General solutions to differential equations

[jonnburton]

Hi, Could anybody point out where I've gone wrong with my solution to this exercise? I've completed a number of questions on finding the general solution to differential equations succesfully, but this is one where I can't see how the correct answer is arrived at:

Find the general solution to: \(\displaystyle \frac{dx}{dt}= e^xe^t\)

This is what I have done:

\(\displaystyle \frac{1}{e^x}\frac{dx}{dt} = e^t\)

\(\displaystyle \int\frac{1}{e^x}dx = \int\frac{e^t}dt\)

\(\displaystyle lne^x = e^t + A\)

\(\displaystyle x = e^t + A\)

However, according to the book the correct answer is \(\displaystyle x = -ln|a - e^t|\) but I can't see where I have gone wrong in my working.

 

[evinda]

\(\displaystyle \frac{dx}{dt} \) = \(\displaystyle e^{x}e^{t} \)
=>\(\displaystyle dx \)=\(\displaystyle e^{x}e^{t}dt \)
=>\(\displaystyle \frac{1}{e^{x}}dx \)= \(\displaystyle e^{t}dt \)
=>\(\displaystyle e^{-x}dx \)= \(\displaystyle e^{t}dt \)
=>\(\displaystyle -e^{-x}+c \) =\(\displaystyle e^{t} \)
=>\(\displaystyle e^{-x}\) = \(\displaystyle -e^{t}+c \)
=>\(\displaystyle lne^{-x} \) = \(\displaystyle ln|c-e^{t}| \)
=>\(\displaystyle x \)= \(\displaystyle -ln|c-e^{t}| \)

 

[jonnburton]

\(\displaystyle \frac{dx}{dt} \) = \(\displaystyle e^{x}e^{t} \)
=>\(\displaystyle dx \)=\(\displaystyle e^{x}e^{t}dt \)
=>\(\displaystyle \frac{1}{e^{x}}dx \)= \(\displaystyle e^{t}dt \)
=>\(\displaystyle e^{-x}dx \)= \(\displaystyle e^{t}dt \)
=>\(\displaystyle -e^{-x}+c \) =\(\displaystyle e^{t} \)
=>\(\displaystyle e^{-x}\) = \(\displaystyle -e^{t}+c \)
=>\(\displaystyle lne^{-x} \) = \(\displaystyle ln|c-e^{t}| \)
=>\(\displaystyle x \)= \(\displaystyle -ln|c-e^{t}| \)

Thanks a lot for your reply, evinda. I have just realised that one of the main reasons I went wrong was in failing to recall that \(\displaystyle \int \frac{1}{e^x} = -e^{-x}\). It completely slipped my mind...

 

[DrPhil]

Thanks a lot for your reply, evinda. I have just realised that one of the main reasons I went wrong was in failing to recall that \(\displaystyle \int \frac{1}{e^x} = -e^{-x}\). It completely slipped my mind...

Something else you have to more careful about is including "dx" or "dt" as part of writing integrals.
\(\displaystyle \int \frac{1}{e^x} \mathrm d x= -e^{-x}\)

 

[HallsofIvy]

Your error is more general than just not recognizing that \(\displaystyle \frac{1}{e^x}= e^{-x}\). You need to understand that, if f(x) is a function of x, you cannot treat if as if it were x itself. That is, \(\displaystyle \int\frac{1}{f(x)}dx\) is NOT "\(\displaystyle ln(f(x))+ C\)". IF you happened to have the derivative of f, f'(x), in the integral, \(\displaystyle \int\frac{1}{f(x)}f'(x)dx\), then you could substitute u= f(x), du= f'(x)dx, so that the integral becomes \(\displaystyle \int \frac{1}{u}du= ln(u)+ C= ln(f(x))+ C\).

Of course, f'(x) has to be in the integral to begin with!

 

[jonnburton]

OK, thanks DrPhil and HallsofIvy. Some more things to concentrate on... Actually, thinking about it now, I don't know how I came up with the idea \(\displaystyle \int\frac{1}{e^x}dx = ln|e^x|+ c\); it must have been part absent-mindedness and part lack of experience...

 

[jonnburton]

Reviewing this question again, I have noticed something that I can't explain. If I add the constant on the other side of the equals sign, the result comes out differently.

Starting at

\(\displaystyle \int e^{-x}dx = \int e^t dt\)

\(\displaystyle -e^{-x} = e^t + c\)

\(\displaystyle -lne^{-x} = ln(e^t +c)\)

\(\displaystyle (-x)(-lne) = ln(e^t+c)\)

\(\displaystyle x = ln(e^t+c)\)

I've done this several times and always come to the same result - as far as I can see there are no mistakes with the minus signs, etc, so how on earth can the solution be different. It stands to reason that I must have made an error, but I can't see it.

