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.

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- Sep 10th 2007, 11:38 PMIdeasmanSeparation of Variables
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. - Sep 11th 2007, 01:18 AMticbol
**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? - Sep 11th 2007, 05:17 AMSoroban
Hello, Ideasman!

Quote:

I have to use sep. of var. to solve the following diff eq:

. .

Multiply by

And we have: .