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FragKrag
United States11567 Posts
"Solve the differential equation by separation of variables." essentially, integrate the function by separating y and x. question: dy/dx = cos(x) * e^(y + sin(x)) I get: sin(x) = u, cos(x)dx = du dy = du * e^(y + u) dy = du * e^(y) * e^(u) e^(-y)dy = e^(u)du => -e^(-y) = e^(u) + C ln(-e^(-y)) = ln (e^(u) + C) y = ln(e^(u) + C) answer is y = - ln(C - e^(u)) Where the hell did I go wrong? :/   | ||
mikeymoo
Canada7170 Posts
e^(-y) = -e^u + C (C is an arbitrary constant anyway) -y = ln(C-e^u) y = -ln(C-e^u) As for where your problem is, give me a second to figure that out. I think it's your last line. ln(-e&(-y)) != y ln(-e^(-y)) = (ln(e^(-y))^(-1) = -y^(-1) ? | ||
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igotmyown
United States4291 Posts
ln(-e^(-y)) = ln (e^(u) + C) is illegal | ||
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FragKrag
United States11567 Posts
thanks!_! So the negative doesn't affect the constant? fuuuuu tytyty | ||
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Turbovolver
Australia2453 Posts
If you do move the negative over, you will get ln(e^(-y)) = ln (-e^(u) + D) (new constant which would be the negative of your old one) And now, you will get their answer. It's really weird though, it feels like it should still technically be "legal" to do it your way. edit: I'm too slow. The negative affects the constant but you just relabel it as a new constant. For all you knew C was negative anyway, so your new one is a positive constant. | ||
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FragKrag
United States11567 Posts
i <3 tl | ||
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coltrane
Chile988 Posts
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evanthebouncy!
United States12816 Posts
here, cheer you up | ||
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