can anyone check my work, please?
dy/dx=1-0.2y, solve for y when y=0, x=0
This is how I solved it dy/(1-0.2y)=dx <=> -5* (-0.2dy/1-0.2y)=dx
integrating the equation, ln(abs) 1-0.2= x+c/-5
1-0.2y=e^(-x/5+c) <=> 1-0.2y=ce^-x/5
My final equation is y=e^(-x/5)-1/-0.2 after plugging in y=0 and x=0
Is it correct?
A quick hint for using MHF - unanswered questions stand out from the rest, so it's not in your interest to answer your own question. Unless you've found the solution, of course.
I think your answer is correct.
- Hollywood
P.S. I interpreted y=e^(-x/5)-1/-0.2 as y=(e^(-x/5)-1)/-0.2. But the order of operations puts division first, so the correct interpretation is y=e^(-x/5)-(1/-0.2).