Differential Equations
Grinshpan
Problem 19, page 48.
The equation
sin(2x)dx + cos(3y)dy = 0,
y(π/2) = π/3,
is obviously separable as the variables are already separated. Write
dy
cos(3y)
= − sin(2x)
dx
or
d 1
d
sin(3y) =
(cos2 x).
dx 3
dx
Integrate both sides with respect to x:
1
sin(3y) = cos2 x + c
3
or
sin(3y) = 3 cos2 x + C.
We just obtained an implicit formula for the general solution. To
determine C set x = π/2 and y = π/3:
sin(π) = 3 cos2 (π/2) + C
0=0+C
C = 0.
Hence the solution of the initial value problem is given implicitly by
sin(3y) = 3 cos2 x.
To solve for y we must exercise a little caution. The answer
y(x) = 31 arcsin(3 cos2 x) is wrong because then y(π/2) = 0, not π/3.
First, note that the right side
√ is nonnegative and should not exceed 1. So
2
cos x ≤ 1/3 or | cos x| ≤ 1/ 3. Since the interval of values of x includes
π/2 we must have:
√
√
arccos(1/ 3) ≤ x ≤ π − arccos(1/ 3).
This is the interval of definition of the solution. It is symmetric about
π/2 ≈ 1.570 :
0.955... ≤ x ≤ 2.186... .
Next, recall that there are infinitely many angles θ with the same value of
sine, sin θ = κ. One such angle is arcsin κ and it lies in the interval
[−π/2, π/2]. The entire collection of angles is described by
θ = arcsin κ + 2πn, π − arcsin κ + 2πn.
2
To match y(π/2) = π/3 we must choose
3y = π − arcsin(3 cos2 x)
or
π 1
− arcsin(3 cos2 x).
3 3
√
The solution is an even function. As x runs from arccos(1/ 3) to π/2,
y(x) goes up from
√ π/6 to its max value π/3. As x runs from π/2 to
π − arccos(1/ 3), y(x) goes back down to π/6.
y=