1. ## 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.

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:

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

Now

$\displaystyle y(x)=x$

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

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

or:

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

Similarly

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

is also a solution so:

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

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

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

So rearranging we find:

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

and:

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

RonL

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

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

Now

$\displaystyle y(x)=x$

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

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

or:

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

Similarly

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

is also a solution so:

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

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

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

So rearranging we find:

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

and:

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

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