In order to solve with we first After where c is the y' we have - 3x² ex 2y - 5 the variables. This gives both sides, we obtain y(0) = 1 (2y – 5)dy = (3x² – eª )dx. y — 5y = x — e to of integration. Setting x= . Thus, we have y — 5y — (2 — e* _3) =0. and

Calculus: Early Transcendentals
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Author:James Stewart
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Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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In order to solve
3x2
y'
et
2y – 5
with
y(0) = 1
we first
the variables. This gives
(2y – 5)dy = (3x² – e* )dx.
After
both sides, we obtain
y? – 5y = x – e² +c
where c is the
of integration. Setting x=
and
we have
. Thus, we have
у? — 5у — (23 — e — 3) — 0.
Transcribed Image Text:In order to solve 3x2 y' et 2y – 5 with y(0) = 1 we first the variables. This gives (2y – 5)dy = (3x² – e* )dx. After both sides, we obtain y? – 5y = x – e² +c where c is the of integration. Setting x= and we have . Thus, we have у? — 5у — (23 — e — 3) — 0.
y? – 5y = x – e" +c
where c is the
of integration. Setting x=
and
we have
Thus, we have
y? – 5y – (x³ – e" – 3) = 0.
By using
13
y(x) :
et
4
In the quadratic formula, negative square root is chosen because of
disintegrate
C=-3
factor
c=1
value
the initial conditions
y=1
constant
c=3
C=-3
y=0
c=0
separate
the quadratic formula
integration by parts
eliminate
x=1
integrating
Transcribed Image Text:y? – 5y = x – e" +c where c is the of integration. Setting x= and we have Thus, we have y? – 5y – (x³ – e" – 3) = 0. By using 13 y(x) : et 4 In the quadratic formula, negative square root is chosen because of disintegrate C=-3 factor c=1 value the initial conditions y=1 constant c=3 C=-3 y=0 c=0 separate the quadratic formula integration by parts eliminate x=1 integrating
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