Analytic solution of this ordinary differentiale equation

[bsodmike]

Homework Statement

y'=4e^{0.8x}-0.5y

This question was obtained from a textbook, where it is used as an example for the application of Heun's method (of ODE integration). They state that it has a 'simple' analytic solution of,

y=\dfrac{4}{1.3}(e^{0.8x}-e^{-0.5x})+2e^{-0.5x}

Homework Equations

It is of the form.
https://www.physicsforums.com/latex_images/20/2024026-2.png

The Attempt at a Solution

Attempted to use an integrating factor, https://www.physicsforums.com/latex_images/20/2024278-0.png
https://www.physicsforums.com/latex_images/20/2024026-6.png

Obtaining,
https://www.physicsforums.com/latex_images/20/2024278-1.png

Hence,
https://www.physicsforums.com/latex_images/20/2024278-2.png

Any ideas ?!?

 

[bsodmike]

It is also given, y(0) = 2. i.e. at x=0, y=2.

 

[HallsofIvy]

Homework Statement

y'=4e^{0.8x}-0.5y

This question was obtained from a textbook, where it is used as an example for the application of Heun's method (of ODE integration). They state that it has a 'simple' analytic solution of,

y=\dfrac{4}{1.3}(e^{0.8x}-e^{-0.5x})+2e^{-0.5x}

Homework Equations

It is of the form.
https://www.physicsforums.com/latex_images/20/2024026-2.png

The Attempt at a Solution

Attempted to use an integrating factor, https://www.physicsforums.com/latex_images/20/2024278-0.png
https://www.physicsforums.com/latex_images/20/2024026-6.png

Obtaining,
https://www.physicsforums.com/latex_images/20/2024278-1.png

Hence,
https://www.physicsforums.com/latex_images/20/2024278-2.png

Any ideas ?!?

Yes, that is correct.

It is also given, y(0) = 2. i.e. at x=0, y=2.

Then since
y(x)= \frac{4}{1.3}e^{.8t}+ Ce^{-.5t}
y(0)= \frac{4}{1.3}+ C= 2
so
C= 2- \frac{4}{1.3}[/itex]<br /> <br /> That is, <br /> y(x)= \frac{4}{1.3}e^{.8t}+ \left(2- \frac{4}{1.3}\right)e^{-.5t}<br /> <br /> y(x)= \frac{4}{1.3}\left(e^{.8t}+ e^{-.5t}\right)+ 2e^{-.5t}<br /> <br /> exactly as given.

 

[Thaakisfox]

This differential equation is quite simple, so its better if we see what's behind this integration factor. Its basically Lagranges method, this "fits" to your hand a bit more:

So we have:

y&#039;+p(x)y=q(x)

As we know the general solution of the DE can be obtained by adding the Y general solution of the homogenous part and a y_0 particular solution of the entire DE.

first of all let's solve the homogenous part, that is:

Y&#039;+p(x)Y=0 \Longrightarrow \frac{dY}{Y}=-p(x)dx \Longrightarrow Y=C\exp\left(-\int^x p(x&#039;)dx&#039;\right)

Now we only have to find a particular solution of the entire DE. Here is the trick, let's consider the constant in the homogenous part to be some function of the free variable, that is: C=C(x)
so we have: y_0=C(x)\exp\left(-\int^x p(x&#039;)dx&#039;\right)

Now plug this back into the DE to get the C(x) function, and then you have the particular and add this to the homogenous part, and then you have the general solution:
y=y_0+Y

So basicaly the integration factor is a solution of the homogenous part of the DE.

 

[bsodmike]

Thanks HallsOfIvy!

I was trying to evaluate the following,

y(x)= \frac{4}{1.3}e^{.8t}+ Ce^{-.5t}

at y(0) and x=0 (in your example, t=0) as all the exp() expressions tend to 1; and tried to substitute back C=-14/13

y(x)= \frac{4}{1.3}e^{.8t} - \dfrac{14}{13}e^{-.5t}

d'oh! Cheers :) :)

 

[bsodmike]

Thanks Thaakisfox!