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OpenStudy (anonymous):
How would I use an exact differential equation to solve dy/dt=((-e)^y)/(t(e^y)+2y)?
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OpenStudy (anonymous):
hey
OpenStudy (anonymous):
get it into M(x,y)dx + N(x,y)dy = 0
OpenStudy (anonymous):
How? I've only ever seen this type of equation as a very simple example already in the form M(x,y)dx + N(x,y)dy = 0. How would I separate this one?
OpenStudy (anonymous):
or M(t,y) dt + N(t,y)dy = 0
OpenStudy (anonymous):
i have dy(t e^y + 2y ) = (-e^y dt
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OpenStudy (anonymous):
so you start it like the solution to a separable function?
OpenStudy (anonymous):
so e^y dt + (te^y + 2y ) dy = 0
OpenStudy (anonymous):
THANK YOU SO MUCH!
OpenStudy (anonymous):
I don't know why I couldn't separate that on my own.
OpenStudy (anonymous):
i have the formula, its a bit long
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OpenStudy (anonymous):
first we have to check that My = Nt ,
OpenStudy (anonymous):
the partial of M(t,y) with respect to y
OpenStudy (anonymous):
deriv wrt to y M(t,y) = e^y and deriv wrt to y N(t,y) = e^y , so its an exact equation
OpenStudy (anonymous):
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