 

[evinda]

You wrote -\(\displaystyle e^{-x} \) = \(\displaystyle e^{t}\) +c=>-\(\displaystyle ln|e^{-x}| \) = \(\displaystyle ln|e^{t}+c| \), that is wrong.
First you have to change the sings because -\(\displaystyle e^{-x}\)<0, and you can only use ln with positive numbers.
It is like that:
\(\displaystyle -e^{-x} \) = \(\displaystyle e^{t} \) + c
=> \(\displaystyle e^{-x} \) = \(\displaystyle -e^{t} \)-c
=>\(\displaystyle ln|e^{-x}| \) = \(\displaystyle ln|-e^{t}-c| \)
=> -x=\(\displaystyle ln|-e^{t}-c| \)
=>x=-\(\displaystyle ln|-e^{t}-c| \) ,where c=-a

Reviewing this question again, I have noticed something that I can't explain. If I add the constant on the other side of the equals sign, the result comes out differently.

Starting at

\(\displaystyle \int e^{-x}dx = \int e^t dt\)

\(\displaystyle -e^{-x} = e^t + c\)

\(\displaystyle -lne^{-x} = ln(e^t +c)\)

\(\displaystyle (-x)(-lne) = ln(e^t+c)\)

\(\displaystyle x = ln(e^t+c)\)

I've done this several times and always come to the same result - as far as I can see there are no mistakes with the minus signs, etc, so how on earth can the solution be different. It stands to reason that I must have made an error, but I can't see it.

 

[jonnburton]

Thanks a lot evinda, that makes sense.

(I've just realised there's a separate thread for differential equations, where I should have posted this!)

You wrote -\(\displaystyle e^{-x} \) = \(\displaystyle e^{t}\) +c=>-\(\displaystyle ln|e^{-x}| \) = \(\displaystyle ln|e^{t}+c| \), that is wrong.
First you have to change the sings because -\(\displaystyle e^{-x}\)<0, and you can only use ln with positive numbers.
It is like that:
\(\displaystyle -e^{-x} \) = \(\displaystyle e^{t} \) + c
=> \(\displaystyle e^{-x} \) = \(\displaystyle -e^{t} \)-c
=>\(\displaystyle ln|e^{-x}| \) = \(\displaystyle ln|-e^{t}-c| \)
=> -x=\(\displaystyle ln|-e^{t}-c| \)
=>x=-\(\displaystyle ln|-e^{t}-c| \) ,where c=-a

 

[DrPhil]

Reviewing this question again, I have noticed something that I can't explain. If I add the constant on the other side of the equals sign, the result comes out differently.

Starting at

\(\displaystyle \int e^{-x}dx = \int e^t dt\)

\(\displaystyle -e^{-x} = e^t + c\)
\(\displaystyle -lne^{-x} = ln(e^t +c)\)....X

\(\displaystyle (-x)(-lne) = ln(e^t+c)\)

\(\displaystyle x = ln(e^t+c)\)

I've done this several times and always come to the same result - as far as I can see there are no mistakes with the minus signs, etc, so how on earth can the solution be different. It stands to reason that I must have made an error, but I can't see it.

You can't take a logarithm of a negative number, so you must change signs first. [Or you could take absolute value of both sides.]

\(\displaystyle e^{-x} = -e^t - c\)

Now since \(\displaystyle c\) is an arbitrary constant, we can give it either sign. It will look better with a +.

 

[jonnburton]

You can't take a logarithm of a negative number, so you must change signs first. [Or you could take absolute value of both sides.]

\(\displaystyle e^{-x} = -e^t - c\)

Now since \(\displaystyle c\) is an arbitrary constant, we can give it either sign. It will look better with a +.

Thanks, Dr Phil.

I don't see why the following would be wrong then (as far as I can see this is taking absolute values):

\(\displaystyle -e^{-x} = e^t+c\)

\(\displaystyle ln|-e^{-x}| = ln|e^t +c|\)


Becuase we've got the modulus of 'e', the left hand side is equivalent to \(\displaystyle -xln(e) = -x\)

\(\displaystyle -x = ln|e^t +c|\)

\(\displaystyle x = -ln|e^t +c|\)


I can't see how \(\displaystyle ln|-e^{-2x}|\) is different from \(\displaystyle ln|-e^t -c|\). The only difference is that in the former we're following it through and bringing the 'x' down, using the absolute value of 'e'.