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¹) Solve the initial value problem yy' + x = To solve this, we should use the substitution U = x^2+y^2 u' help (formulas) help (formulas) dy Enter derivatives using prime notation (e.g., you would enter y' for da 2 2 x² + y² with y(5) = √56. = 2x+2y -). After the substitution above, we obtain the following linear differential equation in x, u, u'. help (equations) The solution to the original initial value problem is described by the following equation in x, y. help (equations)

¹) Solve the initial value problem yy' + x = To solve this, we should use the substitution U = x^2+y^2 u' help (formulas) help (formulas) dy Enter derivatives using prime notation (e.g., you would enter y' for da 2 2 x² + y² with y(5) = √56. = 2x+2y -). After the substitution above, we obtain the following linear differential equation in x, u, u'. help (equations) The solution to the original initial value problem is described by the following equation in x, y. help (equations)

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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') Solve the initial value problem yy' + x = To solve this, we should use the substitution U= x^2+y^2 u' help (formulas) help (formulas) dy Enter derivatives using prime notation (e.g., you would enter y' for da 2 2 x² + y² with y(5) = √56. = 2x+2y -). After the substitution above, we obtain the following linear differential equation in x, u, u'. help (equations) The solution to the original initial value problem is described by the following equation in x, y. help (equations)
Transcribed Image Text:') Solve the initial value problem yy' + x = To solve this, we should use the substitution U= x^2+y^2 u' help (formulas) help (formulas) dy Enter derivatives using prime notation (e.g., you would enter y' for da 2 2 x² + y² with y(5) = √56. = 2x+2y -). After the substitution above, we obtain the following linear differential equation in x, u, u'. help (equations) The solution to the original initial value problem is described by the following equation in x, y. help (equations)
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