Solve the differential equation. Let C represent an arbitrary constant. (Note: In this case, your answer willto have a negative sign in front of the arbitrary C.)
(dz)/(dt) + e^(t+z) = 0
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(dz/dt) + (e^t)(e^z) = 0
(dz/dt) = -(e^t)(e^z)
dz = -(e^t)(e^z)dt
(dz/e^z) = -(e^t)dt
Integral[dz/e^z] = Integral[-(e^t)dt]
Sinh(z)-Cosh(z) = -te^t + C
(e^z - e^-z)/2 - (e^z + e^-z) = -te^t +C
(e^z - e^-z) - (e^z + e^-z) = -2te^t +2C
-2e^-z = -2te^t + 2C
e^-z = te^t - 2C
ln(e^-z) = ln(te^t + 2C)
-z = ln(te^t + 2C)
z = -ln(te^t + 2c)
Answer = incorrect...what am I doing wrong? Thanks!
I don't follow your reasoning in going from
Integral[dz/e^z] = Integral[-(e^t)dt]
to
Sinh(z)-Cosh(z) = -te^t + C
It looks to me like it should be just
e^(-z) = e^t + C
So then solving for z would be:
-z = ln(e^t + C)
z = -ln(e^t + C)
z = -t - ln (C)?
In the question is says there should be a negative in front of the constant C so should it be:
z = -t + ln (C)? I am not sure what it means by this. Thanks.
I agree with z = -ln (e^t + C), but your last step does not follow from that. The log of the sum of two terms is not equal to the sum of the logs. You are thinking of the log of the PRODUCT of two terms.
Since C is any constant, it does not matter if there is a minus sign in front of it or not.
0 answers