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

[cgward]

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)

 

[Matthew Rodman]

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).

 

[tacman]

To solve this differential equation, we can use the method of separation of variables. We start by rearranging the equation to have all the z terms on one side and all the t terms on the other side:

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

Next, we can divide both sides by the exponential term to isolate dz/dt:

dz/dt = -1/e^(t+z)

Now, we can integrate both sides with respect to t, and also include the arbitrary constant C:

∫dz = -∫1/e^(t+z) dt + C

Integrating the left side gives us just z, while for the right side we can use the substitution u = t+z, du = dt, to get:

z = -∫1/e^u du + C

Simplifying the integral and substituting back in t+z for u, we get:

z = e^(-t-z) + C

To solve for z, we can rearrange the equation to have the exponential term on one side and everything else on the other side:

z + e^(-t-z) = C

Now, we can use the fact that e^(-t-z) is the same as 1/e^(t+z), and multiply both sides by e^(t+z):

z*e^(t+z) + 1 = Ce^(t+z)

Finally, we can solve for z to get the general solution to the differential equation:

z = Ce^(t+z) - 1

Therefore, the solution to the given differential equation with an arbitrary constant C is:

z = Ce^(t+z) - 1