How exactly is the Wronskian tied with Cramer's rule in a system of ODEs? The Next CEO of...

Do I need to write [sic] when including a quotation with a number less than 10 that isn't written out?

What does "shotgun unity" refer to here in this sentence?

Inductor and Capacitor in Parallel

Why don't programming languages automatically manage the synchronous/asynchronous problem?

(How) Could a medieval fantasy world survive a magic-induced "nuclear winter"?

Does the Idaho Potato Commission associate potato skins with healthy eating?

Can Sneak Attack be used when hitting with an improvised weapon?

Why is the US ranked as #45 in Press Freedom ratings, despite its extremely permissive free speech laws?

Won the lottery - how do I keep the money?

Man transported from Alternate World into ours by a Neutrino Detector

Is a distribution that is normal, but highly skewed, considered Gaussian?

IC has pull-down resistors on SMBus lines?

Which Pokemon have a special animation when running with them out of their pokeball?

How do you define an element with an ID attribute using LWC?

Strange use of "whether ... than ..." in official text

Small nick on power cord from an electric alarm clock, and copper wiring exposed but intact

Purpose of level-shifter with same in and out voltages

Airplane gently rocking its wings during whole flight

Vector calculus integration identity problem

Is there such a thing as a proper verb, like a proper noun?

In the "Harry Potter and the Order of the Phoenix" video game, what potion is used to sabotage Umbridge's speakers?

What is the process for purifying your home if you believe it may have been previously used for pagan worship?

Help/tips for a first time writer?

Why am I getting "Static method cannot be referenced from a non static context: String String.valueOf(Object)"?



How exactly is the Wronskian tied with Cramer's rule in a system of ODEs?



The Next CEO of Stack OverflowLinear Differential SystemWronskian determinant for $y_2(t)=begin{pmatrix} e^t\ e^t end{pmatrix}$ and $y_1(t)=begin{pmatrix} t^2\ 2t end{pmatrix}$.How can you calculate the derivative of this Wronskian?Prove that the determinant of a given matrix is proportional to the area of the triangle whose corners are the three points.Linear transformations using matricesIs the Wronskian determinant positive or negative?Solving System of Linear Equations with LU Decomposition of $4 times 3$ matrixDetermine independence of functions of r.v.'s and compute joint distributionHow to find the general solution of a system of ODEs?Linear independence of $1, x, x^2$Solving a simple system of equations through matrix operations












1












$begingroup$


If we have a nonhomogeneous system of linear differential equations:
$$frac{dvec{y}}{dx}=Avec{y}+vec{b}$$
where:
$$frac{dvec{y}}{dx}=begin{bmatrix}y_1'\y_2'\y_3'end{bmatrix},vec{y}=begin{bmatrix}y_1\y_2\y_3end{bmatrix},A=begin{bmatrix}a_{11}&a_{12}&a_{13}\a_{21}&a_{22}&a_{23}\a_{31}&a_{32}&a_{33}end{bmatrix}, vec{b}=begin{bmatrix}b_1(x)\b_2(x)\b_3(x)end{bmatrix}$$
the particular solution involving these three functions could be written:
$$y_{particular}=c_1(x)e^{lambda_1x}vec{u_1}+c_2(x)e^{lambda_2x}vec{u_2}+c_3(x)e^{lambda_3x}vec{u_3}$$
where $lambda_i, vec{u_i}$ are the eigenvalues and eigenvectors of the constant matrix $A$ (I chose it constant to make things simpler.)



When trying to evaluate these $c_i(x)$ functions, I was told to evaluate them by:
$$c_i(x)=intfrac{W_i}{W}dx$$
Where $W$ is the Wronskian determinant whose columns are the elements of the vectors $c_i(x)e^{lambda_3i}vec{u_i}$, and $W_i$ is the Wronskian determinant whose $i$th column was replaced by the contents of the $vec{b}$ vector.



My questions are: How does the integral show up? Why does the process look like Cramer's rule? How did we lead up to that result? Where did this 'Wronskian' determinant come from?



If I have missed something, please point it out and I'll add it to the post.



