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
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
Problem 1RQ
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Question
![First, use the substitution \( y_1(x) = e^{-5x} \).
\[
y_2(x) = u(x)y_1(x) = u(x)e^{-5x}
\]
Then, use the product rule to find the first and second derivatives of \( y_2 \).
\[
y_2' = -5ue^{-5x} + u'e^{-5x}
\]
\[
y_2'' = \left(-5u'e^{-5x} + \underline{\hspace{1cm}}\right)ue^{-5x} + (u''e^{-5x} - 5u'e^{-5x})
\]
\[
= u''e^{-5x} - 10u'e^{-5x} + \left(\underline{\hspace{1cm}}\right)ue^{-5x}
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F335131ff-484f-4768-821d-726b05458836%2F1e0697f4-08b4-4c27-adef-0495dec71420%2Fouvtvrc_processed.jpeg&w=3840&q=75)
Transcribed Image Text:First, use the substitution \( y_1(x) = e^{-5x} \).
\[
y_2(x) = u(x)y_1(x) = u(x)e^{-5x}
\]
Then, use the product rule to find the first and second derivatives of \( y_2 \).
\[
y_2' = -5ue^{-5x} + u'e^{-5x}
\]
\[
y_2'' = \left(-5u'e^{-5x} + \underline{\hspace{1cm}}\right)ue^{-5x} + (u''e^{-5x} - 5u'e^{-5x})
\]
\[
= u''e^{-5x} - 10u'e^{-5x} + \left(\underline{\hspace{1cm}}\right)ue^{-5x}
\]
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