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The differential equation has as a solution. Applying reduction order we set y = uxª. x² d²y dy dx² 7x + 16y= 0 da

The differential equation has as a solution. Applying reduction order we set y = uxª. x² d²y dy dx² 7x + 16y= 0 da

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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The differential equation has as a solution. Applying reduction order we set y = ux¹. Then (using the prime notation for the derivatives) y' = x² d²y dx² So, plugging y into the left side of the differential equation, and reducing, we get dy 7x + 16y=0 dx x²y" 7xy + 16y= The reduced form has a common factor of 5 which we can divide out of the equation so that we have xu" + u' = 0. Since this equation does not have any u terms in it we can make the substitution w = u' giving us the first order linear equation xw' + w = 0. If we use "a" as the constant of integration, the solution to this equation is w = Integrating to get u, and using "b" as our second constant of integration we have u = Finally the general solution is y =
Transcribed Image Text:The differential equation has as a solution. Applying reduction order we set y = ux¹. Then (using the prime notation for the derivatives) y' = x² d²y dx² So, plugging y into the left side of the differential equation, and reducing, we get dy 7x + 16y=0 dx x²y" 7xy + 16y= The reduced form has a common factor of 5 which we can divide out of the equation so that we have xu" + u' = 0. Since this equation does not have any u terms in it we can make the substitution w = u' giving us the first order linear equation xw' + w = 0. If we use "a" as the constant of integration, the solution to this equation is w = Integrating to get u, and using "b" as our second constant of integration we have u = Finally the general solution is y =
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