Finding the intergral function (dy/dt = a(q-y))

[sweet springs]

Hi. Relevant integral formula is

\int\frac{1}{x} dx = ln |x|+C

You can easily check validation by differentiating formula. Regards.

 

[miniradman]

Alrighty...

\normalsize \frac{1}{3} \int \frac{1}{u} du

\normalsize \frac{1}{3} ln u

sub \normalsize u = x^3 - 1 So:

\normalsize\frac{1}{3} ln |x^3 - 1|+ c

or

\normalsize\frac{ln |x^3 -1|}{3} + c

Ok, I think I'm ready to tacking the original question using the u substitution (you're probably think "gosh, finally!")

ok.

\normalsize \frac{dy}{dt} = a(q-y)

\normalsize \frac{dy}{q-y} = a dt

let u = (q -y)

So: \normalsize \frac{du}{dy} = 0 - 1

\normalsize du = - dy

\normalsize \int \frac{dy}{u} = \int a dt

\normalsize \int \frac{1}{u} dy = \int a dt

sub in - du

\normalsize - \int \frac{1}{u} du = \int a dt

\normalsize - ln |u| = at + c

\normalsize ln |u| = (-at - c)

sub in the u value q - y

\normalsize - ln |q - y| = at + c (what sweet springs had!)

\normalsize ln |q - y| = -at - c

\normalsize |q - y| = e^{-at-c}

\normalsize |q - y| = e^{-at} × e^{-c}

\normalsize - y = e^{-at} × e^{-c} -q

\normalsize y = (e^{-at} × e^{-c} -q) \div -1

\normalsize y = -e^{-at} × -e^{-c} + q

\normalsize y = e^{-at} × e^{-c} + q (negative multiplied by negative make positive)

Let e^-c = A (A is just what we use at school to conform with the standard exponential form of: y = Ae^{Bt} + C

\normalsize y = Ae^{-at} + q

I'm guessing that's the end of it. YAY!

Aaand I did it all by myself... jokes

I'm sorry for being such a pain to teach and I swear I've never had this much trouble understanding a concept (well, I don't usually have this much trouble anyway )

Thank you; Failexam, RoshanBBQ, Vela, Noorac and sweet springs... miniradman is forever in your debt

Thanks again!

 

[SammyS]

...
Ok, I think I'm ready to tackling the original question using the u substitution (you're probably think "gosh, finally!")

ok.

\normalsize \frac{dy}{dt} = a(q-y)

\normalsize \frac{dy}{q-y} = a dt

let u = (q -y)

So: \normalsize \frac{du}{dy} = 0 - 1

\normalsize du = - dy

\normalsize \int \frac{dy}{u} = \int a dt

\normalsize \int \frac{1}{u} dy = \int a dt

sub in - du

\normalsize - \int \frac{1}{u} du = \int a dt

\normalsize - ln |u| = at + c

\normalsize ln |u| = (-at - c)

sub in the u value q - y

\normalsize - ln |q - y| = at + c (what sweet springs had!)

\normalsize ln |q - y| = -at - c

\normalsize |q - y| = e^{-at-c}

\normalsize |q - y| = e^{-at} × e^{-c}

\normalsize - y = e^{-at} × e^{-c} -q

\normalsize y = (e^{-at} × e^{-c} -q) \div -1

\normalsize y = -e^{-at} × -e^{-c} + q

\normalsize y = e^{-at} × e^{-c} + q (negative multiplied by negative make positive)

Let e^-c = A (A is just what we use at school to conform with the standard exponential form of: y = Ae^{Bt} + C

\normalsize y = Ae^{-at} + q

I'm guessing that's the end of it. YAY!

Aaand I did it all by myself... jokes

I'm sorry for being such a pain to teach and I swear I've never had this much trouble understanding a concept (well, I don't usually have this much trouble anyway )

Thank you; Failexam, RoshanBBQ, Vela, Noorac and sweet springs... miniradman is forever in your debt

Thanks again!

