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S[y1, y2] = =S Hence From the first part we have y' + (y₁ + 3y2) = 0, Using the linear transformation Z1 = y1 + 3y2, we have can write y₁ and y2 in terms of z₁ and z2, provided the matrix Rearranging leads to y1 = dx ((v₁)² + (v₂)² − (y₁ + 3y2)²). S[Z₁, 2] = | Z1 - - 3z2 1 - За = / ₁x (1 dx (( y + 3(y1 + 3y2) = 0. " Z2 = ayı + y2 y2 = 2 2 - 3z2 Sta.al - / + (( =)² + (-³) ²-4) S[Z₁, Z2] = dx 1 3a 3a 1 az1 - Z2 За - 1 - The only terms involving both dependant variables and their derivatives are the zz½ terms: -6z/z2 - 2aziz. Taking a = -3 (note that then 1 - 3a 0) kills these 'cross terms', and we have z² 3 (1³) a how come (z)² + 9(z)² +9(z₁)² + (z₂)² 100 is invertible. ??

S[y1, y2] = =S Hence From the first part we have y' + (y₁ + 3y2) = 0, Using the linear transformation Z1 = y1 + 3y2, we have can write y₁ and y2 in terms of z₁ and z2, provided the matrix Rearranging leads to y1 = dx ((v₁)² + (v₂)² − (y₁ + 3y2)²). S[Z₁, 2] = | Z1 - - 3z2 1 - За = / ₁x (1 dx (( y + 3(y1 + 3y2) = 0. " Z2 = ayı + y2 y2 = 2 2 - 3z2 Sta.al - / + (( =)² + (-³) ²-4) S[Z₁, Z2] = dx 1 3a 3a 1 az1 - Z2 За - 1 - The only terms involving both dependant variables and their derivatives are the zz½ terms: -6z/z2 - 2aziz. Taking a = -3 (note that then 1 - 3a 0) kills these 'cross terms', and we have z² 3 (1³) a how come (z)² + 9(z)² +9(z₁)² + (z₂)² 100 is invertible. ??

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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S[y1, y2] = =/ Hence From the first part we have y₁ + (y₁ + 3y2) = 0, y + 3(y₁ + 3y2) = 0. Using the linear transformation Z1 = y1 + 3y2, we have can write y₁ and y2 in terms of z₁ and z2, provided the matrix Rearranging leads to dx ((v₁)² + (v₂)² − (y1 + 3y2)²) . y1 = S[21, 2] = | Z1 - - 3z2 1 - 3a = [ ₁x (1 dx (( Z2 = ayı + y2 " y2 = 2 2 3z2 az Sta.al - / + (( =)² + (K = ³ ) ² --) S[Z1, Z2] = dx - 1 3a За - 1 az1 - Z2 3a - 1 The only terms involving both dependant variables and their derivatives are the zz½ terms: -6z/z2 - 2azz2₂. Taking a = -3 (note that then 1 – 3a 0) kills these 'cross terms', and we have z₁ 3 (1³) a how come (z₁)² +9(z)² +9(z₁)² + (z₂)² 100 is invertible. ??
Transcribed Image Text:S[y1, y2] = =/ Hence From the first part we have y₁ + (y₁ + 3y2) = 0, y + 3(y₁ + 3y2) = 0. Using the linear transformation Z1 = y1 + 3y2, we have can write y₁ and y2 in terms of z₁ and z2, provided the matrix Rearranging leads to dx ((v₁)² + (v₂)² − (y1 + 3y2)²) . y1 = S[21, 2] = | Z1 - - 3z2 1 - 3a = [ ₁x (1 dx (( Z2 = ayı + y2 " y2 = 2 2 3z2 az Sta.al - / + (( =)² + (K = ³ ) ² --) S[Z1, Z2] = dx - 1 3a За - 1 az1 - Z2 3a - 1 The only terms involving both dependant variables and their derivatives are the zz½ terms: -6z/z2 - 2azz2₂. Taking a = -3 (note that then 1 – 3a 0) kills these 'cross terms', and we have z₁ 3 (1³) a how come (z₁)² +9(z)² +9(z₁)² + (z₂)² 100 is invertible. ??
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