Download Correction to Example with Error from January 30, 2017 Lecture

Correction to Example with Error from January 30, 2017 Lecture
Math 81, Spring 2017
Instructor: Dr. Doreen De Leon
In class on January 30, 2017, the second example for exact equations had an error in it. The
problem should have been
Example 2: Solve the IVP
π (cot(x) + 2y)y 0 = y csc2 (x), y
= 1.
2
It is done correctly below.
Solution: First, write the DE in the proper form:
−y csc2 (x) + (cot(x) + y)y 0 = 0,
and so
M (x, y) = −y csc2 (x) and N (x, y) = cot(x) + 2y.
• Check if the equation is exact.
My (x, y) = − csc2 (x)
Nx (x, y) = − csc2 (x) = My (x, y).
Therefore, the equation is exact.
• Find the general solution.
Z
F (x, y) =
N (x, y) dy
Z
=
(cot(x) + 2y) dy
F (x, y) = y cot(x) + y 2 + h(x).
Find h(x) using Fx (x, y) = M (x, y).
∂
y cot(x) + y 2 + h(x)
∂x
= −y csc2 (x) + h0 (x).
Fx (x, y) =
Fx (x, y) = M (x, y) gives
−y csc2 (x) + h0 (x) = −y csc2 (x)
h0 (x) = 0
Z
h(x) = 0 dx = 0 + c1 (c1 ∈ R).
So,
F (x, y) = y cot(x) + y 2 + c1 .
General solution: Set F (x, y) = c (c ∈ R), so
y cot(x) + y 2 = c (c ∈ R).
1
• Find c using the initial condition y
π 2
= 1: plug in x =
1 · cot
π 2
+ (1)2 = c,
so c = 1.
Answer:
y cot(x) + y 2 = 1.
2
π
and y = 1 to obtain
2