Asked by Anon
Find the function y = y(x) which solves the initial value problem: 4sec(x)*dy/dx = e^(y+sin(x)) where y(0) = -5
Answers
Answered by
Steve
4sec(x) dy/dx = e^(y+sin(x))
4secx dy = e^y e^sinx dx
e^-y dy = 1/4 e^sinx cosx dx
-e^-y = 1/4 e^sinx + c
e^-y = -1/4 e^sinx + c
-y = ln(c - e^sinx/4)
y = -ln(c - e^sinx/4)
y(0) = -5, so
-5 = -ln(c - 1/4)
5 = ln(c - 1/4)
e^5 = c - 1/4
c = 1/4 + e^5
y = -ln(1/4 + e^5 - e^sinx/4)
4secx dy = e^y e^sinx dx
e^-y dy = 1/4 e^sinx cosx dx
-e^-y = 1/4 e^sinx + c
e^-y = -1/4 e^sinx + c
-y = ln(c - e^sinx/4)
y = -ln(c - e^sinx/4)
y(0) = -5, so
-5 = -ln(c - 1/4)
5 = ln(c - 1/4)
e^5 = c - 1/4
c = 1/4 + e^5
y = -ln(1/4 + e^5 - e^sinx/4)
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