Using the Wronskian to Solve Differential Equations with Non-Constant Solutions

[mathmari]

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)??

 

[I like Serena]

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]

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??

 

[I like Serena]

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]

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?

 

[I like Serena]

$d_2=\frac{f'(0)-\frac{f(0)v'_1(0)}{v_1(0)}}{v'_2(0)-\frac{v_2(0)v'_1(0)}{v_1(0)}}$

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]

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?

 

[I like Serena]

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?

 

[mathmari]

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?

 

[I like Serena]

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]

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]

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 $B=d_2$?