# Help with First Order Differential

• Tom Hardy
In summary, the conversation was about solving a differential equation involving the term ##mc_vkt##. The expert suggests first solving the equation without the term and then using the solution as a basis to build the solution for the full equation. The process involves dividing the equation through by ##mc_v## and using an integrating factor to solve the resulting constant coefficient linear DE. The asker is unsure about using the integrating factor due to a constant term, but the expert encourages them to try it and see what happens. In the end, the approach is successful.
Tom Hardy

## 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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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.)

wabbit said:
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

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

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
LCKurtz said:
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.

Tom Hardy said:
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
SammyS said:
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?

Tom Hardy said:
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
LCKurtz said:
Why don't you just try it and see what happens?

Fine, I'll give it a go now.

LCKurtz said:
Why don't you just try it and see what happens?
Ah, it worked out, thank you for your help.

## 1. What is a first order differential equation?

A first order differential equation is an equation that contains a function and its first derivative. It relates the rate of change of a dependent variable to the value of an independent variable. It is typically written in the form dy/dx = f(x), where y is the dependent variable and x is the independent variable.

## 2. How do you solve a first order differential equation?

To solve a first order differential equation, you can use various methods such as separation of variables, integrating factor, or substitution. The specific method used depends on the form of the equation and the initial conditions given. It is important to note that not all first order differential equations are solvable analytically and may require numerical methods.

## 3. What are initial conditions in a first order differential equation?

Initial conditions refer to the values of the dependent and independent variables at a specific point, typically denoted as x0 and y0, respectively. These values are used to find the particular solution to a differential equation, as they help determine the value of the arbitrary constant that arises during the integration process.

## 4. Can I use a first order differential equation to model real-world phenomena?

Yes, first order differential equations can be used to model various real-world phenomena, such as population growth, chemical reactions, and electrical circuits. However, it is important to note that these models are simplifications of the real-world systems and may not always accurately represent the behavior of the system.

## 5. Are there any applications of first order differential equations in other fields?

Yes, first order differential equations have numerous applications in various fields such as physics, biology, economics, and engineering. They can be used to describe and predict the behavior of systems in these fields, making them a valuable tool for scientific research and problem-solving.

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