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
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter7: Analytic Trigonometry
Section7.6: The Inverse Trigonometric Functions
Problem 94E
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Please (re)solve the questions below in some detail
![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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9d5b509b-2aec-4740-924b-bdebd5fee3b4%2F51d117bf-9fbc-4c06-9b4a-ded06b787215%2Fyouyjnf_processed.jpeg&w=3840&q=75)
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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9d5b509b-2aec-4740-924b-bdebd5fee3b4%2F51d117bf-9fbc-4c06-9b4a-ded06b787215%2Fbu116sr_processed.jpeg&w=3840&q=75)
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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