Welcome to our community

Be a part of something great, join today!

Logan's question at Yahoo! Answers involving an IVP with a linear 1st order ODE

  • Thread starter
  • Admin
  • #1

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
  • Thread starter
  • Admin
  • #2

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Hello Logan,

We are given the first order linear IVP to solve:

\(\displaystyle 2\frac{dy}{dx}-4xy=8x\) where \(\displaystyle y(0)=12\)

To find the general solution to the ODE, I would first divide through by 2 to obtain:

\(\displaystyle \frac{dy}{dx}-2xy=4x\)

Next, let's compute the integrating factor:

\(\displaystyle \mu(x)=e^{-2\int x\,dx}=e^{-x^2}\) which gives us:

\(\displaystyle e^{-x^2}\frac{dy}{dx}-2xe^{-x^2}y=4xe^{-x^2}\)

Now we may rewrite the left side as the differentiation of a product:

\(\displaystyle \frac{d}{dx}\left(e^{-x^2}y \right)=4xe^{-x^2}\)

Integrate with respect to $x$:

\(\displaystyle \int\,d\left(e^{-x^2}y \right)=-2\int e^{-x^2}(-2x\,dx)\)

\(\displaystyle e^{-x^2}y=-2e^{-x^2}+C\)

Multiply through by \(\displaystyle e^{x^2}\):

\(\displaystyle y(x)=-2+Ce^{x^2}\)

Now, use initial conditions to determine the parameter $C$:

\(\displaystyle y(0)=-2+Ce^{0^2}=12\implies C=14\)

and so the solution satisfying the IVP is:

\(\displaystyle y(x)=14e^{x^2}-2\)

To Logan and any other guests viewing this topic, I invite and encourage you to post other ODE problems in our Differential Equations forum.

Best Regards,

Mark.
 

LLand314

Member
Mar 26, 2013
77
Could you explain how you computed the integral factor some more? For example how did you know and come up with μ(x)=e−2∫xdx=e−x2 ?

Thanks
 
  • Thread starter
  • Admin
  • #4

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
It is the standard method used to compute the integrating factor, which makes it possible to rewrite the left side as the product of a differentiation.

Consider the linear ODE in standard form:

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

Now, if we multiply through by the integrating factor \(\displaystyle \mu(x)=e^{\int P(x)\,dx}\) we get:

\(\displaystyle e^{\int P(x)\,dx}\frac{dy}{dx}+e^{\int P(x)\,dx}P(x)y=e^{\int P(x)\,dx}Q(x)\)

Observe now that:

\(\displaystyle \frac{d}{dx}\left(e^{\int P(x)\,dx}y \right)=e^{\int P(x)\,dx}\frac{dy}{dx}+e^{\int P(x)\,dx}P(x)y\)

and so the ODE may be written:

\(\displaystyle \frac{d}{dx}\left(e^{\int P(x)\,dx}y \right)=e^{\int P(x)\,dx}Q(x)\)

Integrating with respect to $x$, we have:

\(\displaystyle \int\,d\left(e^{\int P(x)\,dx}y \right)=\int e^{\int P(x)\,dx}Q(x)\,dx\)

\(\displaystyle e^{\int P(x)\,dx}y=\int e^{\int P(x)\,dx}Q(x)\,dx\)

Solving for $y$, we obtain:

\(\displaystyle y(x)=e^{-\int P(x)\,dx}\int e^{\int P(x)\,dx}Q(x)\,dx\)
 

LLand314

Member
Mar 26, 2013
77
Ok so if its in this form
dydx+P(x)y=Q(x)

I will always compute the integral with
μ(x)=e∫P(x)dx ?

Or only then because of P(x) in the problem

Thanks so much!
 
  • Thread starter
  • Admin
  • #6

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Ok so if its in this form
dydx+P(x)y=Q(x)

I will always compute the integral with
μ(x)=e∫P(x)dx ?

Or only then because of P(x) in the problem

Thanks so much!
Without using $\LaTeX$, I would write the integrating factor as either:

μ(x)=exp(∫P(x)dx ) or μ(x)=e^(∫P(x)dx)

Yes, this is how it is computed because of the special nature of the derivative of the exponential function with $e$ as the base. Did you understand how it allows the left side of the ODE in standard linear form to be simplified as the derivative with respect to $x$ of the product $y(x)\cdot\mu(x)$?
 

LLand314

Member
Mar 26, 2013
77
Without using $\LaTeX$, I would write the integrating factor as either:

μ(x)=exp(∫P(x)dx ) or μ(x)=e^(∫P(x)dx)

Yes, this is how it is computed because of the special nature of the derivative of the exponential function with $e$ as the base. Did you understand how it allows the left side of the ODE in standard linear form to be simplified as the derivative with respect to $x$ of the product $y(x)\cdot\mu(x)$?

O ok sorry I just tried to copy and paste the information, I haven't learned Latex yet. I think I kind of understand it just need to do some more problems.

moderator edit: I have moved your new question into its own topic here:

http://www.mathhelpboards.com/f17/solving-first-order-linear-initial-value-problem-4051/
 
Last edited by a moderator:
  • Thread starter
  • Admin
  • #8

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Hey no worries, I was just showing you a better way to write the integrating factor, just for clarity. :D

I am happy you have registered and are participating here! (Yes)
 
Last edited:

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,404
Just in case you're wondering why the Integrating Factor is \(\displaystyle \displaystyle e^{P(x)\,dx}\), let's look at a general first order linear DE in more detail...

\(\displaystyle \frac{dy}{dx} + P\,y = Q\), where P, Q and y are functions of x.

What we would like is to write the LHS as a single derivative, because then we would be able to solve for y by integrating. To do this, we make use of the Product Rule \(\displaystyle \displaystyle \frac{d}{dx} \left[ f\,g \right] = f\,\frac{dg}{dx} + g\,\frac{df}{dx}\). At the moment, the LHS doesn't look like a product rule expansion, but it's possible to multiply by a function to make this so. Call this function I. Then

\(\displaystyle I\,\frac{dy}{dx} + I\,P\,y = I\,Q\)

Now in order for the LHS to be a product rule expansion, then that would mean

\(\displaystyle \begin{align*} \frac{dI}{dx} &= I\,P \\ \frac{1}{I}\,\frac{dI}{dx} &= P \\ \int{\frac{1}{I}\,\frac{dI}{dx}\,dx} &= \int{P\,dx} \\ \int{\frac{1}{I} \,dI } &= \int{P\,dx} \\ \ln{|I|} &= \int{P\,dx} \\ I &= A\,e^{\int{P\,dx}} \end{align*} \)

So any value of the constant A will give an integrating factor that works, so we usually just choose the principal value with A = 1. So if we multiply both sides of the DE by \(\displaystyle \displaystyle e^{\int{P\,dx}}\) we find

\(\displaystyle \displaystyle \begin{align*} e^{\int{P\,dx}}\,\frac{dy}{dx} + e^{\int{P\,dx}}\,P\,y &= e^{\int{P\,dx}}\,Q \\ \frac{d}{dx} \left( e^{\int{P\,dx}}\,y \right) &= e^{\int{P\,dx}}\,Q \\ e^{\int{P\,dx}}\,y &= \int{e^{\int{P\,dx}}\,Q\,dx} \\ y &= e^{-\int{P\,dx}}\int{e^{\int{P\,dx}}\,Q\,dx} \end{align*}\)

Hope that helps...