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

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17 January, 13:29

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

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Cristopher Cardenas · 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