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Here is a link to the question:Initial value problem?

2(dy/dx) - 4xy = 8x. y(0)=12

Help please

Initial value problem? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

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Here is a link to the question:Initial value problem?

2(dy/dx) - 4xy = 8x. y(0)=12

Help please

Initial value problem? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

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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.

- Mar 26, 2013

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Thanks

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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:

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

- Mar 26, 2013

- 77

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!

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Without using $\LaTeX$, I would write the integrating factor as either:

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!

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 $y(x)\cdot\mu(x)$?

- 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:

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