I have to use sep. of var. to solve the following diff eq:
e^(t)*y dy/dt = e^(-y) + e^(-2t - y)
MY WORK:
e^(t)*y dy = (e^(-y) + e^(-2t - y)) dt
y dy = (e^(-y) + e^(-2t - y))/(e^(t)) dt
I'm not sure how to get the y to the LHS.
e^(t)*y dy/dt = e^(-y) + e^(-2t - y)
ye^t dy/dt = e^(-y) + e^(-(2t +y))
Put them all in positive exponents,
ye^t dy/dt = 1/(e^y) + 1/(e^(2t +y))
ye^t dy/dt = 1/(e^y) + 1/[(e^(2t))(e^y)]
Make the two fractions in the RHS into one fraction only,
common denominator is e^(2t) *e^y,
ye^t dy/dt = [e^(2t) +1] / [(e^(2t))(e^y)]
Clear the fraction, multiply both sides by [(e^(2t))(e^y)],
[(e^(2t))(e^y)][ye^t] dy/dt = e^(2t) +1
(ye^y)(e^(3t)) dy/dt = e^(2t) +1
(ye^y) dy/dt = [e^(2t) +1] /[e^(3t)]
(ye^y) dy = ([e^(2t) +1] / [e^(3t)])dt
(ye^y)dy = {(e^(2t)/(e^(3t)) +1/(e^(3t))}dt
(ye^y)dy = {1/(e^t) +1/(e^(3t))}dt
You want to continue from here?
Hello, Ideasman!
Multiply by $\displaystyle e^y\!:\;\;\displaystyle{e^y\,e^t\,y\,\frac{dy}{dt} \:=\:1 + e^{-2t}}$I have to use sep. of var. to solve the following diff eq:
. . $\displaystyle e^t\,y\,\frac{dy}{dt} \: =\: e^{-y} + e^{-2t - y}$
Multiply by $\displaystyle e^{-t}\!:\;\;e^y\,y\,\frac{dy}{dt} \:=\:e^{-t} + e^{-3t}$
And we have: .$\displaystyle y\,e^y\,dy \;=\;\left(e^{-t} + e^{-3t}\right)\,dt$