Simplifying an ODE into explicit form

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SUMMARY

The discussion focuses on simplifying ordinary differential equations (ODEs) into explicit form, specifically addressing two separable equations. The first equation, \(\frac{dy}{dx} - \frac{x+e^{-x}}{y+e^{y}} = 0\), leads to the implicit solution \(\frac{y^{2}}{2}+e^{y} = \frac{x^{2}}{2}-e^{-x}+C\), which cannot be expressed explicitly for \(y\) using standard functions. The second equation, \(\frac{dx}{dt} = te^{x+t}\), can be solved for \(x\) using logarithmic functions, yielding the implicit solution \(-e^{-x}+C = te^{t}-e^{t}\.

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  • Understanding of ordinary differential equations (ODEs)
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  • Study methods for solving separable ordinary differential equations
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Students and educators in mathematics, particularly those studying differential equations, as well as anyone seeking to deepen their understanding of solving and simplifying ODEs.

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Homework Statement


So i think i found the general solutions to both these separable equations, but I am not sure if I am suppose to simplify any further to get it in explicit form, and how i can even do that.

Homework Equations





The Attempt at a Solution



1. [itex]\frac{dy}{dx}[/itex] - [itex]\frac{x+e^{-x}}{y+e^{y}}[/itex] = 0

2. [itex]\frac{dx}{dt}[/itex] = te[itex]^{x+t}[/itex]

For 1), i get [itex]\frac{y^{2}}{2}[/itex]+e[itex]^{y}[/itex] = [itex]\frac{x^{2}}{2}[/itex]-e[itex]^{-x}[/itex]+C

and for 2) i get:

-e[itex]^{-x}[/itex]+C = te[itex]^{t}[/itex]-e[itex]^{t}[/itex]

Are these right? and is there anyway to simplify them into explicit form? Thanks
 
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They are both correct. You can't solve the first one for y explicitly with the usual functions. You could solve the second for x if you wanted to using logarithms.
 

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