BoyceODEch2s4p30

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Boyce & DiPrima ODEs 10e: Section 2.4 - Problem 30 Page 1 of 2
Problem 30
In each of Problems 28 through 31, the given equation is a Bernoulli equation. In each case solve
it by using the substitution mentioned in Problem 27(b).
y′ = εy − σy3, ε > 0 and σ > 0.
This equation occurs in the study of the stability of fluid flow.
Solution
Bring εy to the left side.
y′ − εy = −σy3
Divide both sides by y3.
y−3y′ − εy−2 = −σ (1)
Make the substitution u = y−2. Now differentiate both sides of it with respect to t to find y′ in
terms of this new variable.
du
dt
= (−2)y−3 · dy
dt
Divide both sides by −2.
−1
2
du
dt
= y−3y′
Substitute this result and u = y−2 into equation (1).
−1
2
du
dt
− εu = −σ
Multiply both sides by −2.
du
dt
+ 2εu = 2σ
This ODE can be solved by multiplying both sides by an integrating factor I.
I = exp
(ˆ t
2ε ds
)
= e2εt
Proceed with the multiplication.
e2εt
du
dt
+ 2εe2εtu = 2σe2εt
The left side can be written as d/dt(Iu) by the product rule.
d
dt
(e2εtu) = 2σe2εt
Integrate both sides with respect to t.
e2εtu =
σ
ε
e2εt + C
Divide both sides by e2εt.
u(t) =
σ
ε
+ Ce−2εt
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Boyce & DiPrima ODEs 10e: Section 2.4 - Problem 30 Page 2 of 2
Now that u is solved for, replace it with y−2 and solve for y.
y−2 =
σ
ε
+ Ce−2εt
y2 =
1
σ
ε + Ce−2εt
Therefore,
y(t) = ±
√
ε
σ + Cεe−2εt
.
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