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
cos(3y)dy
dx =−sin(2x)
or d
dx 1
3sin(3y)=d
dx (cos2x).
Integrate both sides with respect to x:
1
3sin(3y) = cos2x+c
or
sin(3y) = 3 cos2x+C.
We just obtained an implicit formula for the general solution. To
determine Cset 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 cos2x.
To solve for ywe must exercise a little caution. The answer
y(x) = 1
3arcsin(3 cos2x) 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
cos2x≤1/3 or |cos x| ≤ 1/√3. Since the interval of values of xincludes
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 θwith the same value of
sine, sin θ=κ. One such angle is arcsin κand 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.
