# First Order Linear Equations

Yay!

I was sick with some kind of food poisoning so I wasn't able to go to class today (Monday), and now I'm attempting to struggle through section 9.2 -- First Order Linear Equations!

Alright, so here's what I have so far:

Problems begin in various forms, and the first step is to get each equations into the following form:

$\frac{dy}{dx} + P(x)y = Q(x)$

Now, given this form, we have to find something called the "Integrating Factor", which will be factored INTO every term of the equation. This factor is generally given by the notation v(x) in the book. So in other words after this factor is found and distributed, the above equation would look like this:

$v(x)\frac{dy}{dx} + v(x)P(x)y = v(x)Q(x)$

Why do this? Honestly, I don't know, but the general consensus from the internet seems to be that by multiplying this factor through the problem, you are able to put the equation into a form that is integrable on both sides, allowing you to solve for y. Cool huh? Whatever, let's do this.

After a lot of digging, I found the formula for the "Integrating Factor" to be

$v(x) = e^{\int_{}^{}{}}P(x)dx$

This all looks like nothing without context, so I'm going to post the first problem from my homework and maybe I'll figure it out somewhere along the way.

$x\frac{dy}{dx} + y = e^x$, where x > 0

1) Put it in standard form of

$\frac{dy}{dx} + P(x)y = Q(x)$

After some algebraic noodling, I ended up with

$\frac{dy}{dx} + \frac{1}{x}y = \frac{e^x}{x}$

Okay, great. I cheated and it looks like I'm on the right track. Now, onto the "Integrating Factor" bit.

2) Multiply the whole equation by the "Integrating Factor", defined as

$v(x) = e^{\int_{}^{}{}}P(x)dx$

Based on the standard form, we know the following:

$\large P(x) = \frac{1}x{}$

Given this, we know that    $\large v(x) = e^{\int_{}^{}{}}\frac{1}{x}dx$, and we have to solve the integral to get the value of v(x). That's a pretty easy integral, so you'll get      $\large e^{ln(x)}$     and    $\large e^{ln}$  simplifies to 1, giving you  $\large x$  for the value of v(x) or the "Integrating Factor".

3) Multiply the equation in standard form by the "Integrating Factor".

$\large (x)\frac{dy}{dx} + (x)\frac{1}{x}y = (x)\frac{e^x}{x}$

Now the weird thing here is I'm not sure what happens to the dy/dx.  The x's cancel to leave y and e^x, but I'm not sure about the math with the x(dy/dx). All I know is that after you find the "Integrating Factor", you should end up with the following form:

$\large y = \frac{\int v(x)Q(x)dx + C}{v(x)}$

If you can't, you can simply plug in the known v(x) or the "Integrable Factor", the known Q(x) as per the standard form, solve the integral, and lastly do the algebra.

Okay, so to cobble this together:

1) Put the equation in standard form to determine P(x) and Q(x)

2) Find the "Integrable Factor"

3) Use the form of the solution above to take the integral, clean up the algebra and solve for y.

I'm going to try and practice now, but there's a huge portion of this that I'm missing.

I'm a student, I like games, I like music, I like learning.
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### 4 Responses to First Order Linear Equations

1. Darrell says:

Here’s a good video that outlines the process:

2. Darrell says:

Okay, so to cobble this together:

1) Put the equation in standard form to determine P(x) and Q(x)

2) Find the “Integrable Factor”

3) Use the form of the solution above to take the integral, clean up the algebra and solve for y.

The key thing I’m missing is that immediately after distributing v(x) to both sides, you have to integrate both sides immediately–with respect to x (most of the time). This is confusing because on the “left” side of the equation, you’re integrating y’, which simply integrates to y–WHICH YOU CAN SOLVE FOR. The bizarre part of this is that after you integrate both sides and solve for y, you are dividing both sides by what is your “Integrating Factor”, which holds to the form of the solution (found in the post and in the Formula page as well). Re-read this a lot of times to really understand it.

3. Darrell says:

“Now the weird thing here is I’m not sure what happens to the dy/dx. The x’s cancel to leave y and e^x, but I’m not sure about the math with the x(dy/dx).”

It looks weird because I’m using Leibniz notation. If I had written y’ instead, it would be clear that I am integrating y’ into y.

4. Jarratt says:

First post!