Solve Separable DiffEQ: Find General Solution & Functions | Help Needed!

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Can't figure out this seperable diffEQ! :(

Hello everyone 'ive been trying to figure out this easy looking Differential Equation and yet its wrong! weee!
Here is the problem:
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/58/6217226076d5fd259f53ad1e3ed4071.png
has an implicit general solution of the form F(x,y) = K.
In fact, because the differential equation is separable, we can define the solution curve implicitly by a function in the form
F(x,y) = G(x) + H(y) =K.

Find such a solution and then give the related functions requested.
F(x,y) = G(x) + H(y) = ?

I submitted:
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/5a/1bcf742ffddc932f739864bc6d3e4a1.png
which was wrong.


here is my work:
http://img202.imageshack.us/img202/9317/lastscan6uc.jpg

Any help would be fantastical. <--yeah I'm pretty sure that's not a word.:bugeye:
 
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Go back and insert a pair of parens on the left hand side of the second line that it would read

\int e^y(\sin y+9\cos y) dy = \int(14x+5)dx
 
Know also that

\int e^y(\sin y+9\cos y) dy = e^y(5\sin y+4\cos y)

and in closing note that fantastical is now a word by extension, since, for example antideparameterizationism is likwise wise a word, for I am very much opposed to not eliminating a perfectly useful paramter from my solutions (especially once they've been obtained by inserting it cleverly into the givens,) and I am thus a practioner of antideparameterizationalism. :smile:
 
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hah i like your way of thinking.
I submitted
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/08/74e463c2a0d2a1740eacdf142dadd41.png
and they told me to f off, i was quite distraught. Any ideas what i did now
?
 
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try writing it in the order (fcn. of x) + (fcn. of y), e.g. G(x) + H(y), such as

-7*x^2-5*x + 5*exp(y)*sin(y)+4*exp(y)*cos(y)
 
ur the man!
For some reason, i had to multiply through -1 to make it work, it loved this answer:http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/e6/513e4938c92e397902bfbf80c0e9551.png
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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