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4.4 Linear second order homogenous equation with constant coefficient (x). dx" d"-ly dx"-1 dy · +...+a₁(x)————+a₁(x)y = F(x) dx Have there are three cases 1.real roots r1=r2=r y=(c₁exc2 xex) (general solution) 2.if r1 r2 y=(c₁e 1x + c2e2x) (general solution) 2.if r1&2 are complex number y=ex [c1 cos ẞx + c2 sin ẞx] a = the real part & B = The imaginary part Example 1// Solve the equation d²y dy + 2y = 0 dx² dx Solution/ (D²+D-2)y=0 (r²+1-2)=0+ (r+2)(r-1)=0 r₁=-2 &r2=1(r₁#r₂) y=(c₁e¹¹x + c2 e¹²x) = (c1e¯²x + c2 e¹x) Example 2//Solve Solution / y+4y+4y=0 (D²+4D+4)y=0 (r²+4r+4)=0 →> (r+2)(r+2)=0 r1=-2 &r2=-2 (r1=r2) y=(c₁e** + c2 xe**) = (c1e−2x + c2 xe¯2x) Example 3// Solve d² + +w²y = 0 dx² Solution / (r²+w2)=0 (D²+w²)y=0 → r² = −w² → r = iw a=0 & ẞ= w y = eax [c1 cos ẞx + c2 sin ẞx] y = ex [c1 cos wx + c2 sin wx] y+y+y=0 Example 4//Solve Solution / (D²+D+1)y=0 (r²+r+1)=0 → -b±√√b2-4ac −1 ± √1-4*1*1 r = 2a 2*1 -1±√√-3 -1±i√√3 2 2 απ 글을 y = eax [c1 cos ẞx + c2 sin ẞx] y = e²²* [c1 cos √ ³3 x + c2 sin е 2 √3

4.4 Linear second order homogenous equation with constant coefficient (x). dx" d"-ly dx"-1 dy · +...+a₁(x)————+a₁(x)y = F(x) dx Have there are three cases 1.real roots r1=r2=r y=(c₁exc2 xex) (general solution) 2.if r1 r2 y=(c₁e 1x + c2e2x) (general solution) 2.if r1&2 are complex number y=ex [c1 cos ẞx + c2 sin ẞx] a = the real part & B = The imaginary part Example 1// Solve the equation d²y dy + 2y = 0 dx² dx Solution/ (D²+D-2)y=0 (r²+1-2)=0+ (r+2)(r-1)=0 r₁=-2 &r2=1(r₁#r₂) y=(c₁e¹¹x + c2 e¹²x) = (c1e¯²x + c2 e¹x) Example 2//Solve Solution / y+4y+4y=0 (D²+4D+4)y=0 (r²+4r+4)=0 →> (r+2)(r+2)=0 r1=-2 &r2=-2 (r1=r2) y=(c₁e** + c2 xe**) = (c1e−2x + c2 xe¯2x) Example 3// Solve d² + +w²y = 0 dx² Solution / (r²+w2)=0 (D²+w²)y=0 → r² = −w² → r = iw a=0 & ẞ= w y = eax [c1 cos ẞx + c2 sin ẞx] y = ex [c1 cos wx + c2 sin wx] y+y+y=0 Example 4//Solve Solution / (D²+D+1)y=0 (r²+r+1)=0 → -b±√√b2-4ac −1 ± √1-4*1*1 r = 2a 2*1 -1±√√-3 -1±i√√3 2 2 απ 글을 y = eax [c1 cos ẞx + c2 sin ẞx] y = e²²* [c1 cos √ ³3 x + c2 sin е 2 √3

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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4.4 Linear second order homogenous equation with constant coefficient (x). dx" d"-ly dx"-1 dy · +...+a₁(x)————+a₁(x)y = F(x) dx Have there are three cases 1.real roots r1=r2=r y=(c₁exc2 xex) (general solution) 2.if r1 r2 y=(c₁e 1x + c2e2x) (general solution) 2.if r1&2 are complex number y=ex [c1 cos ẞx + c2 sin ẞx] a = the real part & B = The imaginary part Example 1// Solve the equation d²y dy + 2y = 0 dx² dx Solution/ (D²+D-2)y=0 (r²+1-2)=0+ (r+2)(r-1)=0 r₁=-2 &r2=1(r₁#r₂) y=(c₁e¹¹x + c2 e¹²x) = (c1e¯²x + c2 e¹x) Example 2//Solve Solution / y+4y+4y=0 (D²+4D+4)y=0
Transcribed Image Text:4.4 Linear second order homogenous equation with constant coefficient (x). dx" d"-ly dx"-1 dy · +...+a₁(x)————+a₁(x)y = F(x) dx Have there are three cases 1.real roots r1=r2=r y=(c₁exc2 xex) (general solution) 2.if r1 r2 y=(c₁e 1x + c2e2x) (general solution) 2.if r1&2 are complex number y=ex [c1 cos ẞx + c2 sin ẞx] a = the real part & B = The imaginary part Example 1// Solve the equation d²y dy + 2y = 0 dx² dx Solution/ (D²+D-2)y=0 (r²+1-2)=0+ (r+2)(r-1)=0 r₁=-2 &r2=1(r₁#r₂) y=(c₁e¹¹x + c2 e¹²x) = (c1e¯²x + c2 e¹x) Example 2//Solve Solution / y+4y+4y=0 (D²+4D+4)y=0
(r²+4r+4)=0 →> (r+2)(r+2)=0 r1=-2 &r2=-2 (r1=r2) y=(c₁e** + c2 xe**) = (c1e−2x + c2 xe¯2x) Example 3// Solve d² + +w²y = 0 dx² Solution / (r²+w2)=0 (D²+w²)y=0 → r² = −w² → r = iw a=0 & ẞ= w y = eax [c1 cos ẞx + c2 sin ẞx] y = ex [c1 cos wx + c2 sin wx] y+y+y=0 Example 4//Solve Solution / (D²+D+1)y=0 (r²+r+1)=0 → -b±√√b2-4ac −1 ± √1-4*1*1 r = 2a 2*1 -1±√√-3 -1±i√√3 2 2 απ 글을 y = eax [c1 cos ẞx + c2 sin ẞx] y = e²²* [c1 cos √ ³3 x + c2 sin е 2 √3
Transcribed Image Text:(r²+4r+4)=0 →> (r+2)(r+2)=0 r1=-2 &r2=-2 (r1=r2) y=(c₁e** + c2 xe**) = (c1e−2x + c2 xe¯2x) Example 3// Solve d² + +w²y = 0 dx² Solution / (r²+w2)=0 (D²+w²)y=0 → r² = −w² → r = iw a=0 & ẞ= w y = eax [c1 cos ẞx + c2 sin ẞx] y = ex [c1 cos wx + c2 sin wx] y+y+y=0 Example 4//Solve Solution / (D²+D+1)y=0 (r²+r+1)=0 → -b±√√b2-4ac −1 ± √1-4*1*1 r = 2a 2*1 -1±√√-3 -1±i√√3 2 2 απ 글을 y = eax [c1 cos ẞx + c2 sin ẞx] y = e²²* [c1 cos √ ³3 x + c2 sin е 2 √3
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