Thanks in advance.










share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    If we have a nonhomogeneous system of linear differential equations:
    $$frac{dvec{y}}{dx}=Avec{y}+vec{b}$$
    where:
    $$frac{dvec{y}}{dx}=begin{bmatrix}y_1'\y_2'\y_3'end{bmatrix},vec{y}=begin{bmatrix}y_1\y_2\y_3end{bmatrix},A=begin{bmatrix}a_{11}&a_{12}&a_{13}\a_{21}&a_{22}&a_{23}\a_{31}&a_{32}&a_{33}end{bmatrix}, vec{b}=begin{bmatrix}b_1(x)\b_2(x)\b_3(x)end{bmatrix}$$
    the particular solution involving these three functions could be written:
    $$y_{particular}=c_1(x)e^{lambda_1x}vec{u_1}+c_2(x)e^{lambda_2x}vec{u_2}+c_3(x)e^{lambda_3x}vec{u_3}$$
    where $lambda_i, vec{u_i}$ are the eigenvalues and eigenvectors of the constant matrix $A$ (I chose it constant to make things simpler.)



    When trying to evaluate these $c_i(x)$ functions, I was told to evaluate them by:
    $$c_i(x)=intfrac{W_i}{W}dx$$
    Where $W$ is the Wronskian determinant whose columns are the elements of the vectors $c_i(x)e^{lambda_3i}vec{u_i}$, and $W_i$ is the Wronskian determinant whose $i$th column was replaced by the contents of the $vec{b}$ vector.



    My questions are: How does the integral show up? Why does the process look like Cramer's rule? How did we lead up to that result? Where did this 'Wronskian' determinant come from?



    If I have missed something, please point it out and I'll add it to the post.



    Thanks in advance.










    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      If we have a nonhomogeneous system of linear differential equations:
      $$frac{dvec{y}}{dx}=Avec{y}+vec{b}$$
      where:
      $$frac{dvec{y}}{dx}=begin{bmatrix}y_1'\y_2'\y_3'end{bmatrix},vec{y}=begin{bmatrix}y_1\y_2\y_3end{bmatrix},A=begin{bmatrix}a_{11}&a_{12}&a_{13}\a_{21}&a_{22}&a_{23}\a_{31}&a_{32}&a_{33}end{bmatrix}, vec{b}=begin{bmatrix}b_1(x)\b_2(x)\b_3(x)end{bmatrix}$$
      the particular solution involving these three functions could be written:
      $$y_{particular}=c_1(x)e^{lambda_1x}vec{u_1}+c_2(x)e^{lambda_2x}vec{u_2}+c_3(x)e^{lambda_3x}vec{u_3}$$
      where $lambda_i, vec{u_i}$ are the eigenvalues and eigenvectors of the constant matrix $A$ (I chose it constant to make things simpler.)



      When trying to evaluate these $c_i(x)$ functions, I was told to evaluate them by:
      $$c_i(x)=intfrac{W_i}{W}dx$$
      Where $W$ is the Wronskian determinant whose columns are the elements of the vectors $c_i(x)e^{lambda_3i}vec{u_i}$, and $W_i$ is the Wronskian determinant whose $i$th column was replaced by the contents of the $vec{b}$ vector.



      My questions are: How does the integral show up? Why does the process look like Cramer's rule? How did we lead up to that result? Where did this 'Wronskian' determinant come from?



      If I have missed something, please point it out and I'll add it to the post.



      Thanks in advance.










      share|cite|improve this question











      $endgroup$




      If we have a nonhomogeneous system of linear differential equations:
      $$frac{dvec{y}}{dx}=Avec{y}+vec{b}$$
      where:
      $$frac{dvec{y}}{dx}=begin{bmatrix}y_1'\y_2'\y_3'end{bmatrix},vec{y}=begin{bmatrix}y_1\y_2\y_3end{bmatrix},A=begin{bmatrix}a_{11}&a_{12}&a_{13}\a_{21}&a_{22}&a_{23}\a_{31}&a_{32}&a_{33}end{bmatrix}, vec{b}=begin{bmatrix}b_1(x)\b_2(x)\b_3(x)end{bmatrix}$$
      the particular solution involving these three functions could be written:
      $$y_{particular}=c_1(x)e^{lambda_1x}vec{u_1}+c_2(x)e^{lambda_2x}vec{u_2}+c_3(x)e^{lambda_3x}vec{u_3}$$
      where $lambda_i, vec{u_i}$ are the eigenvalues and eigenvectors of the constant matrix $A$ (I chose it constant to make things simpler.)



