y' = xy + y² x² " y(−1) g the steps: e substitution u = - 6 Y and simpl X aution - calculate u' carefully!! hat the new equation is separab the substitution to solve for y initial condition to solve for the on we've been given an intial value here a < 0, so this only determ responding to the left half of th the part of the graph correspo

Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)
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ISBN:9781337111348
Author:Bruce Crauder, Benny Evans, Alan Noell
Publisher:Bruce Crauder, Benny Evans, Alan Noell
Chapter2: Graphical And Tabular Analysis
Section2.1: Tables And Trends
Problem 1TU: If a coffee filter is dropped, its velocity after t seconds is given by v(t)=4(10.0003t) feet per...
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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) =
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.
Transcribed Image Text: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) = 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.
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