Diff Eq dy/dx = (ycos (x)) / (1+y^2) initial condition is y (0) = 1 work below ... ?

Question

Mathematics

Vincent Krause

29 January, 21:19

Diff Eq dy/dx = (ycos (x)) / (1+y^2) initial condition is y (0) = 1 work below ... ?

+4

Answers (1)

Know the Answer?

Jean Daugherty · Answer 1

Dy/dx = (ycos (x)) / (1 + y²)

(1 + y²) / y dy = cos (x) dx

(1/y + y) dy = cos (x) dx

Integrating:

ln (y) + y²/2 = sin (x) + c

ln (1) + 1/2 = sin (0) + c

c = 1/2

Thus,

ln (y) + y²/2 = sin (x) + 1/2