- #1

- 5

- 0

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

I cant seem to figure it out i wind up getting z=-2e^(t^2)+e^(C)

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- Thread starter cgward
- Start date

- #1

- 5

- 0

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

I cant seem to figure it out i wind up getting z=-2e^(t^2)+e^(C)

- #2

- 66

- 0

Ok, divide by [tex]e^z[/tex] on both sides to get

[tex]e^{-z} \frac{dz}{dt} = -e^t [/tex]

then integrate with respect to t,

[tex]-e^{-z} = k - e^t [/tex]

re-arrange to get

[tex]z(t) = - \ln{(k + e^t)}[/tex].

(k is a constant).

[tex]e^{-z} \frac{dz}{dt} = -e^t [/tex]

then integrate with respect to t,

[tex]-e^{-z} = k - e^t [/tex]

re-arrange to get

[tex]z(t) = - \ln{(k + e^t)}[/tex].

(k is a constant).

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