Another Separable Differential Equation

[BarackObama]

Homework Statement

dz/dt + e^(t+z) = 0

Homework Equations

The Attempt at a Solution

dz/dt = -e^te^z
integral(dz/e^z) = integral(-e^tdt)

let u = 1/e^z
dv = dz
du = -e^-zdz v= z

integral(udv)
= z/e^z + integral(ze^-zdz)

 

[tiny-tim]

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integral(dz/e^z) = integral(-e^tdt)

So far so good …

Now just rewrite ∫ dz/e^z as ∫ e^-z dz = … ?

 

[BarackObama]

Good morning!

∫e^-zdz = ∫-e^tdt
-e^-z = -e^t + C
e^-z = e^t + C

Is there any way to bring the variables down?

Thanks!

 

[Mark44]

Take ln of both sides, but keep in mind that the log of a sum doesn't simplify.

 

[BarackObama]

z = -ln(e^t + C)

... not exactly easy to verify

 

[Mark44]

What's so hard about it? Note the -ln(e^t + C) = ln(1/(e^t + C))