3 years ago

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

1 answer:

Volgvan3 years ago

5 0

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

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