Solve the exact differential: (e^(x)siny+tany)dx + (e^(x)cosy + xsec^()y)dy =
I get h'(y) = dy, h(y) = y +C and final answer: e^(x)siny+tanyx+y=C, Book says e^(x)siny+tanyx = C
Answer
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I’m going to examine the book’s answer first.
e^x sin(y) + x tan(y) = C
The “exact differential” of the left side is
d[ e^x sin(y) + x tan(y) ]/dx times dx + d[e^x sin(y) + x tan(y)]/dy times dy.
The derivatives are partials, I just can’t type them that way.
They come out to
[e^x sin(y) + tan(y) ] dx + [e^x cos(y) + x sec^(y)] dy
Obviously the “exact differential” of the right side is zero.
So the book’s answer does match the given differential equation!
Next I’ll examine your answer.
The exact differential of the expression on the left side of your result is
[e^x sin(y) + tan(y)] dx + [e^x cos(y) + x sec^(y) + ] dy
which does not match the given differential equation.
The book is right, and you are wrong. I don’t know your “h” method,
sorry I can’t comment on why it introduced an error.