Section 1.2, Problem 3
Solve
y'=ky.
Rearranging we get:
y'/y=k
Integrating we have:
∫dy/y = ∫kdx
ln y = kx + C
Section 1.2, Problem 4
Solve
y' = sec y
Rearranging we get:
y' cos y = 1
Integrating we have:
sin y = x + C
y = arc sin x + C
Section 1.2, Problem 5
Solve
y' = y cot x
Rearranging we get:
Integrating we have:
ln |y| = ln |sin x| + C
y = C |sin x|
No comments:
Post a Comment