Second Order DE (non homogenous)

**padawan** · August 13th 2010, 03:36 AM

Hello!

I having problems getting started with this problem:

Using subn $t = e^x$ and $z(t) = y(x)$ rewrite:

$\frac{d^2y}{dx^2} - \frac {dy}{dx} + e^{2x}y = xe^2x -1$

hence find all solutions.

When I perform the substitution using $t^2 = e^{2x}$ and $x = lnt$ , I get:

$\frac{d^2z}{dt^2} - \frac{dz}{dt} + t^2z = t^2lnt +1$

I can't find any complimentary functions and probably due to my undergrad engineering major, I tried a Laplace transform, but end up with a horrible function owing to the $t^2z$ part that would leave a $\frac{d^2s}{dt^2}$ on the LHS- I suspect this is not be the correct method as alluded to within the question - there must be a reason for the subn.....

I can't use an order reduction or variation of parameters as I don't have an already known solution- unless I use a trial and error! If anyone can kick me in the right direction, I will be most greatful.... I can't seem to see through it right now!

Thanks for looking!

**Prove It** · August 13th 2010, 03:48 AM

Should this actually be

$\frac{d^2 y}{dx^2} - \frac{dy}{dx} + e^{2x}y = xe^{2x} -1$ ?

If so...

You plan to make the substitutions $t = e^x$ and $z(t) = y(x)$ .

This would mean $\frac{dt}{dx} = e^x = t, t^2 = e^{2x}, x = \ln{(t)}$ .

It would also mean

$\frac{dy}{dx} = \frac{dz}{dx}$

$= \frac{dt}{dx}\frac{dz}{dt}$

$= t\,\frac{dz}{dt}$

and

$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(t\,\frac{dz}{dt}\right)$

$= \frac{dt}{dx}\frac{dz}{dt} + t\,\frac{d}{dx}\left(\frac{dz}{dt}\right)$

$= t\,\frac{dz}{dt} + t\left[\frac{dt}{dx}\,\frac{d}{dt}\left(\frac{dz}{dt}\rig ht)\right]$

$= t\,\frac{dz}{dt} + t\,\frac{dt}{dx}\,\frac{d^2z}{dt^2}$

$= t\,\frac{dz}{dt} + t^2\,\frac{d^2z}{dt^2}$ .

So finally we can substitute into the original DE...

$\frac{d^2 y}{dx^2} - \frac{dy}{dx} + e^{2x}y = xe^{2x} -1$

$t\,\frac{dz}{dt} + t^2\,\frac{d^2z}{dt^2} - t\,\frac{dz}{dt} + t^2z = t^2\ln{(t)} - 1$

$t^2\,\frac{d^2z}{dt^2} + t^2z = t^2\ln{(t)} - 1$

$t^2\left(\frac{d^2z}{dt^2} + z\right) = t^2\ln{(t)} - 1$

$\frac{d^2z}{dt^2} + z= \ln{(t)} - t^{-2}$ .

You should be able to solve this now...

**padawan** · August 13th 2010, 04:13 AM

Yes it is.... I'll change the original, also missed off a'y'..... My latex errors!

**Prove It** · August 13th 2010, 04:16 AM

Oh dear, I just solved the original DE without the extra $y$ , see my edit above...

Well, I think you can see the technique anyway, will edit further...

Edited again... You now have a second order DE that you should be able to solve...

**padawan** · August 13th 2010, 04:22 AM

Thanks Prove it!

No need to edit if you havent done so... leaves me something to do. !

Big thanks for pointing the technique to me!

**padawan** · August 13th 2010, 07:33 AM

I am getting to a difficult integral on the $\int w_1/w$

The aux eqn is $m^2 + 1 = 0$

giving roots of $+i or -i$ and giving compliemntary function:

$z = A cos(t) + B sin(t)$

Evaluating the wronskian determinents:

$w_1 = det \begin{bmatrix} 0 & sin(t) \\ ln(t) - t^{-2} & cos(t) \end{bmatrix}$

$= -sin(t) [ln(t) - t^{-2}]$

$w_2 = det \begin{bmatrix} cos(t) & 0 \\ -sin(t) & ln(t) - t^{-2} \end{bmatrix}$

$= cos(t) [ln(t) - t^{-2}]$

$w = det \begin{bmatrix} cos(t) & sin(t) \\ -sin(t) & cos(t) \end{bmatrix}$

which drops to be $1$

this leaves

$u_1 = \int \frac {w_1}{w} = \int -sin(t) [ln(t) - t^{-2}] dt$

and similarly for $w_2$ with $-sin(t)$ replaced by $cos(t)$

It does not seem to possible to integrate $t^{-2}sin(t)$ .... have I made an error in my working or gone off track somewhere?

**Danny** · August 13th 2010, 12:47 PM

I might add that

$\frac{d^2z}{dt^2} + z= \ln{(t)} - t^{-2}$

has the particular solution $z_p = \ln t$ .

**padawan** · August 13th 2010, 01:04 PM

Cheers danny.... I see that works via subn in and evaluating the DE.... was that by observation or just through some other way?

**Danny** · August 13th 2010, 01:12 PM

Originally Posted by padawan

Cheers danny.... I see that works via subn in and evaluating the DE.... was that by observation or just through some other way?

Actually, when I saw your ODE

$\dfrac{d^2y}{dx^2} - \dfrac {dy}{dx} + e^{2x}y = xe^{2x} -1$

I saw

$\dfrac{d^2y}{dx^2} - \left(\dfrac {dy}{dx}-1\right) + e^{2x}\left(y - x\right) = 0$

and $y = x$ identically satisfies your DE.

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