      When trying to evaluate these $c_i(x)$ functions, I was told to evaluate them by:
      $$c_i(x)=intfrac{W_i}{W}dx$$
      Where $W$ is the Wronskian determinant whose columns are the elements of the vectors $c_i(x)e^{lambda_3i}vec{u_i}$, and $W_i$ is the Wronskian determinant whose $i$th column was replaced by the contents of the $vec{b}$ vector.



      My questions are: How does the integral show up? Why does the process look like Cramer's rule? How did we lead up to that result? Where did this 'Wronskian' determinant come from?



      If I have missed something, please point it out and I'll add it to the post.



      Thanks in advance.







      linear-algebra ordinary-differential-equations systems-of-equations






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 17 at 17:13









      J. W. Tanner

      4,1961320




      4,1961320










      asked Mar 17 at 17:03









      Gradient EntropyGradient Entropy

      204




      204






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          Write the general solution of the homogeneous linear ODE
          $$
          tag{1}
          frac{mathrm{d}vec{y}}{mathrm{d}x} = A(x) vec{y}
          $$

          in the matrix form as
          $$
          Phi(x; x_0) vec{c},
          $$

          where $vec{c} = mathrm{col}(c_1, dots, c_n)$ is a constant matrix (that is, independent of $x$), and $Phi$ is the state-transition matrix such that $Phi(x_0, x_0) = I$ (the identity matrix).



          Now we are looking for a general solution of the nonhomogeneous linear ODE
          $$
          tag{2}
          frac{mathrm{d}vec{y}}{mathrm{d}x} = A(x) vec{y} + vec{b}(x)
          $$

          in the form
          $$
          Phi(x, x_0) vec{c}(x).
          $$

          Incidentally, here is the source of the term "variation of constants". We look for conditions for the above function of $x$ to be a solution of $(2)$, that is, to have
          $$
          frac{mathrm{d}}{mathrm{d}x}(Phi(x, x_0) vec{c}(x)) = A(x) (Phi(x, x_0) vec{c}(x)) + vec{b}(x).
          $$

          But
          $$
          frac{mathrm{d}}{mathrm{d}x}(Phi(x, x_0) vec{c}(x)) = frac{mathrm{d}}{mathrm{d}x} Phi(x, x_0) cdot vec{c}(x) + Phi(x, x_0) cdot vec{c}'(x)
          $$

          and
          $$
          frac{mathrm{d}}{mathrm{d}x} Phi(x, x_0) = A(x) Phi(x, x_0),
          $$

          so after cancellation we obtain
          $$
          Phi(x, x_0) cdot vec{c}'(x) = vec{b}(x),
          $$

          which is, for each fixed $x$, a Cramer (because $det{Phi(x, x_0)} = W(x, x_0) ne 0$) system of $n$ linear equations with $n$ unknowns, $c'_1(x), dots, c'_n(x)$. The unknowns are given by the Cramer rule. We have thus obtained the derivative of the matrix function $vec{c}(x)$, so we have to integrate that derivative. And that is the source of the integral.



          The "Wrońskian" determinant was named after a nineteenth-century Polish philosopher and mathematician, Józef Maria Hoene-Wroński.



          EDIT: I changed slightly the terminology and provided a reference to a Wikipedia entry.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Thank you for your answer, but I am having difficulty in understanding what this ' fundamental matrix ' is, as well as its notation. What is that semicolon's purpose?
            $endgroup$
            – Gradient Entropy
            Mar 17 at 21:53










          • $begingroup$
            I replaced the semicolons with commas, to keep the notation in line with the Wikipedia entry.
            $endgroup$
            – user539887
            Mar 17 at 22:00










          • $begingroup$
            One last question. I'm quite unfamiliar with control theory, and I don't know how this state transition matrix is composed of. I assume it's a 3x3 matrix (according to my example) and it's invertible, because the wronskian determinant is non zero. This means we can write: $vec{c}'=Phi^{-1}(x,x_0)vec{b}$, then integrate to find the vector c. Is this correct? If so, how is that done? That is why I'm asking about the elements of the state transition matrix.
            $endgroup$
            – Gradient Entropy
            Mar 17 at 23:01












