# Identities of Wronskian

#### mathmari

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

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

Since $$\frac{v_{1}}{v_{2}}$$ 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
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
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
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
$d_2=\frac{f'(0)-\frac{f(0)v'_1(0)}{v_1(0)}}{v'_2(0)-\frac{v_2(0)v'_1(0)}{v_1(0)}}$
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
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
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

##### Well-known member
MHB Site Helper
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
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
These two relations are not given.
However, we have just proven that they hold.
A ok. 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
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 $B=d_2$?