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In Problems 1 through 16, a homogeneous second-order lin- ear differential equation, two functions y₁ and y2, and a pair of initial conditions are given. First verify that y₁ and y₂ are solutions of the differential equation. Then find a particular solution of the form y = C₁y1 + C2y2 that satisfies the given initial conditions. Primes denote derivatives with respect to x. 13. x²y" - 2xy' + 2y = 0; y₁ = x, y2 = x²; y(1) : = 3, y'(1) = 1

In Problems 1 through 16, a homogeneous second-order lin- ear differential equation, two functions y₁ and y2, and a pair of initial conditions are given. First verify that y₁ and y₂ are solutions of the differential equation. Then find a particular solution of the form y = C₁y1 + C2y2 that satisfies the given initial conditions. Primes denote derivatives with respect to x. 13. x²y" - 2xy' + 2y = 0; y₁ = x, y2 = x²; y(1) : = 3, y'(1) = 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 15CR
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In Problems 1 through 16, a homogeneous second-order lin- ear differential equation, two functions y₁ and y2, and a pair of initial conditions are given. First verify that y₁ and y₂ are solutions of the differential equation. Then find a particular solution of the form y = C₁y1 + C2y2 that satisfies the given initial conditions. Primes denote derivatives with respect to x. 13. x²y" - 2xy' + 2y = 0; y₁ = x, y2 = x²; y(1) : = 3, y'(1) = 1
Transcribed Image Text:In Problems 1 through 16, a homogeneous second-order lin- ear differential equation, two functions y₁ and y2, and a pair of initial conditions are given. First verify that y₁ and y₂ are solutions of the differential equation. Then find a particular solution of the form y = C₁y1 + C2y2 that satisfies the given initial conditions. Primes denote derivatives with respect to x. 13. x²y" - 2xy' + 2y = 0; y₁ = x, y2 = x²; y(1) : = 3, y'(1) = 1
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