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BROOKDALE COMMUNITY COLLEGE / Math / Mat 274 / What makes an equation homogeneous?

What makes an equation homogeneous?

What makes an equation homogeneous?

Description

School: Brookdale Community College
Department: Math
Course: Elementary Differential Equations
Professor: Steven hiamang
Term: Fall 2016
Tags:
Cost: 25
Name: Section 4.4 NOTES
Description: These notes cover what is going to be on our next exam. Quiz on 10/25/16
Uploaded: 10/22/2016
3 Pages 182 Views 0 Unlocks
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Section 4.4 2nd - order non - homogenous DE’S - method of undetermined coefficientsWe also discuss several other topics like mth 234

We will now turn our attention to non - homogeneous 2nd - order linear equations; equations with a standard form:

        Y” + P(x) y’ + Q(x)y = g(x); g(x) 0

Each such non - homogeneous equation has a corresponding homogeneous equation:Don't forget about the age old question of Was Freud good at assessing dysfunctionality?

        Y” + P(x) y’ + Q(x)y = 0We also discuss several other topics like How many sensory receptors are there in the body?

We will focus on non - homogeneous 2nd order DE’S with constant coefficients:Don't forget about the age old question of the Brain's Reward Circuits / TH 10/20 o What is a drug?

        Ay” + by’ + cy = g(x); g(x) 0

And the corresponding homogeneous equation:

        Ay” + by’ cy = 0

Yc = complementary solution of the homogeneous equationIf you want to learn more check out How to solve the rational inequality?

Up = particular solution of the non homogeneous equation

Thus, the general solution is given by: y = yc + Yp and, yc = c1y1 + c2y2 so, y = c1y1 + c2y2 + YpIf you want to learn more check out Which aspects of Christian belief strongly influenced the rise in capitalism?

The complementary solution can be found from the roots of the auxiliary / characteristic equation

        Am2 + bm + c = 0

  1. Yc = c1em1x + c2em2x
  2. Yc = c1em1x + c2xem1x
  3. Yc = e∝x(c1 cos bx + c2 sin bx)

Section 4.4 notes continued

We are left to find the particular solution, Yp, which is any one function that satisfies the given nonhomogeneous equation

**Yp depends on g(x)

There are 2 general approaches to find Yp:

Section 4.4) 1 The method of undetermined coefficients

Section 4.5) 2 Variation of parameters

To find Yp, we use the method of Judicious guessing to determine the general form of YP

**These functions will be used in the method of undetermined coefficients:

  • Exponentials, - polynomial, - sine and cosine

Steps:

  1. Find yc
  2. Guess a Tp depending on g(x) and write Tp with constants (A, B, C, D…)
  3. Find Yp and Yp” and substitute Yp, Up’, Up” into the DE
  4. Solve for the constant (A,B,C,D…)
  5. Plug yc and Yp into y = yc + Yp to get the answer

Solution 4.4 notes continued

Example:

Y” - 2’ - 3y = 5cos 2x

M2 - 2m - 3 = 0        Yp = Acos 2x + Bsin 2x

(m - 3) (m + 1) = 0        Yp = -2Asin 2x + 2B cos 2x

M1 = 3, m2 = -1        Yp = -4Acos 2x - 4B sin2x

Yc = C1e3x + C2e-x

        -4Acos 2x - 4B sin 2x - 2 (-2Asin 2x + 2Bcos 2x) -3 (Acos 2x + Bsin 2x) = 5cos 2x - 4Acos 2x - 4Bsin 2x + 4Asin 2x - 4B cos 2x - 3Acos 2x - 3Bsin 2x = 5 cos 2x

Equate coefficients:

Cos2x: -4A - 4B - 3A = 5 → -7A -4B = 5

Sin2x: -4B + 4A - 3B = 0 → 4A - 7B = 0 → 4A = 7B → A = 7/4B

        - 7 (7/4B) - 4B = 5 → 49/4B - 4B = 5

→49B - 16B = 20 → 65B = 20 → B = 4/13

A = 7/4 (-4/13) → A = -4/13

Yp = -7/13 cos2x -4/13 sin2x

Y = C1e3x+C2e-x - 7/13cos2x -4/13sin2x

Section 4.4 Notes Continued

When g(x) is the sum of several terms; g(x) = g1(s) + g2(x) + g3(x) +...gn(x)

Then,

Y” + P(x)y’ + Q(x)y = g1(x) + g2(x) + g3(x) + ...gn(x)

So, y” + P(x)y1 + Q(x)y = g1(x)

Y’+ P(x)y1 + Q(x)y = g2(x)

Y’+ P(x)y1 + Q(x)y = g3(x)

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