Use Wronskian method in solving the given second order differential equation

[chwala]

Homework Statement

see attached-

Solve $y{''} +11y{'} +24 y=0$

$y(0)=0, y^{'}(0)=-7$

Relevant Equations

Wronskian method

I am looking at this link;


https://tutorial.math.lamar.edu/Classes/DE/RealRoots.aspx

$y{''} +11y{'} +24 y=0, y(0)=0, y^{'}(0)=-7$

Now the general approach of applying boundary conditions directly is quite straightforward to me. I am interested in using an alternative approach, that is the Wronskian method, to determine the coefficients and solve the problem for a particular solution.

The question then is whether the Wronskian is a powerful tool for use in both homogeneous and inhomogeneous second-order differential equations. Certainly, the solutions must be independent, as a requirement.


Now, Using Wronskian method,... gives me the following; Note that the matrices shown below are just but a combination of the solutions and the boundary conditions. I substituted that directly...

$c_1 = \begin{pmatrix} 0 & 1 \\ -7 & -3 \\ \end{pmatrix}\div \begin{pmatrix} 1 & 1 \\ -8 & -3 \\ \end{pmatrix}=\dfrac{7}{-3--8}=\dfrac{7}{35}$

and

$c_2 = \begin{pmatrix} 1 & 0 \\ -8 & -7 \\ \end{pmatrix}\div \begin{pmatrix} 1 & 1 \\ -8 & -3 \\ \end{pmatrix}=\dfrac{-7}{-3--8}=-\dfrac{7}{35}$

having found the constants, then the particular solution is realized immediately.


insight is welcome on highlighted part.

 

[pasmith]

I'm not sure how you divide two matrices to get a number. I'm also not sure how <br /> \frac{7}{-3 --8} = \frac{7}{35}

 

[chwala]

I'm not sure how you divide two matrices to get a number. I'm also not sure how <br /> \frac{7}{-3 --8} = \frac{7}{35}

sorry typo;

it is supposed to be:

$c_1 = \begin{pmatrix} 0 & 1 \\ -7 & -3 \\ \end{pmatrix}\div \begin{pmatrix} 1 & 1 \\ -8 & -3 \\ \end{pmatrix}=\dfrac{7}{-3--8}=\dfrac{7}{5}$

and

$c_2 = \begin{pmatrix} 1 & 0 \\ -8 & -7 \\ \end{pmatrix}\div \begin{pmatrix} 1 & 1 \\ -8 & -3 \\ \end{pmatrix}=\dfrac{-7}{-3--8}=-\dfrac{7}{5}$

having found the constants, then the particular solution is realized immediately.

 

[chwala]

$c_1= \dfrac{ \begin{pmatrix} 0 & 1 \\ -7 & -3 \\ \end{pmatrix} }{ \begin{pmatrix} 1 & 1 \\ -8 & -3 \\ \end{pmatrix} }=\dfrac{7}{5}$

and

$c_2=\dfrac{ \begin{pmatrix} 1 & 0 \\ -8 & -7 \\ \end{pmatrix} }{ \begin{pmatrix} 1 & 1 \\ -8 & -3 \\ \end{pmatrix} }=\dfrac{-7}{5}$

using Wronskian. This should be clear now.

 

[Gavran]

The question then is whether the Wronskian is a powerful tool for use in both homogeneous and inhomogeneous second-order differential equations. Certainly, the solutions must be independent, as a requirement.

In general, the Wronskian is a powerful tool to determine linear independence or dependence of a set of functions at a point or over an interval.

 

[chwala]

In general, the Wronskian is a powerful tool to determine linear independence or dependence of a set of functions at a point or over an interval.

I agree and fully understand that.

It can also be used to determine the constants for differential equations, as I mentioned. The independence or dependence is determined by the determinant matrix of the solutions. If the determinant of the solutions is equal to 0, then this may imply that the solutions are dependent.

My question is whether using the Wronskian to determine the constants is as powerful as other conventional methods, or if I should explore this with other types of second order ODEs.

 

[Gavran]

My question is whether using the Wronskian to determine the constants is as powerful as other conventional methods, or if I should explore this with other types of second order ODEs.

You have already answered the question in the original post. The Wronskian can be the second choice but not the first choice except in the case of inhomogeneous differential equations which can not be solved by using the method of undetermined coefficients. See https://en.wikipedia.org/wiki/Variation_of_parameters.

 

[Mark44]

$c_1= \dfrac{ \begin{pmatrix} 0 & 1 \\ -7 & -3 \\ \end{pmatrix} }{ \begin{pmatrix} 1 & 1 \\ -8 & -3 \\ \end{pmatrix} }=\dfrac{7}{5}$

and

$c_2=\dfrac{ \begin{pmatrix} 1 & 0 \\ -8 & -7 \\ \end{pmatrix} }{ \begin{pmatrix} 1 & 1 \\ -8 & -3 \\ \end{pmatrix} }=\dfrac{-7}{5}$

using Wronskian. This should be clear now.

I believe you have confused some who responded in the thread who thought that the above were matrices. Writing the LaTeX like so emphasizes that they are determinants.
$c_1= \dfrac{ \left|\begin{matrix} 0 & 1 \\ -7 & -3 \\ \end{matrix} \right|}{ \left|\begin{matrix} 1 & 1 \\ -8 & -3 \\ \end{matrix}\right| }=\dfrac{7}{5}$
And similar for $c_2$. You can click on my script to see what I did.