          • $begingroup$
            It is only the name "state-transition matrix" that is used in Control Theory. It is one of the most fundamental concepts in the theory of linear systems of ODEs: it is the fundamental matrix that takes value $I$ at $x_0$. In other words, if $Psi(x)$ is any fundamental matrix of $(1)$ then the state-transition matrix of $(1)$ equals $Phi(x,x_0)=Psi(x)Psi^{-1}(x_0)$. Another name (when $x_0$ is fixed): principal fundamental matrix, see, e.g., Linear Differential System.
            $endgroup$
            – user539887
            Mar 18 at 7:18












          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3151773%2fhow-exactly-is-the-wronskian-tied-with-cramers-rule-in-a-system-of-odes%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          Write the general solution of the homogeneous linear ODE
          $$
          tag{1}
          frac{mathrm{d}vec{y}}{mathrm{d}x} = A(x) vec{y}
          $$

          in the matrix form as
          $$
          Phi(x; x_0) vec{c},
          $$

          where $vec{c} = mathrm{col}(c_1, dots, c_n)$ is a constant matrix (that is, independent of $x$), and $Phi$ is the state-transition matrix such that $Phi(x_0, x_0) = I$ (the identity matrix).



          Now we are looking for a general solution of the nonhomogeneous linear ODE
          $$
          tag{2}
          frac{mathrm{d}vec{y}}{mathrm{d}x} = A(x) vec{y} + vec{b}(x)
          $$

          in the form
          $$
          Phi(x, x_0) vec{c}(x).
          $$

          Incidentally, here is the source of the term "variation of constants". We look for conditions for the above function of $x$ to be a solution of $(2)$, that is, to have
          $$
          frac{mathrm{d}}{mathrm{d}x}(Phi(x, x_0) vec{c}(x)) = A(x) (Phi(x, x_0) vec{c}(x)) + vec{b}(x).
          $$

          But
          $$
          frac{mathrm{d}}{mathrm{d}x}(Phi(x, x_0) vec{c}(x)) = frac{mathrm{d}}{mathrm{d}x} Phi(x, x_0) cdot vec{c}(x) + Phi(x, x_0) cdot vec{c}'(x)
          $$

          and
          $$
          frac{mathrm{d}}{mathrm{d}x} Phi(x, x_0) = A(x) Phi(x, x_0),
          $$

          so after cancellation we obtain
          $$
          Phi(x, x_0) cdot vec{c}'(x) = vec{b}(x),
          $$

          which is, for each fixed $x$, a Cramer (because $det{Phi(x, x_0)} = W(x, x_0) ne 0$) system of $n$ linear equations with $n$ unknowns, $c'_1(x), dots, c'_n(x)$. The unknowns are given by the Cramer rule. We have thus obtained the derivative of the matrix function $vec{c}(x)$, so we have to integrate that derivative. And that is the source of the integral.



          The "Wrońskian" determinant was named after a nineteenth-century Polish philosopher and mathematician, Józef Maria Hoene-Wroński.



          EDIT: I changed slightly the terminology and provided a reference to a Wikipedia entry.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Thank you for your answer, but I am having difficulty in understanding what this ' fundamental matrix ' is, as well as its notation. What is that semicolon's purpose?
            $endgroup$
            – Gradient Entropy
            Mar 17 at 21:53










          • $begingroup$
            I replaced the semicolons with commas, to keep the notation in line with the Wikipedia entry.
            $endgroup$
            – user539887
            Mar 17 at 22:00










          • $begingroup$
            One last question. I'm quite unfamiliar with control theory, and I don't know how this state transition matrix is composed of. I assume it's a 3x3 matrix (according to my example) and it's invertible, because the wronskian determinant is non zero. This means we can write: $vec{c}'=Phi^{-1}(x,x_0)vec{b}$, then integrate to find the vector c. Is this correct? If so, how is that done? That is why I'm asking about the elements of the state transition matrix.
            $endgroup$
            – Gradient Entropy
            Mar 17 at 23:01












