Help with First Order Differential

[Tom Hardy]

Homework Statement

The Attempt at a Solution

The first part is fairly simple I think. It's just rate of accumulation = rate of generation - rate of output(losses)

I'm not too sure how to solve this differential equation. I divide the whole equation through by mc and rearrange but I keep getting an extra constant and I have no idea what to do with it.

I get someone in the form dtheta/dt + P(t)theta = Q(t) + constant and I have no idea how to solve that. Any help will be appreciated, thank you.
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[wabbit]

The term that is messing things up is $mc_vkt$. So, first solve the differential equation without that term. You'll get a solution of the form Constant*Something. This gives you an idea of how the solution might look like : try the form C(t)*Something (same Something, now C(t) is what you solve for.)

 

[Tom Hardy]

The term that is messing things up is $mc_vkt$. So, first solve the differential equation without that term. You'll get a solution of the form Constant*Something. This gives you an idea of how the solution might look like : try the form C(t)*Something (same Something, now C(t) is what you solve for.)

Hey, thanks for replying. Surely if I just remove a term from the differential I won't be solving what I'm supposed to or am I missing something else? :S

 

[wabbit]

Of course you won't, but it's like a warm up lap. The solution to the simplified equation serves you as a basis from which you build the solution to the full equation - this is a very common and useful strategy in diff equations (and not only there)

I told you how in first post

 

[LCKurtz]

Divide your equation through by $mc_v$ and put it in the form$$
\frac{d\theta}{dt}+\alpha \theta = \alpha \theta_w + kt$$That is a constant coefficient linear DE. You can solve it using constant coefficient methods for homogeneous and non-homogeneous equations, or use an integrating factor $e^{\alpha t}$.

 

[Tom Hardy]

Divide your equation through by $mc_v$ and put it in the form$$
\frac{d\theta}{dt}+\alpha \theta = \alpha \theta_w + kt$$That is a constant coefficient linear DE. You can solve it using constant coefficient methods for homogeneous and non-homogeneous equations, or use an integrating factor $e^{\alpha t}$.

Hey thanks for answering, I thought about that but surely that's not in the correct form to use an integrating factor? From my notes on the left it should just be Q(t) but we have a constant term in there as well.

 

[SammyS]

Hey thanks for answering, I thought about that but surely that's not in the correct form to use an integrating factor? From my notes on the left it should just be Q(t) but we have a constant term in there as well.

What do you mean by "it" in ... on the left it should just be Q(t) ?

That is the correct integrating factor.

What is the derivative of $\theta\,e^{\alpha\, t}\$?

 

[Tom Hardy]

What do you mean by "it" in ... on the left it should just be Q(t) ?

That is the correct integrating factor.

What is the derivative of $\theta\,e^{\alpha\, t}\$?

No I get that, it's just that I thought in order to use the integrating factor differentials had to be in the form dy/dx + p(x)y = Q(x)

and LCKurtz reply mentioned dividing through by mc to get dy/dx + p(x)y = Q(x) + constant
I'm just wondering is it still okay to use the IF here?

 

[LCKurtz]

No I get that, it's just that I thought in order to use the integrating factor differentials had to be in the form dy/dx + p(x)y = Q(x)

and LCKurtz reply mentioned dividing through by mc to get dy/dx + p(x)y = Q(x) + constant
I'm just wondering is it still okay to use the IF here?

Why don't you just try it and see what happens?

 

[Tom Hardy]

Why don't you just try it and see what happens?

Fine, I'll give it a go now.

 

[Tom Hardy]

Why don't you just try it and see what happens?

Ah, it worked out, thank you for your help.