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Identities of Wronskian

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,047
Hey! :)
I need some help at the following exercise:
Let [tex]v_{1}[/tex], [tex]v_{2}[/tex] solutions of the differential equation [tex]y''+ay'+by=0[/tex] (where a and b real constants)so that [tex]\frac{v_{1}}{v_{2}}[/tex] is not constant.If [tex]y=f(x)[/tex] any solution of the differential equation ,use the identities of the Wronskian to show that there are constants [tex]d_{1}[/tex], [tex]d_{2}[/tex] so that:[tex]d_{1}v_{1}(0)+d_{2}v_{2}(0)=f(0)[/tex] (1), [tex]d_{1}v_{1}'(0) +d_{2}v_{2}'(0)=f'(0)[/tex] (2) and that each solution of the differential equation has the form:[tex]y=d_{1}v_{1}(x)+d_{2}v_{2}(x)[/tex].

Since [tex]\frac{v_{1}}{v_{2}}[/tex] is not constant, the Wronskian is not equal to zero, right? But how can I continue to show the relations (1) and (2)??
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
Hey! :)
I need some help at the following exercise:
Let [tex]v_{1}[/tex], [tex]v_{2}[/tex] solutions of the differential equation [tex]y''+ay'+by=0[/tex] (where a and b real constants)so that [tex]\frac{v_{1}}{v_{2}}[/tex] is not constant.If [tex]y=f(x)[/tex] any solution of the differential equation ,use the identities of the Wronskian to show that there are constants [tex]d_{1}[/tex], [tex]d_{2}[/tex] so that:[tex]d_{1}v_{1}(0)+d_{2}v_{2}(0)=f(0)[/tex] (1), [tex]d_{1}v_{1}'(0) +d_{2}v_{2}'(0)=f'(0)[/tex] (2) and that each solution of the differential equation has the form:[tex]y=d_{1}v_{1}(x)+d_{2}v_{2}(x)[/tex].

Since [tex]\frac{v_{1}}{v_{2}}[/tex] is not constant, the Wronskian is not equal to zero, right? But how can I continue to show the relations (1) and (2)??
Hi! :)

What do you get if you solve equations (1) and (2) for $d_1$ and $d_2$?
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,047
Hi! :)

What do you get if you solve equations (1) and (2) for $d_1$ and $d_2$?
Solving for $d_2$ I get, $d_2=\frac{f'(0)-\frac{f(0)v'_1(0)}{v_1(0)}}{v'_2(0)-\frac{v_2(0)v'_1(0)}{v_1(0)}}$.

And $d_1=\frac{f(0)-d_2v_2(0)}{v_1(0)}$, where $d_2$ is the above relation.

But am I not asked to show the equations (1) and (2) by using the identities of Wronskian??
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
Solving for $d_2$ I get, $d_2=\frac{f'(0)-\frac{f(0)v'_1(0)}{v_1(0)}}{v'_2(0)-\frac{v_2(0)v'_1(0)}{v_1(0)}}$.

And $d_1=\frac{f(0)-d_2v_2(0)}{v_1(0)}$, where $d_2$ is the above relation.

But am I not asked to show the equations (1) and (2) by using the identities of Wronskian??
Just a sec...
Suppose you multiply both the numerator and the denominator by $v_1(0)$, what do you get?
In other words, exactly when is the system of those 2 equations solvable?
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,047
Just a sec...
Suppose you multiply both the numerator and the denominator by $v_1(0)$, what do you get?
In other words, exactly when is the system of those 2 equations solvable?
$d_2=\frac{f'(0)-\frac{f(0)v'_1(0)}{v_1(0)}}{v'_2(0)-\frac{v_2(0)v'_1(0)}{v_1(0)}}$

$v_1(0)v'_2(0)-v_2(0)v'_1(0)!=0 $ ,that we have from the Wronskian.right?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
$d_2=\frac{f'(0)-\frac{f(0)v'_1(0)}{v_1(0)}}{v'_2(0)-\frac{v_2(0)v'_1(0)}{v_1(0)}}$
That should be:
$$d_2=\frac{f'(0)v_1(0)-f(0)v'_1(0)}{v_1(0)v'_2(0)-v'_1(0)v_2(0)}
= \frac{f'(0)v_1(0)-f(0)v'_1(0)}{W(v_1, v_2)(0)}$$


$v_1(0)v'_2(0)-v_2(0)v'_1(0)!=0 $,that we have from the Wronskian.right?
Yes.
One of the identities of the Wronskian is, that if the two functions are independent (their quotient is well-defined and is not constant), then the Wronskian is different from zero.
Therefore the equation has a guaranteed solution! ;)
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,047
That should be:
$$d_2=\frac{f'(0)v_1(0)-f(0)v'_1(0)}{v_1(0)v'_2(0)-v'_1(0)v_2(0)}
= \frac{f'(0)v_1(0)-f(0)v'_1(0)}{W(v_1, v_2)(0)}$$




Yes.
One of the identities of the Wronskian is, that if the two functions are independent (their quotient is well-defined and is not constant), then the Wronskian is different from zero.
Therefore the equation has a guaranteed solution! ;)
And what does this mean? Have we shown now that there are $d_1$ and $d_2$ that satisfy the two relations? Do these two constants exist because the denominator is different from zero?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
And what does this mean? Have we shown now that there are $d_1$ and $d_2$ that satisfy the two relations? Do these two constants exist because the denominator is different from zero?
Together with the solution for $d_1$, which is similar, we have shown that there are constants $d_1$, $d_2$ so that: $d_1v_1(0)+d_2v_2(0)=f(0)$ (1), $d_1v′_1(0)+d_2v′_2(0)=f′(0)$ (2).

Indeed, these solutions only exist because the denominator for both $d_1$ and $d_2$ are different from zero. Otherwise they would be undefined. That is, a solution would not exist.

Btw, how do you feel about picking an avatar picture? :eek:
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,047
Together with the solution for $d_1$, which is similar, we have shown that there are constants $d_1$, $d_2$ so that: $d_1v_1(0)+d_2v_2(0)=f(0)$ (1), $d_1v′_1(0)+d_2v′_2(0)=f′(0)$ (2).

Indeed, these solutions only exist because the denominator for both $d_1$ and $d_2$ is different from zero. Otherwise they would be undefined. That is, a solution would not exist.
These two relations are given, or do I have to prove them?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
These two relations are given, or do I have to prove them?
These two relations are not given.
However, we have just proven that they hold.
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,047
These two relations are not given.
However, we have just proven that they hold.
A ok. :D

How can I show that the general solution is $y=d_1v_1(x)+d_2v_2(x)$?
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,047
Since $v_1$ and $v_2$ are solutions of $y$, the general solution is $y=Av_1+Bv_2$. But how could we show that $A=d_1$ and $B=d_2$?