Boyce & DiPrima ODEs 10e: Section 2.6 - Problem 30
Page 1 of 3
Problem 30
In each of Problems 25 through 31, find an integrating factor and solve the given equation.
[4(x3 /y 2 ) + (3/y)] + [3(x/y 2 ) + 4y]y 0 = 0
Solution
This ODE is not exact at the moment because
3
∂
x
x3
x
3
3
∂
3
4 2 +
= −8 3 − 2 6=
3 2 + 4y = 2 .
∂y
y
y
y
y
∂x
y
y
To solve it, we seek an integrating factor µ = µ(x, y) such that when both sides are multiplied by
it, the ODE becomes exact.
3
3
x
x
µ + µ 3 2 + 4y y 0 = 0
4 2 +
y
y
y
Since the ODE is exact now,
∂
∂y
3
x
3
∂
x
4 2 +
µ =
µ 3 2 + 4y .
y
y
∂x
y
Expand both sides.
3
x3
3
x
3 ∂µ
∂µ
x
3
−8 3 − 2 µ + 4 2 +
=
3 2 + 4y + µ
y
y
y
y ∂y
∂x
y
y2
Assume that µ is only dependent on y: µ = µ(y).
3
x3
3
x
3 dµ
3
−8 3 − 2 µ + 4 2 +
=µ
y
y
y
y dy
y2
3
x3
6
x
3 dµ
−8 3 − 2 µ + 4 2 +
=0
y
y
y
y dy
3
3
2
3
3 dµ
x
x
−
4 2 +
µ+ 4 2 +
=0
y
y
y
y
y dy
2
dµ
=0
− µ+
y
dy
dµ
2
= µ
dy
y
Solve this ODE by separating variables.
dµ
2
= dy
µ
y
Integrate both sides.
ln µ = 2 ln y + C
ln µ = ln y 2 + C
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Boyce & DiPrima ODEs 10e: Section 2.6 - Problem 30
Page 2 of 3
Exponentiate both sides.
µ = (y 2 )eC
Taking eC to be 1, an integrating factor is
µ = y2.
Multiply both sides of the original ODE by y 2 .
(4x3 + 3y) + (3x + 4y 3 )y 0 = 0
Because it’s exact, there exists a potential function ψ = ψ(x, y) that satisfies
∂ψ
= 4x3 + 3y
∂x
∂ψ
= 3x + 4y 3 .
∂y
(1)
(2)
Integrate both sides of equation (1) partially with respect to x to get ψ.
ψ(x, y) = x4 + 3xy + f (y)
Here f (y) is an arbitrary function of y. Differentiate both sides with respect to y.
ψy (x, y) = 3x + f 0 (y)
Comparing this to equation (2), we see that
f 0 (y) = 4y 3
→
f (y) = y 4 .
As a result, a potential function is
ψ(x, y) = x4 + 3xy + y 4 .
Notice that by substituting equations (1) and (2), the ODE can be written as
∂ψ ∂ψ dy
+
= 0.
∂x
∂y dx
(3)
Recall that the differential of ψ(x, y) is defined as
dψ =
∂ψ
∂ψ
dx +
dy.
∂x
∂y
Dividing both sides by dx, we obtain the fundamental relationship between the total derivative of
ψ and its partial derivatives.
dψ
∂ψ ∂ψ dy
=
+
dx
∂x
∂y dx
With it, equation (3) becomes
dψ
= 0.
dx
Integrate both sides with respect to x.
ψ(x, y) = C1
Therefore,
x4 + 3xy + y 4 = C1 .
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Boyce & DiPrima ODEs 10e: Section 2.6 - Problem 30
Page 3 of 3
This figure illustrates several solutions of the family. In red, orange, yellow, green, blue, and
purple are C1 = −1, C1 = 0, C1 = 1, C1 = 4, C1 = 8, and C1 = 20, respectively.
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