# Math Help - No idea how to solve this differential equation

1. ## No idea how to solve this differential equation

dy/dx + y = x

hmmmmmmmm

2. Hello,
Originally Posted by Kaitosan
dy/dx + y = x

hmmmmmmmm
Integrating factor is $e^x$

so multiply the equation by $e^x$ :

$\frac{dy}{dx}\cdot e^x +ye^x=xe^x$

$\frac{d}{dx} \left(ye^x\right)=xe^x$

So $ye^x=\int xe^x ~dx$

$\Rightarrow y=\frac{\int xe^x ~dx}{e^x}$

(use an integration by parts to calculate the integral)

3. Geez..... I don't know what you did. I've only been doing basic differential equations. I know it's a lot to ask but could you go a bit deeper? That method... integrating factor....? It's all new to me. Also, how can d(ye^x)/dx = e^x(dy/dx +y)?

4. Originally Posted by Kaitosan
Geez..... I don't know what you did. I've only been doing basic differential equations. I know it's a lot to ask but could you go a bit deeper? That method... integrating factor....? It's all new to me. Also, how can d(ye^x)/dx = e^x(dy/dx +y)?

It's a method based on the product rule of differentiation

Okay, let's take a general example :
$y'+p(x)y=q(x)$

The integrating factor is defined as $\exp\left(\int p(x) ~dx\right)$ (it is not necessary to have boundaries)

Why is it useful ? Because the product rule will help you eliminate y'.

We indeed have, after multiplying by the integrating factor :
$\exp\left(\int p(x) ~dx\right) \cdot y'+p(x) \exp\left(\int p(x) ~dx\right) y=q(x)\exp\left(\int p(x) ~dx\right) \quad {\color{red}\star}$

But look at the LHS.
You can see (or you know) that it's the derivative of $y\exp\left(\int p(x) ~dx\right)$
---> apply the product rule (along with the chain rule) to check this :

\begin{aligned}
\text{derivative} &=y'\exp\left(\int p(x) ~dx\right)+y \left(\exp\left(\int p(x) ~dx\right)\right)' \\
&=y'\exp\left(\int p(x) ~dx\right)+\left(\int p(x) ~dx\right)' \exp\left(\int p(x) ~dx\right) y \\
&=y'\exp\left(\int p(x) ~dx\right)+p(x) \exp\left(\int p(x) ~dx\right) y \end{aligned}

So finally, $\color{red}\star$ becomes :
$\left(y \exp\left(\int p(x) ~dx\right)\right)'=p(x) \exp\left(\int p(x) ~dx\right)$

And in order to get y, integrate both sides of the equation. And then divide by $\exp\left(\int p(x) ~dx\right)$

In this case, it's "easy", because p(x)=1. And thus the integrating factor is $\exp\left(\int dx\right)=e^x$

Am I speakig Chinese to you ?
Anyway, you can find lots of reference in google, looking for "integrating factor"

Originally Posted by Moo

It's a method based on the product rule of differentiation

Okay, let's take a general example :
$y'+p(x)y=q(x)$
How in the world does this example relate to my problem? All I see is a jumble of letters/functions. Can you please use a better example consisted of only variables "x" and "y" and one that shows how to recognize an equation as one requiring the integrating factor?

Originally Posted by Moo
The integrating factor is defined as $\exp\left(\int p(x) ~dx\right)$ (it is not necessary to have boundaries)
Do you mean e^(integ of x)? If so, is this the only integrating factor? Are there various kinds of integrating factors? If so, how do I determine the appropriate integrating factor?

Originally Posted by Moo
$

\frac{dy}{dx}\cdot e^x +ye^x=xe^x
$

$

\frac{d}{dx} \left(ye^x\right)=xe^x
$

I'm sorry, but I'm not a pro in "calgebra." Please explicitly show the algebraic steps.

6. Originally Posted by Kaitosan

How in the world does this example relate to my problem? All I see is a jumble of letters/functions. Can you please use a better example consisted of only variables "x" and "y" and one that shows how to recognize an equation as one requiring the integrating factor?

Do you mean e^(integ of x)? If so, is this the only integrating factor? Are there various kinds of integrating factors? If so, how do I determine the appropriate integrating factor?

I'm sorry, but I'm not a pro in "calgebra." Please explicitly show the algebraic steps.
Why don't you tell us what you do know so we can pitch the answers at your level.

So far you seem to be indicating that you do not know the product rule for derivatives (google for "product rule calculus"), and are not at home with functional notation. If so I doubt we can help you, go back to your instructor and ask for help.

CB

7. edit: nvm. I had a mental block throughout this but I think I got it.