Almost:
\displaystyle (e^{-at} × e^{-c})\div(-1)=-e^{-at} × e^{-c}\ .

There is no distributive law for division over multiplication. (A positive times a positive divided by a negative results in a negative.)

Also, use the given conditions to evaluate c, or A. In particular, y(0) =0, and t ≥ 0 .

 

[miniradman]

So then shouldn't the final function be ?

y = -Ae^{-at} + q

 

[RoshanBBQ]

I don't know, but I do want to bring up an important point. Since you didn't know U-substitution, you were probably supposed to do it another way. The only way I can see you doing it is to just think it out. Ask yourself, "What could I differentiate to equal 1/(q-y)"? If you 1.) remember that the derivative of ln(x) is 1/x and 2.) remember chain rule, you can brute force the answer using your human mind.

Is that what they're teaching you right now?

 

[miniradman]

Well, we've learned the u substitution, but we've never applied it to a question like this.

But I must queery, why don't we use the u substitution method for Newton's law of cooling questions?

\frac {dT}{dt} = k (T - T_s)

 

[RoshanBBQ]

Well, we've learned the u substitution, but we've never applied it to a question like this.

But I must queery, why don't we use the u substitution method for Newton's law of cooling questions?

\frac {dT}{dt} = k (T - T_s)

From what I'm seeing, I think you would use u substitution (in one of the simplest cases possible with u = T - T_s; du = dT) after you separate the variables and integrate both sides.

 

[miniradman]

That's the thing... I never differentiate u (or T - T_s)?

Considering that the question I posted is almost (but not exactly) the same as the rule for Newton's Law of Cooling. Hence the reason why I went straight from

\normalsize\int\frac {dy}{(q-y)}

to

ln |q-y|

because I've done many Newtons law of cooling questions, none of which did I differentiate "u" (in fact, I didn't even use u)

 

[RoshanBBQ]

That's the thing... I never differentiate u (or T - T_s)?

Considering that the question I posted is almost (but not exactly) the same as the rule for Newton's Law of Cooling. Hence the reason why I went straight from

\normalsize\int\frac {dy}{(q-y)}

to

ln |q-y|

because I've done many Newtons law of cooling questions, none of which did I differentiate "u" (in fact, I didn't even use u)

I'm a bit unsure what you're asking. At the least, do you see how to do

\int \frac{dT}{T-T_s}

using u substitution? It's likely you never used u substitution to do it, because it's so basic that you can just think out what the answer is. But if you ever get stuck, you can rely on the systematic application of certain tools they teach you like u substitution.

 

[miniradman]

\int \frac{dT}{T-T_s}
It's likely you never used u substitution to do it, because it's so basic that you can just think out what the answer is.

oh, well I just tried doing it with the du = d(whatever) method and I found that du = dT. Which means that I didn't have to adjust the formula at all... it all makes sense now.

 

[RoshanBBQ]

oh, well I just tried doing it with the du = d(whatever) method and I found that du = dT. Which means that I didn't have to adjust the formula at all... it all makes sense now.

Yeah. You just get

\int \frac{du}{u}

That's what makes it probably the simplest (yet still useful) u substitution possible. The simplest is to let u = x.

 

[Mark44]

That's what makes it probably the simplest (yet still useful) u substitution possible. The simplest is to let u = x.

We can infer from what you said that the substitution u = x is not useful, which is true. Using the substitution is not useful, because all you would be doing is changing a letter in the problem.

 

[sweet springs]

Hi.

So then shouldn't the final function be ?
y = -Ae^{-at} + q

Back to your first post

Find the intergral function of:
dy/dt=a(q-y) where t ≥0, y(0)=0 a and q are constants.

So A=q then y = q(1-e^-at). If y is temperature of a body for example, from zero it is warmed up to the temperature of heat bath q. Regards.

 

[miniradman]

Hmmm, I've completely forgotten about that condition... it all makes sense now!

Thanks again sweet springs!