Solve the initial value problem: xy + y² " x² y' = y(−1) = by following the steps: Y (1) make the substitution u = 6 and simplify the X equation. Caution - calculate u' carefully!! (2) notice that the new equation is separable and solve for u (3) reverse the substitution to solve for y (4) use the initial condition to solve for the constant of integration y(x) = n(x) + 6 Note that we've been given an intial value of the form alla) h where so this only determines X

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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Solve the initial value problem:
v=xv+2²³, x-1) = -—
y'
xy
by following the steps:
Y
(1) make the substitution u =
and simplify the
X
equation. Caution - calculate u' carefully!!
(2) notice that the new equation is separable and
solve for u
(3) reverse the substitution to solve for y
(4) use the initial condition to solve for the constant
of integration
y(x) = n(x) + 6
Note that we've been given an intial value of the form
y(a) = b where a < 0, so this only determines a
solution corresponding to the left half of the graph of
In(|x|), i.e., the part of the graph corresponding to
negative values of x. Therefore, we should write
In(-x) instead of ln(x), since the left half of the
graph is not determined by the initial condition given.
X
Transcribed Image Text:Solve the initial value problem: v=xv+2²³, x-1) = -— y' xy by following the steps: Y (1) make the substitution u = and simplify the X equation. Caution - calculate u' carefully!! (2) notice that the new equation is separable and solve for u (3) reverse the substitution to solve for y (4) use the initial condition to solve for the constant of integration y(x) = n(x) + 6 Note that we've been given an intial value of the form y(a) = b where a < 0, so this only determines a solution corresponding to the left half of the graph of In(|x|), i.e., the part of the graph corresponding to negative values of x. Therefore, we should write In(-x) instead of ln(x), since the left half of the graph is not determined by the initial condition given. X
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