First, use the substitution y₁(x) = e-5x. Y₂(x) = u(x)y₁(x) u(x)e-5x = Then, use the product rule to find the first and second derivatives of y2. Y₂' = - 5ue-5x + u'e-5x Y2" = (-5u'e-5x + + = u'e-5x - 10u'e-5x + Jue-5x) + (u'e-5x -5u'e-5x ) Jue-5x

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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First, use the substitution y₁(x) = e-5x.
y₂(x) = u(x)y₁(x)
u(x)e-5x
=
Then, use the product rule to find the first and second derivatives of y2.
Y₂' = - 5ue-5x
Y2" = (-5u'e-5x +
+
=
+ u'e-5x
u"e-5x - 10u'e-5x +
Jue-5x) +
+ (u'e-5x - 5u'e-5x )
Jue-5x
Transcribed Image Text:First, use the substitution y₁(x) = e-5x. y₂(x) = u(x)y₁(x) u(x)e-5x = Then, use the product rule to find the first and second derivatives of y2. Y₂' = - 5ue-5x Y2" = (-5u'e-5x + + = + u'e-5x u"e-5x - 10u'e-5x + Jue-5x) + + (u'e-5x - 5u'e-5x ) Jue-5x
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