          • $begingroup$
            It is only the name "state-transition matrix" that is used in Control Theory. It is one of the most fundamental concepts in the theory of linear systems of ODEs: it is the fundamental matrix that takes value $I$ at $x_0$. In other words, if $Psi(x)$ is any fundamental matrix of $(1)$ then the state-transition matrix of $(1)$ equals $Phi(x,x_0)=Psi(x)Psi^{-1}(x_0)$. Another name (when $x_0$ is fixed): principal fundamental matrix, see, e.g., Linear Differential System.
            $endgroup$
            – user539887
            Mar 18 at 7:18
















          1












          $begingroup$

          Write the general solution of the homogeneous linear ODE
          $$
          tag{1}
          frac{mathrm{d}vec{y}}{mathrm{d}x} = A(x) vec{y}
          $$

          in the matrix form as
          $$
          Phi(x; x_0) vec{c},
          $$

          where $vec{c} = mathrm{col}(c_1, dots, c_n)$ is a constant matrix (that is, independent of $x$), and $Phi$ is the state-transition matrix such that $Phi(x_0, x_0) = I$ (the identity matrix).



          Now we are looking for a general solution of the nonhomogeneous linear ODE
          $$
          tag{2}
          frac{mathrm{d}vec{y}}{mathrm{d}x} = A(x) vec{y} + vec{b}(x)
          $$

          in the form
          $$
          Phi(x, x_0) vec{c}(x).
          $$

          Incidentally, here is the source of the term "variation of constants". We look for conditions for the above function of $x$ to be a solution of $(2)$, that is, to have
          $$
          frac{mathrm{d}}{mathrm{d}x}(Phi(x, x_0) vec{c}(x)) = A(x) (Phi(x, x_0) vec{c}(x)) + vec{b}(x).
          $$

          But
          $$
          frac{mathrm{d}}{mathrm{d}x}(Phi(x, x_0) vec{c}(x)) = frac{mathrm{d}}{mathrm{d}x} Phi(x, x_0) cdot vec{c}(x) + Phi(x, x_0) cdot vec{c}'(x)
          $$

          and
          $$
          frac{mathrm{d}}{mathrm{d}x} Phi(x, x_0) = A(x) Phi(x, x_0),
          $$

          so after cancellation we obtain
          $$
          Phi(x, x_0) cdot vec{c}'(x) = vec{b}(x),
          $$

          which is, for each fixed $x$, a Cramer (because $det{Phi(x, x_0)} = W(x, x_0) ne 0$) system of $n$ linear equations with $n$ unknowns, $c'_1(x), dots, c'_n(x)$. The unknowns are given by the Cramer rule. We have thus obtained the derivative of the matrix function $vec{c}(x)$, so we have to integrate that derivative. And that is the source of the integral.



          The "Wrońskian" determinant was named after a nineteenth-century Polish philosopher and mathematician, Józef Maria Hoene-Wroński.



          EDIT: I changed slightly the terminology and provided a reference to a Wikipedia entry.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Thank you for your answer, but I am having difficulty in understanding what this ' fundamental matrix ' is, as well as its notation. What is that semicolon's purpose?
            $endgroup$
            – Gradient Entropy
            Mar 17 at 21:53










          • $begingroup$
            I replaced the semicolons with commas, to keep the notation in line with the Wikipedia entry.
            $endgroup$
            – user539887
            Mar 17 at 22:00










          • $begingroup$
            One last question. I'm quite unfamiliar with control theory, and I don't know how this state transition matrix is composed of. I assume it's a 3x3 matrix (according to my example) and it's invertible, because the wronskian determinant is non zero. This means we can write: $vec{c}'=Phi^{-1}(x,x_0)vec{b}$, then integrate to find the vector c. Is this correct? If so, how is that done? That is why I'm asking about the elements of the state transition matrix.
            $endgroup$
            – Gradient Entropy
            Mar 17 at 23:01












