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• Muhammad Rizwan

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ORDINARY DIFFERENTIAL EQUATION REVIEW SHEET I [28/10/2015] IMPORTANT EQUATIONS R R

• Explicit Dependent variable can be separated, x + 2y = 0

udv = uv − vdu d(lnx) = x1 dx d (sin x) = cos(x) dx d (cos x) = − sin(x) dx d (tan x) = sec2 (x) dx d (csc x) = − csc(x) cot(x)x dx d (sec x) = sec(x) tan(x) dx d (cot x) = − csc2 (x) dx 0

g ( fg )0 = f g−f g2 Leibniz Rule:

• Singular Solution: ???

0

• Partial Differential Equation One or more functions of two or more independent variables.

Order of DE dy 3 + 5( dx ) − 4y = ex  4  ∂ u ∂2u 2 4 + =0 ∂x ∂t2

dy dx

Dependent variable y and its

dy , ... dx

g(x): constant, polynomial, exponential, sin, cos

Method of Undetermined Coefficients Example: 10, x2 − 5x, 15x + 8e− x, sin(3x) − 5x cos(2x), xe−x (3x2 − 1) Not Applicable: ln x, x1 , tan(x), arcsin(x) • 1: A

LINEAR EQUATIONS

depend at most on the

dy (1 − y) dx + 2y = ex

+ sin(y) = 0 + y2 = 0

Homogeneous & Non-Homogeneous DE Linear, Separable & Exact DE Solutions of DE nth

A solution of order DE is a function φ that possesses atleast ‘n’ derivative. F (x, φ(x), φ0 (x), ..., φ( n)(x)) = 0 for all x in I. • Trivial: A solution of an nth-order ordinary differential equation is a function ’phi’ that possesses at least ’n’ derivatives. • Implicit Dependent variable cannot be separated, sin(x + ey ) = 3y

• 5x + 7 : Ax + B

+ a0 (x)y = g(x)

• 3x2 − 2 : Ax2 + Bx + C

Property

• sin(4x) : A cos(4x) + B sin(4x)

1. Standard Form: y 0 + P (x)y = f (x) 2. Integrating Factor: I(x) = e

R

• cos(4x) : A cos(4x) + B sin(4x)

P (x)dx

Variation of Parameters

3. Differential Equation: (I(x)y)0 = I(x)f (x) R 4. Integration: I(x)y = I(x)f (x)dx + C R  1 I(x)f (x)dx + C 5. Solution: y = I(x)

• Complementary Solution: yc = c1 .y1 + c2 .y2 • Wronskian: W (y1 , y2 ) • f (x) from Standard Form: y 00 + P y 0 + qy = f (x)

Solution: y = yc (homogeneous) + yp (non − homogeneous) Autonomous DE: When a1 , a0 &g(x) are constants. Initial Value Problem: Has unique solution. Interval: P &f (x) are continuous Singular Points: Values of x for which a1 (x) = 0

Linear Equations dn y dn−1 y dy SOLVE:an (x) dx n +an−1 (x) dxn−1 +...a1 (x) dx +a0 (x)y 0 n−1 IC: y(x0 ) = y0 , y (x0 ) = y1 , ..., y (x0 ) = yn−1

= g(x)

Initial Value Problem • Unique Sol: an (x), an−1 , ..., a0 (x) and g(x) continuous on interval I and an (x) 6= 0 for every x in this interval. Boundary Value Problem

• Real & Equal:y = c1 em1 x + c2 xem2 x

Particular Solution

4. Constant Solution:

Non-Linear DE d2 y dx2 d4 y dx4

= g(x)f (y)

1 1. Separate x & y: f (y) dy = g(x)dx R 1 R 2. Integral f (y) dy = g(x)dx + C

3. Solution:

are first degree

• Real & Distinct Roots: y = c1 em1 x + c2 em2 x • Complex & Conjugate: y = eax (c1 cos βx + c2 sin βx), where m1 = α + ιβ,m2 = α − ιβ

SEPARABLE EQUATIONS

Linear DE The coefficient of y and independet variable x.

= 0.2xy

dy a1 (x) dx dy , ... derivative dx

ay00 + by0 + cy = 0 Let y = emx , Auxiliary Equation: am2 + bm + c = 0

Independent variable doesn’t appear explicitly. (Dont change dy = 1 − y2 over time) Autonomous, dx dy dx

• A constant multiple is also a solution of homogeneous DE.

Second Order Equations

Directional Fields Autonomous DE

Non-autonomous,

Superposition Principle: y = c1 y1 (x) + c2 y2 (x) + ... + ck yk (x) Notes:

• Homogeneous linear DE always posses trivial solution y = 0.

INITIAL VALUE PROBLEM SOLUTION CURVES

• Ordinary Differential Equation One or more function with respect to a single independent variable.

d y dx2

• No Sol

• Families of Solution: One parameter family of solution, n-parameter family of solution, particular solution.

Types of Differential Equations

Order 2: Order 4:

• Unique Sol

• Particular

DEFINITION

2

• Infinite Many Sol

• Determine u1 & u2 , where u01 =

w1 w

& u02 =

w2 w

• Particular Solution: yp = u1 .y1 + u2 .y2 • Complete Solution: y = yc + yp

Green’s Theorem R yp (x) = Steps:

x xo

G(x, t)f (t)dt, G(x, t) =

y1 (t)y2 (x)−y1 (x)y2 (t) W (t)

• Complementary Solution: y1 & y2 • w(t): • Green Function: G(x, t) = • Particular Solution: yp (x)

y1 (t).y2 (x)−y1 (x)y2 (t) w(t) Rx = x G(x, t)f (t)dt 0

Linear System • X 0 = AX + F

• Unique Solution: A(t), F (t) be continuous on interval I that contains t0

c 2015 RiZ Copyright