2nd order ODE

• Feb 14th 2006, 09:46 PM
2nd order ODE
Hi ,
I am stuck with the following problem (to turn tomorrow):

Find a second order linear homogeneous equation having the pair as a fundamental set of solutions:
y1(x)=x , y2(x)=x*ln(x).

My problem here is that I dont have the exponential form for the proposed solutions.

B.
• Feb 15th 2006, 05:31 AM
CaptainBlack
Quote:

Hi ,
I am stuck with the following problem (to turn tomorrow):

Find a second order linear homogeneous equation having the pair as a fundamental set of solutions:
y1(x)=x , y2(x)=x*ln(x).

My problem here is that I dont have the exponential form for the proposed solutions.

B.

A second order linear homogeneous ODE has the form:

$
\ddot y + f(x) \dot y+g(x) y=0
$

Now

$
y(x)=x
$

is a solution so differentiating twice and substituting the derivatives of $y$
into the ODE gives:

$
f(x)+xg(x)=0
$

or:

$
g(x)=-\frac{f(x)}{x}\ \ \ ...(1)
$

Similarly

$
y(x)=x \ln(x)
$

is also a solution so:

$
\frac{1}{x}+(\ln (x)+1)f(x)+x \ln (x)g(x)=0
$

Now substituting $g(x)$ from $(1)$ into this last
equation gives:

$
\frac{1}{x}+(\ln (x)+1)f(x)- \ln (x)f(x)=0
$

So rearranging we find:

$
f(x)=-\frac{1}{x}
$

and:

$
g(x)=\frac{1}{x^2}
$

RonL
• Feb 15th 2006, 05:50 AM
Quote:

Originally Posted by CaptainBlack
A second order linear homogeneous ODE has the form:

$
\ddot y + f(x) \dot y+g(x) y=0
$

Now

$
y(x)=x
$

is a solution so differentiating twice and substituting the derivatives of $y$
into the ODE gives:

$
f(x)+xg(x)=0
$

or:

$
g(x)=-\frac{f(x)}{x}\ \ \ ...(1)
$

Similarly

$
y(x)=x \ln(x)
$

is also a solution so:

$
\frac{1}{x}+(\ln (x)+1)f(x)+x \ln (x)g(x)=0
$

Now substituting $g(x)$ from $(1)$ into this last
equation gives:

$
\frac{1}{x}+(\ln (x)+1)f(x)- \ln (x)f(x)=0
$

So rearranging we find:

$
f(x)=-\frac{1}{x}
$

and:

$
g(x)=\frac{1}{x^2}
$

RonL

Thank you very much Captain,
Because we use a different form in class, the exponential form for the solution (y=e(rx)), I was a little bit confused by this problem . Now I get it!

B