          • $begingroup$
            It is only the name "state-transition matrix" that is used in Control Theory. It is one of the most fundamental concepts in the theory of linear systems of ODEs: it is the fundamental matrix that takes value $I$ at $x_0$. In other words, if $Psi(x)$ is any fundamental matrix of $(1)$ then the state-transition matrix of $(1)$ equals $Phi(x,x_0)=Psi(x)Psi^{-1}(x_0)$. Another name (when $x_0$ is fixed): principal fundamental matrix, see, e.g., Linear Differential System.
            $endgroup$
            – user539887
            Mar 18 at 7:18














          1












          1








          1





          $begingroup$

          Write the general solution of the homogeneous linear ODE
          $$
          tag{1}
          frac{mathrm{d}vec{y}}{mathrm{d}x} = A(x) vec{y}
          $$

          in the matrix form as
          $$
          Phi(x; x_0) vec{c},
          $$

          where $vec{c} = mathrm{col}(c_1, dots, c_n)$ is a constant matrix (that is, independent of $x$), and $Phi$ is the state-transition matrix such that $Phi(x_0, x_0) = I$ (the identity matrix).



          Now we are looking for a general solution of the nonhomogeneous linear ODE
          $$
          tag{2}
          frac{mathrm{d}vec{y}}{mathrm{d}x} = A(x) vec{y} + vec{b}(x)
          $$

          in the form
          $$
          Phi(x, x_0) vec{c}(x).
          $$

          Incidentally, here is the source of the term "variation of constants". We look for conditions for the above function of $x$ to be a solution of $(2)$, that is, to have
          $$
          frac{mathrm{d}}{mathrm{d}x}(Phi(x, x_0) vec{c}(x)) = A(x) (Phi(x, x_0) vec{c}(x)) + vec{b}(x).
          $$

          But
          $$
          frac{mathrm{d}}{mathrm{d}x}(Phi(x, x_0) vec{c}(x)) = frac{mathrm{d}}{mathrm{d}x} Phi(x, x_0) cdot vec{c}(x) + Phi(x, x_0) cdot vec{c}'(x)
          $$

          and
          $$
          frac{mathrm{d}}{mathrm{d}x} Phi(x, x_0) = A(x) Phi(x, x_0),
          $$

          so after cancellation we obtain
          $$
          Phi(x, x_0) cdot vec{c}'(x) = vec{b}(x),
          $$

          which is, for each fixed $x$, a Cramer (because $det{Phi(x, x_0)} = W(x, x_0) ne 0$) system of $n$ linear equations with $n$ unknowns, $c'_1(x), dots, c'_n(x)$. The unknowns are given by the Cramer rule. We have thus obtained the derivative of the matrix function $vec{c}(x)$, so we have to integrate that derivative. And that is the source of the integral.



          The "Wrońskian" determinant was named after a nineteenth-century Polish philosopher and mathematician, Józef Maria Hoene-Wroński.



          EDIT: I changed slightly the terminology and provided a reference to a Wikipedia entry.






          share|cite|improve this answer











          $endgroup$



          Write the general solution of the homogeneous linear ODE
          $$
          tag{1}
          frac{mathrm{d}vec{y}}{mathrm{d}x} = A(x) vec{y}
          $$

          in the matrix form as
          $$
          Phi(x; x_0) vec{c},
          $$

          where $vec{c} = mathrm{col}(c_1, dots, c_n)$ is a constant matrix (that is, independent of $x$), and $Phi$ is the state-transition matrix such that $Phi(x_0, x_0) = I$ (the identity matrix).



          Now we are looking for a general solution of the nonhomogeneous linear ODE
          $$
          tag{2}
          frac{mathrm{d}vec{y}}{mathrm{d}x} = A(x) vec{y} + vec{b}(x)
          $$

          in the form
          $$
          Phi(x, x_0) vec{c}(x).
          $$

          Incidentally, here is the source of the term "variation of constants". We look for conditions for the above function of $x$ to be a solution of $(2)$, that is, to have
          $$
          frac{mathrm{d}}{mathrm{d}x}(Phi(x, x_0) vec{c}(x)) = A(x) (Phi(x, x_0) vec{c}(x)) + vec{b}(x).
          $$

