Differential eqn dx/dt = (10 - x)/5 (don't understand soln)

[Monkeyseat]

Hi,

Question

Solution

What I don't understand is, why is it neccessary to multiply both sides of the equation by 5 and get t not as a fraction (see where the red arrow is)? If you don't multiply through by five, you don't get the correct value of the constant, but I don't know why using t/5 to find the constant gives a wrong answer.

For a different differential equation question, I was at this stage:

tan(y) = (xe[sup:kw4qnrsk]-3x[/sup:kw4qnrsk]/-3) - (e[sup:kw4qnrsk]-3x[/sup:kw4qnrsk]/9) + C

I didn't multiply through to get rid of the fractions there, and I got the correct answer... So why do you need to multiply by 5 in the question I highlighted?

Thanks.

 

[BigGlenntheHeavy]

Re: Differential equation

\(\displaystyle \frac{dx}{dt} \ = \ \frac{10-x}{5}\)

\(\displaystyle \int\frac{dx}{10-x} = \int\frac{dt}{5}\)

\(\displaystyle -ln|10-x| \ = \ \frac{t}{5}+C\)

\(\displaystyle ln|10-x| \ = \ \frac{-t}{5}-C\)

\(\displaystyle |10-x| \ = \ (e^{-t/5})(e^{-C})\)

\(\displaystyle 10-x \ = \ Ae^{-t/5}, \ A \ = \ \pm e^{-C} \ or \ zero.\)

\(\displaystyle x(t) \ = \ 10-Ae^{-t/5}\)

\(\displaystyle x(0) \ = \ 1 \ = \ 10-A, \ A \ = \ 9\)

\(\displaystyle Hence, \ x(t) \ = \ 10-9e^{-t/5}.\)

\(\displaystyle Check: \ \frac{dx}{dt} \ = \ \frac{9e^{-t/5}}{5} \ = \ \frac{10-x}{5}, \ 9e^{-t/5} \ = \ 10-x.\)

 

[Monkeyseat]

Re: Differential equation

Thanks for the reply. I know my answer is correct. What I meant was, at this stage:

-ln(10 - x) = t/5 + C

If I substitute values in for x and t now:

-ln9 = 0/5 + C

Thus, C = -ln9.

But that is not the correct value of the constant. You have to multiply by 5 and get -5ln(10 - x) = t + A, so when you substitute values in for x and t you get A = -5ln9.

I wanted to know, in the working I gave, how come if I don't multiply by 5 to get t on its own the answer is wrong?

Thanks.

 

[Deleted member 4993]

Re: Differential equation

Thanks for the reply. I know my answer is correct. What I meant was, at this stage:

-ln(10 - x) = t/5 + C

If I substitute values in for x and t now:

-ln9 = 0/5 + C

Thus, C = -ln9.

But that is not the correct value of the constant. You have to multiply by 5 and get -5ln(10 - x) = t + A, so when you substitute values in for x and t you get A = -5ln9.

I wanted to know, in the working I gave, how come if I don't multiply by 5 to get t on its own the answer is wrong? <<< No - look at the total expression for (x) [or (t)]. Whether you multiply by '5' or not - the solution is same.Thanks.

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[BigGlenntheHeavy]

Your solution is t as a function of x, t(x), which is correct, but the problem wants the solution of x as a function of t, x(t), my solution. Solve dx/dt, not dt/dx

 

[Monkeyseat]

Re: Differential equation

I wanted to know, in the working I gave, how come if I don't multiply by 5 to get t on its own the answer is wrong? <<< No - look at the total expression for (x) [or (t)]. Whether you multiply by '5' or not - the solution is same.Thanks.

I've realised that now. Thanks.

Your solution is t as a function of x, t(x), which is correct, but the problem wants the solution of x as a function of t, x(t), my solution. Solve dx/dt, not dt/dx

Oh ok. I didn't think it mattered whether the answer was in the form t = f(x) or x = f(t) - I thought either was acceptable as the question didn't specify what form to leave it in. How should I know what form to leave it in? For the book I have, for one question that is "dx/dt = ..." it leaves the answer in the form t = f(x), yet for another question that is "dx/dt = ..." it leaves the answer in the form x = f(t).

 

[BigGlenntheHeavy]

No biggie, small potatos.