Asked by Lauren
solve the differential equation that satisfies the given condition:
dy/dx=(ycos(x))/(6+y^2), y(0)=1
dy/dx=(ycos(x))/(6+y^2), y(0)=1
Answers
Answered by
Steve
y' = (y cosx)/(6+y^2)
(6+y^2)/y dy = cosx dx
6lny + 1/2 y^2 = sinx + c
y^2 + 12lny = 2sinx + c
I don't see any way to express y explicitly as a function of x, using elementary functions.
(6+y^2)/y dy = cosx dx
6lny + 1/2 y^2 = sinx + c
y^2 + 12lny = 2sinx + c
I don't see any way to express y explicitly as a function of x, using elementary functions.
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