Question

dy/dx-y=ex/x given that y(e)=0

Answer 1

The trick is to multiply this thing by a function to help you undo the product rule.

\[y' -y = \frac{e^x}{x} \]
Multiply by some function, for some reason people really like to do this with this greek letter \(\mu\) but it's just some function of x.

\[\mu y' -\mu y = \mu\frac{e^x}{x} \]
Now this is where we undo the product rule, by replacing \(-\mu = \mu'\)

\[\mu y' +\mu' y = \mu\frac{e^x}{x} \]\[(\mu y)' = \mu\frac{e^x}{x} \]

Since we made up \(\mu\) we solve this new, simpler differential equation we made up and pick the simplest solution, \(-\mu = \mu'\) in other words you can throw away the constant of integration and then plug that in and keep solving for y.

Answer 2

Thanks

Answer 3

\[compare ~with~\frac{ dy }{ dx }+Py=Q,where~P~and~Q~are~functions~of~x.\]
P=-1
\[Integrating ~factor=e ^{\int\limits Pdx}=e ^{\int\limits -1 dx}=e ^{-x}\]
complete solution is
\[y.e ^{-x}=\int\limits \frac{ e^x }{ x }e ^{-x}dx+c=\ln \left| x \right|+c\]
when x=e
y=0
\[0=\ln e+c,0=1+c,c=-1\]
?