2 d² y dy dx R 7x + 16y = 0 dx2 has x* as a solution. Applying reduction order we set y, = ux. Then (using the prime notation for the derivatives) %3D %3D So, plugging y2 into the left side of the differential equation, and reducing, we get x²y – 7xy, + 16y2 = The reduced form has a common factor of x 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 = This equation has integrating factor for x > 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 y2 = and the general solution is (of course it has same form with y2 )
2 d² y dy dx R 7x + 16y = 0 dx2 has x* as a solution. Applying reduction order we set y, = ux. Then (using the prime notation for the derivatives) %3D %3D So, plugging y2 into the left side of the differential equation, and reducing, we get x²y – 7xy, + 16y2 = The reduced form has a common factor of x 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 = This equation has integrating factor for x > 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 y2 = and the general solution is (of course it has same form with y2 )
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
Transcribed Image Text:dy
+ 16y = 0
7x
dx
Rema
dx2
has x* as a solution.
Applying reduction order we set y2 = ux.
Then (using the prime notation for the derivatives)
=
%3D
So, plugging y2 into the left side of the differential equation, and reducing, we get
x²y – 7xy, + 16y2 =
The reduced form has a common factor of x 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.
This equation has integrating factor
for x > 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 y2 =
and the general solution is (of course it has same form with y2)
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