Solve the differential equation. Let C represent an arbitrary constant. (Note: In this case, WebAssign expects your answer to have a negative sign in front of the arbitrary C.)
(dz)/(dt) + e^(t+z) = 0
To solve the given differential equation:
(dz)/(dt) + e^(t+z) = 0
we first need to separate the variables.
Rearranging the equation, we have:
(dz)/(dt) = -e^(t+z)
Now, we can multiply both sides by dt to separate the variables:
dz = -e^(t+z) dt
Next, we integrate both sides:
∫ dz = ∫ -e^(t+z) dt
Integrating the left side with respect to z gives us z as the result.
For the right side, we need to apply a u-substitution to integrate -e^(t+z) dt.
Let u = t + z, then du = dt + dz. Rearranging this equation, we have dt = du - dz.
Substituting this back into the integral, we get:
∫ -e^(t+z) dt = ∫ -e^u (du - dz)
Simplifying the right side gives us:
∫ -e^u (du - dz) = ∫ -e^u du + ∫ e^u dz
Integrating both parts separately gives us:
= -∫ e^u du + ∫ e^u dz
= -e^u + ∫ e^u dz
= -e^(t+z) + ∫ e^u dz
= -e^(t+z) + e^u + C
Finally, substituting back u = t + z and rearranging, we have:
z = -e^(t+z) + e^(t+z) + C
Simplifying further, we get:
z = C
Therefore, the solution to the given differential equation is z = C, where C represents an arbitrary constant.