Finding the Second Order Linear Homogeneous Equation from the given Fundamental Pair

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• Feb 11th 2011, 07:06 PM
Jenkins
Finding the Second Order Linear Homogeneous Equation from the given Fundamental Pair
The problem instructions read like so:

Show that the given functions are linearly independent on the interval I and ﬁnd
a second-order linear homogeneous equation having the pair as a fundamental set of
solutions.

$y_1=x$
$y_2=x^2$
$I=(\infty,-\infty)$

I'm able to prove that they are linearly independent easily by simply using the Wronskian Matrix to get

$W(x)=3x^3\neq0\in I$

However, I don't know how to find the equation based off of what I'm given.
I know the answer to the question (it's in the book) so I will post it for reference of what I'm looking for.

$x^2 y\prime\prime - 2x y\prime +2y = 0$

please show me the method to find this equation based off of the fundamental set given!
• Feb 11th 2011, 08:17 PM
adkinsjr
Quote:

Originally Posted by Jenkins
The problem instructions read like so:

Show that the given functions are linearly independent on the interval I and ﬁnd
a second-order linear homogeneous equation having the pair as a fundamental set of
solutions.

$y_1=x$
$y_2=x^2$
$I=(\infty,-\infty)$

I'm able to prove that they are linearly independent easily by simply using the Wronskian Matrix to get

$W(x)=3x^3\neq0\in I$

However, I don't know how to find the equation based off of what I'm given.
I know the answer to the question (it's in the book) so I will post it for reference of what I'm looking for.

$x^2 y\prime\prime - 2x y\prime +2y = 0$

please show me the method to find this equation based off of the fundamental set given!

You should be able to set up three equations with three unknowns:

$a_2(x)y''_1+a_1(x)y'_1+a_o(x)y_1=0$

$a_2(x)y''_2+a_1(x)y'_2+a_o(x)y_2=0$

$a_2(x)(y_1+y_2)''+a_1(x)(y_1+y_2)'+a_o(x)(y_1+y_2) =0$

Try that...
• Feb 11th 2011, 08:33 PM
Jenkins
Quote:

Originally Posted by adkinsjr
You should be able to set up three equations with three unknowns:

$a_2(x)y''_1+a_1(x)y'_1+a_o(x)y_1=0$

$a_2(x)y''_2+a_1(x)y'_2+a_o(x)y_2=0$

$a_2(x)(y_1+y_2)''+a_1(x)(y_1+y_2)'+a_o(x)(y_1+y_2) =0$

Try that...

I tried it and got that
$a_2=0$
$a_1=0$
$a_o=0$

which makes sense considering equations 1 and 2 both equal 0 and equation 3 is a linear combination of the two.

Thanks for your help though (Happy)
• Feb 11th 2011, 08:44 PM
adkinsjr
Yeah, I was just about to edit that. It doesn't work.
• Feb 11th 2011, 09:23 PM
adkinsjr
Ok, I think I know what to do now, lol. I was on the right track... Just define $P(x)=\frac{a_1(x)}{a_2(x)}$ and $Q(x)=\frac{a_o(x)}{a_2(x)}$, basically divide the first two equations (from my failed system) by the coefficient of y'' to get:

$y_1''+P(x)y'_1+Q(x)y_1=0$

$y_2''+P(x)y'_2+Q(x)y_2=0$

From the first:

$P(x)=-Q(x)x$

Now substitute into the second:

$2-Q(x)2x^2+Q(x)x^2=0$

$Q(x)=\frac{2}{x^2}$

That should help... (Giggle)

Remember we defined $Q(x)=\frac{a_o(x)}{a_2(x)}$.
• Feb 11th 2011, 10:01 PM
Jenkins
Quote:

Originally Posted by adkinsjr
Ok, I think I know what to do now, lol. I was on the right track... Just define $P(x)=\frac{a_1(x)}{a_2(x)}$ and $Q(x)=\frac{a_o(x)}{a_2(x)}$, basically divide the first two equations (from my failed system) by the coefficient of y'' to get:

$y_1''+P(x)y'_1+Q(x)y_1=0$

$y_2''+P(x)y'_2+Q(x)y_2=0$

From the first:

$P(x)=-Q(x)x$

Now substitute into the second:

$2-Q(x)2x^2+Q(x)x^2=0$

$Q(x)=\frac{2}{x^2}$

That should help... (Giggle)

Remember we defined $Q(x)=\frac{a_o(x)}{a_2(x)}$.

you are magical (Party)