apply this operator series to xm. Since D³(x²) = 0, the series will terminate.
Form 4: R(x) = eax · V(x) (where V is any function, e.g., sin(x) or x²)
Rule (Shift Theorem): 1. "Shift" the eax to the front. 2. Replace every 'D' in the operator with '(D + a)'. 3. Apply the new operator [ 1 / f(D+a) ] to V(x) using one of the other rules.
Form 5: R(x) = x · V(x) (where V is sin(x), cos(x), etc.)
Rule: This is a general rule. 1. First find the PI for V alone: W = [ 1 / f(D) ] · V. 2. Then the full PI is: yp = x·W - [ f'(D) / (f(D))² ] · V
5. Formation of Partial Differential Equations (PDEs)
A PDE is an equation involving partial derivatives. We form them by eliminating arbitrary elements (constants or functions).
Standard Notation:
z = f(x, y)
p = ∂z/∂x
q = ∂z/∂y
1. By Elimination of Arbitrary Constants
If an equation has arbitrary constants, you must differentiate partially w.r.t. x and y to get enough equations to eliminate the constants.
Example: Form a PDE from z = ax + by + ab
- ∂z/∂x = a => p = a
- ∂z/∂y = b => q = b
- Substitute 'a' and 'b' back into the original equation.
2. By Elimination of Arbitrary Functions
If an equation has an arbitrary function (e.g., f(...)), you must eliminate the *function itself*.
Example: Form a PDE from z = f(x + y)
- Let u = x + y. So z = f(u).
- Find p and q using the chain rule:
- p = ∂z/∂x = (df/du) · (∂u/∂x) = f'(u) · 1 = f'(u)
- q = ∂z/∂y = (df/du) · (∂u/∂y) = f'(u) · 1 = f'(u)
- We have p = f'(u) and q = f'(u). This means p and q are equal.
6. Solution of Partial Differential Equations (Lagrange's Method)
This section covers solving first-order, linear PDEs, which are in Lagrange's Form:
Lagrange's Method of Solution:
- Form the "Subsidiary Equations" (or "Auxiliary Equations"):
This is a set of symmetric ordinary differential equations.dx / P = dy / Q = dz / R - Find two independent solutions:
Take any two ratios (e.g., dx/P = dy/Q) and solve them. This gives a first solution:
u(x, y, z) = c1
Then, take another pair (or use multipliers) to find a *second*, *independent* solution:
v(x, y, z) = c2 - Write the General Solution:
The general solution can be written in two equivalent forms, where F (or Φ) is an arbitrary function.F(u, v) = 0
or
u = F(v)
- Grouping: If a ratio involves only two variables (e.g., dx/x = dy/y), you can solve it directly (ln(x) = ln(y) + c => x/y = c1).
- Multipliers: Sometimes you need to choose multipliers (l, m, n) such that lP + mQ + nR = 0. If so, then l·dx + m·dy + n·dz = 0 is also a valid solution.