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

**Jenkins** · February 11th 2011, 06:06 PM

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!

**adkinsjr** · February 11th 2011, 07:17 PM

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...

**Jenkins** · February 11th 2011, 07:33 PM

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

**adkinsjr** · February 11th 2011, 07:44 PM

Yeah, I was just about to edit that. It doesn't work.

**adkinsjr** · February 11th 2011, 08:23 PM

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...

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

**Jenkins** · February 11th 2011, 09:01 PM

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...

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

you are magical

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