Y'' - y' = e^x [2nd order nonhomogenous diff Eq]

  • Thread starter Thread starter JeffNYC
  • Start date Start date
  • Tags Tags
    Diff eq E^x
JeffNYC
Messages
26
Reaction score
0
I have an equation I need to solve by using undetermined coefficients:

y'' - y' = ex

The auxiliary equation is:

r2- r = 0 , so 2 real roots (R1=0, R2 = 1)

So, yc(x) = C1 + C2ex

Now for the particular solution:

I can try Aex but this is already present in the complementary solution. Do I use:

yp(x) = xAex

Is this the right move at this point in the problem?
 
Physics news on Phys.org
Yes, since r=1 is a root of the auxiliary equation, your PI should be Axex
 
You could also note that you have a first order ODE in disguise: set v = y' and you have

v' - v = e^x

If you multiply by e^(-x) you get

e^(-x)v' - e^(-x)v = (e^(-x)v)' = 1

Which means e^(-x)v = x + C => y' = xe^(x) + Ce^(x); integrate once more to get y.
 
Hi there!

there's also another way of finding the particular solution, but it works, iff the inhomogeneous part is of the form e^(ax) where a is any complex number (i.e. it can be also real or pure imaginary)

so, here's the general formula:
y''+py'+qy=be^{\alpha x}

using the differential operator D=\frac{d}{dx}, we obtain:

(D^2+pD+q)y=be^{\alpha x}

considering D^2+pD+q as the characteristic polynomial and rewriting it as p(D):

p(D)y=be^{\alpha x}

so far so good: Now every particular solution has the following form:

y_p=\frac{be^{\alpha x}}{p(\alpha)}

! IF it turns out that p(\alpha)=0, we take:

y_p=\frac{bxe^{\alpha x}}{p'(\alpha)}

! NOW, if also p'(\alpha)=0, we take:

y_p=\frac{bx^2e^{\alpha x}}{p''(\alpha)}
 
In this example:

y''-y'=e^x
using the differential operator D=\frac{d}{dx}, we obtain:
(D^2-D)y=e^x

Now, consider D^2-D as the characteristic polynomial and write p(D) instead:

p(D)y=e^x

we try the particular solution:

y_p=\frac{be^{\alpha x}}{p(\alpha)}=\frac{e^x}{p(1)}

but 1 is a root of the char. polynomial(i.e. p(1)=0), so we take:

y_p=\frac{bxe^{\alpha x}}{p'(\alpha)}=\frac{xe^x}{p'(1)}

p'(1)=2(1)-1=1 and we obtain for our y_p:

y_p=xe^x

*** Note that trig functions sinx and cosx could be represented as imaginary or real part of the complex exponential - then this formula would also provide us with the correct y_p :)

Marin
 
Great answers all - I didn't realize there were several ways to attack this type of problem.

Jeff
 
There is the following linear Volterra equation of the second kind $$ y(x)+\int_{0}^{x} K(x-s) y(s)\,{\rm d}s = 1 $$ with kernel $$ K(x-s) = 1 - 4 \sum_{n=1}^{\infty} \dfrac{1}{\lambda_n^2} e^{-\beta \lambda_n^2 (x-s)} $$ where $y(0)=1$, $\beta>0$ and $\lambda_n$ is the $n$-th positive root of the equation $J_0(x)=0$ (here $n$ is a natural number that numbers these positive roots in the order of increasing their values), $J_0(x)$ is the Bessel function of the first kind of zero order. I...
Are there any good visualization tutorials, written or video, that show graphically how separation of variables works? I particularly have the time-independent Schrodinger Equation in mind. There are hundreds of demonstrations out there which essentially distill to copies of one another. However I am trying to visualize in my mind how this process looks graphically - for example plotting t on one axis and x on the other for f(x,t). I have seen other good visual representations of...
Back
Top