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gingermom
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Homework Statement
Find the particular solution of the differential equation that satisfies the initial condition.
dx/dy = e^x+y x(1) = 0
Homework Equations
it does specifically state dx/dy not dy/dx
The Attempt at a Solution
dx/dy= e^x *e^y
1/e^x dx=e^y dy
integrate both sides
-e^-x =e^y +C
e^-x=-e^y+C
-x=ln(-e^y +c)
x=-ln (-e^y+c) but I can't take a ln of -e so how do I find c?
0=-ln(-e^1+c)
0=ln(C-e)
1=c-e Can I do this? Since 0 raised to any power is still zero, I am not sure about whether I can get rid of the Ln this way.
then c= 1+C
If so the particular solution would be
x(y)=-ln(-e^y+1+e)