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4. Linear system of equations Find the general solution of the following system of equations. (a). x' = ( - -2 2 -2 x To find the eigenvalues r₁ and r2, 3 -r-2 = (3)(r+2)+4= r²r-2 = (r− 2)(r+ 1) = 0 -2-r r1 = 2, r2 = −1 Now we find the corresponding eigenvectors § (1) and § (2). For r₁ = 2, we know 3-11 -2 2 -2-11 §(1) = 0 - (3) (= 9) 0 Note that here we use superscripts to index eigenvectors, and use subscripts to index coordinates. Since (1) — 2(¹) = 0, - by fixing x(1) = = 1 we know that (1) = 1. The first eigenvector is 3( =

4. Linear system of equations Find the general solution of the following system of equations. (a). x' = ( - -2 2 -2 x To find the eigenvalues r₁ and r2, 3 -r-2 = (3)(r+2)+4= r²r-2 = (r− 2)(r+ 1) = 0 -2-r r1 = 2, r2 = −1 Now we find the corresponding eigenvectors § (1) and § (2). For r₁ = 2, we know 3-11 -2 2 -2-11 §(1) = 0 - (3) (= 9) 0 Note that here we use superscripts to index eigenvectors, and use subscripts to index coordinates. Since (1) — 2(¹) = 0, - by fixing x(1) = = 1 we know that (1) = 1. The first eigenvector is 3( =

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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4. Linear system of equations Find the general solution of the following system of equations. (a). x' = ( - -2 2 -2 x To find the eigenvalues r₁ and r2, 3 -r-2 = (3)(r+2)+4= r²r-2 = (r− 2)(r+ 1) = 0 -2-r r1 = 2, r2 = −1 Now we find the corresponding eigenvectors § (1) and § (2). For r₁ = 2, we know 3-11 -2 2 -2-11 §(1) = 0 - (3) (= 9) 0 Note that here we use superscripts to index eigenvectors, and use subscripts to index coordinates. Since (1) — 2(¹) = 0, - by fixing x(1) = = 1 we know that (1) = 1. The first eigenvector is 3( =
Transcribed Image Text:4. Linear system of equations Find the general solution of the following system of equations. (a). x' = ( - -2 2 -2 x To find the eigenvalues r₁ and r2, 3 -r-2 = (3)(r+2)+4= r²r-2 = (r− 2)(r+ 1) = 0 -2-r r1 = 2, r2 = −1 Now we find the corresponding eigenvectors § (1) and § (2). For r₁ = 2, we know 3-11 -2 2 -2-11 §(1) = 0 - (3) (= 9) 0 Note that here we use superscripts to index eigenvectors, and use subscripts to index coordinates. Since (1) — 2(¹) = 0, - by fixing x(1) = = 1 we know that (1) = 1. The first eigenvector is 3( =
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