Solving Differential Equation (dz/dt) + e^(t+z) = 0 with Arbitrary Constant C

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    Calc 2
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SUMMARY

The differential equation (dz/dt) + e^(t+z) = 0 can be solved by first dividing by e^z, resulting in e^{-z} (dz/dt) = -e^t. Integrating with respect to t yields -e^{-z} = k - e^t, where k is a constant. Rearranging this equation leads to the solution z(t) = -ln(k + e^t), with C representing an arbitrary constant in the context of the solution.

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cgward
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I have to solve the differential equation and let C represent an arbitrary constant.

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

I can't seem to figure it out i wind up getting z=-2e^(t^2)+e^(C)
 
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Ok, divide by e^z on both sides to get

e^{-z} \frac{dz}{dt} = -e^t

then integrate with respect to t,

-e^{-z} = k - e^t

re-arrange to get

z(t) = - \ln{(k + e^t)}.

(k is a constant).
 
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