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Find the solution to the differential equation \[\frac{ dy }{ dx }+\frac{ y }{ 5 }=0 \] with the initial condition y(0)=8

Mathematics
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So, ive done separation of variables, and integrated dy/(-y) amd dx/5 and gotten -ln(y) = x/5+C, but I am not sure what to do next.
try to isolate y
if ln a= b then a=e^b

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Other answers:

when x=0 y=8 plug that solve for c and yeah isolate y
ok, so \[-\ln(y)=\frac{ x }{ 5 }+C\] becomes \[-y=e^(\frac{ x }{ 5 }+C)\]
so -8=e^(0+c), -8=e^c?
c=-ln(8) ?
yup, c=-ln 8 so -ln y +ln 8 = x/5 ln (y/8) = -x/5 got this ?
property of log, log A-log B =log (A/B)
ahhhh yes I see that
y=8*e^(x/5)
y=8*e^(-x/5)
Haha yes, making every mistake posssible on this one. Thanks for the help!
welcome ^_^

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