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

Solve the initial value problem yy + x = x² + y with y(5) = 56. To solve this, we should use the substitution u = x^2+y^2 help (formulas) u' = 2(x+yy') help (formulas) Enter derivatives using prime notation (e.g., you would enter y for 2). dx After the substitution above, we obtain the following linear differential equation in x, u, u'. du/2sqrtu=dx 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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ASolve the initial value problem yy + x = /x + y with y(5) = v56. To solve this, we should use the substitution и 3D х^2+у^2 help (formulas) u' = 2(x+yy') help (formulas) dy Enter derivatives using prime notation (e.g., you would enter y for ). After the substitution above, we obtain the following linear differential equation in x, u, u'. du/2sqrtu=dx help (equations) The solution to the original initial value problem is described by the following equation in x, y. help (equations)
Transcribed Image Text:ASolve the initial value problem yy + x = /x + y with y(5) = v56. To solve this, we should use the substitution и 3D х^2+у^2 help (formulas) u' = 2(x+yy') help (formulas) dy Enter derivatives using prime notation (e.g., you would enter y for ). After the substitution above, we obtain the following linear differential equation in x, u, u'. du/2sqrtu=dx 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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