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In each of Problems 38 through 42, a differential equation and one solution yı are given. Use the method of reduction of or- der as in Problem 37 to find a second linearly independent solution y2. 38. х? у" + ху'— 9у — 0 (х > 0); у1(х) — х3 39. 4y" — 4y' +у %3D 0; у1(х) %3D еx/2 40. х? у" — х (х + 2)y' + (х + 2)у %3D0 (x > 0); у1 (х) — 41. (х + 1)у" —- (х + 2)у' + у - 0 (x> —1); у1 (х) — е* 42. (1 — х?)у"+ 2хy' — 2у — 0 (-1

In each of Problems 38 through 42, a differential equation and one solution yı are given. Use the method of reduction of or- der as in Problem 37 to find a second linearly independent solution y2. 38. х? у" + ху'— 9у — 0 (х > 0); у1(х) — х3 39. 4y" — 4y' +у %3D 0; у1(х) %3D еx/2 40. х? у" — х (х + 2)y' + (х + 2)у %3D0 (x > 0); у1 (х) — 41. (х + 1)у" —- (х + 2)у' + у - 0 (x> —1); у1 (х) — е* 42. (1 — х?)у"+ 2хy' — 2у — 0 (-1

Calculus For The Life Sciences
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter11: Differential Equations
Section11.CR: Chapter 11 Review
Problem 12CR
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help for 43 please 

In each of Problems 38 through 42, a differential equation and one solution yı are given. Use the method of reduction of or- der as in Problem 37 to find a second linearly independent solution y2. 38. х? у" + ху'— 9у — 0 (х > 0); у1(х) — х3 39. 4y" — 4y' +у %3 0; у1(х) %3D еx/2 40. х? у" — х (х + 2)y' + (х + 2)у %3D0 (x > 0); у1 (х) — 41. (х + 1)у" —- (х + 2)у' + у - 0 (x> —1); у1 (х) — е* 42. (1 – x²)y" + 2xy' – 2y = 0 (–1 < x < 1); yı(x) = x 43. First note that y1(x) equation of order 1, | = x is one solution of Legendre's
Transcribed Image Text:In each of Problems 38 through 42, a differential equation and one solution yı are given. Use the method of reduction of or- der as in Problem 37 to find a second linearly independent solution y2. 38. х? у" + ху'— 9у — 0 (х > 0); у1(х) — х3 39. 4y" — 4y' +у %3 0; у1(х) %3D еx/2 40. х? у" — х (х + 2)y' + (х + 2)у %3D0 (x > 0); у1 (х) — 41. (х + 1)у" —- (х + 2)у' + у - 0 (x> —1); у1 (х) — е* 42. (1 – x²)y" + 2xy' – 2y = 0 (–1 < x < 1); yı(x) = x 43. First note that y1(x) equation of order 1, | = x is one solution of Legendre's
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