I've been trying to solve and find the solution of this problem by method of exact equations but eventually I got stuck. Thank you in advance. [y^(x) Cos(2x) -2e^(xy) Sin(2x) + 2x]dx + [xe^(xy) Cos(2x) -3]dy = 0
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Originally Posted by lorenzo I've been trying to solve and find the solution of this problem by method of exact equations but eventually I got stuck. Thank you in advance. [y^(x) Cos(2x) -2e^(xy) Sin(2x) + 2x]dx + [xe^(xy) Cos(2x) -3]dy = 0 May I ask, is the a chance that the term above in red is $\displaystyle y\,e^{xy} \cos 2x $?
Originally Posted by Danny May I ask, is the a chance that the term above in red is $\displaystyle y\,e^{xy} \cos 2x $? I checked my notes again to confirm. But, it is: $\displaystyle ye^{x} \cos 2x$
Originally Posted by lorenzo I checked my notes again to confirm. But, it is: $\displaystyle ye^{x} \cos 2x$ You'll notice that you've changed the term from your first post. When you say notes, are you refering to classroom notes that you took?
Sorry my mistake. is y^(x) Cos 2x Yes the notebook I used to work on.
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