Supposedly simple differential equation

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Homework Help Overview

The discussion revolves around solving a differential equation of the form dy/dt = 1/(e^y - t) with the initial condition y(1) = 0. The subject area is differential equations, specifically focusing on separability and variable manipulation.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss attempts to separate the equation and explore substitutions to simplify the problem. Questions arise about the non-separability of the equation and the implications of the initial condition.

Discussion Status

Several participants have offered different approaches, including variable substitution and switching variables to find t(y). There is an ongoing exploration of methods without a clear consensus on the best approach.

Contextual Notes

Participants note the challenge posed by the non-separable nature of the equation and the initial condition provided. There is an acknowledgment of the difficulty in manipulating the equation due to the presence of both e^y and t.

popo902
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Homework Statement


Find the solution of
dy/dt = 1/ (e^y -t), y(1) = 0


Homework Equations





The Attempt at a Solution


i tried separating the equation, but the subtraction gets in the way
well this is what i have
y = t - 1 + C/e^t
i solved for t then i put that into the single order ODE formula
but the answer is y = arccosht
...so...
can someone give me a hint on how to start this?
 
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Yes, the fact that an equation is not separable does make it hard to separate!

Here's what I would try. Since it is the [itex]e^y- t[/itex] that is the problem, let [itex]u= e^y- t[/itex] so that
[tex]\frac{dy}{dt}= \frac{1}{u}[/tex]
and since [itex]e^y= u+ t[/itex]
[tex]e^y\frac{dy}{dt}= \frac{du}{dt}+ 1[/tex]
[tex](u+ t)\frac{1}{u}= \frac{du}{dt}+ 1[/tex]
[tex]\frac{du}{dt}= \frac{u+t}{u}- 1= \frac{t}{u}[/tex]
which is separable.
 
Well, you can switch the variables and try to find t(y). The ODE which results

[tex]\frac{dt(y)}{dy} + t(y) = e^y[/tex]

should be very easy to integrate, right ?
 
Well, if you want to do it the easy way! :-p
 

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