          But
          $$
          frac{mathrm{d}}{mathrm{d}x}(Phi(x, x_0) vec{c}(x)) = frac{mathrm{d}}{mathrm{d}x} Phi(x, x_0) cdot vec{c}(x) + Phi(x, x_0) cdot vec{c}'(x)
          $$

          and
          $$
          frac{mathrm{d}}{mathrm{d}x} Phi(x, x_0) = A(x) Phi(x, x_0),
          $$

          so after cancellation we obtain
          $$
          Phi(x, x_0) cdot vec{c}'(x) = vec{b}(x),
          $$

          which is, for each fixed $x$, a Cramer (because $det{Phi(x, x_0)} = W(x, x_0) ne 0$) system of $n$ linear equations with $n$ unknowns, $c'_1(x), dots, c'_n(x)$. The unknowns are given by the Cramer rule. We have thus obtained the derivative of the matrix function $vec{c}(x)$, so we have to integrate that derivative. And that is the source of the integral.



          The "Wrońskian" determinant was named after a nineteenth-century Polish philosopher and mathematician, Józef Maria Hoene-Wroński.



          EDIT: I changed slightly the terminology and provided a reference to a Wikipedia entry.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Mar 17 at 21:59

























          answered Mar 17 at 21:17









          user539887user539887

          1,7811414




          1,7811414








          • 1




            $begingroup$
            Thank you for your answer, but I am having difficulty in understanding what this ' fundamental matrix ' is, as well as its notation. What is that semicolon's purpose?
            $endgroup$
            – Gradient Entropy
            Mar 17 at 21:53










          • $begingroup$
            I replaced the semicolons with commas, to keep the notation in line with the Wikipedia entry.
            $endgroup$
            – user539887
            Mar 17 at 22:00










          • $begingroup$
            One last question. I'm quite unfamiliar with control theory, and I don't know how this state transition matrix is composed of. I assume it's a 3x3 matrix (according to my example) and it's invertible, because the wronskian determinant is non zero. This means we can write: $vec{c}'=Phi^{-1}(x,x_0)vec{b}$, then integrate to find the vector c. Is this correct? If so, how is that done? That is why I'm asking about the elements of the state transition matrix.
            $endgroup$
            – Gradient Entropy
            Mar 17 at 23:01












          • $begingroup$
            It is only the name "state-transition matrix" that is used in Control Theory. It is one of the most fundamental concepts in the theory of linear systems of ODEs: it is the fundamental matrix that takes value $I$ at $x_0$. In other words, if $Psi(x)$ is any fundamental matrix of $(1)$ then the state-transition matrix of $(1)$ equals $Phi(x,x_0)=Psi(x)Psi^{-1}(x_0)$. Another name (when $x_0$ is fixed): principal fundamental matrix, see, e.g., Linear Differential System.
            $endgroup$
            – user539887
            Mar 18 at 7:18














          • 1




            $begingroup$
            Thank you for your answer, but I am having difficulty in understanding what this ' fundamental matrix ' is, as well as its notation. What is that semicolon's purpose?
            $endgroup$
            – Gradient Entropy
            Mar 17 at 21:53










          • $begingroup$
            I replaced the semicolons with commas, to keep the notation in line with the Wikipedia entry.
            $endgroup$
            – user539887
            Mar 17 at 22:00










          • $begingroup$
            One last question. I'm quite unfamiliar with control theory, and I don't know how this state transition matrix is composed of. I assume it's a 3x3 matrix (according to my example) and it's invertible, because the wronskian determinant is non zero. This means we can write: $vec{c}'=Phi^{-1}(x,x_0)vec{b}$, then integrate to find the vector c. Is this correct? If so, how is that done? That is why I'm asking about the elements of the state transition matrix.
            $endgroup$
            – Gradient Entropy
            Mar 17 at 23:01












          • $begingroup$
            It is only the name "state-transition matrix" that is used in Control Theory. It is one of the most fundamental concepts in the theory of linear systems of ODEs: it is the fundamental matrix that takes value $I$ at $x_0$. In other words, if $Psi(x)$ is any fundamental matrix of $(1)$ then the state-transition matrix of $(1)$ equals $Phi(x,x_0)=Psi(x)Psi^{-1}(x_0)$. Another name (when $x_0$ is fixed): principal fundamental matrix, see, e.g., Linear Differential System.
            $endgroup$
            – user539887
            Mar 18 at 7:18








          1




          1




          $begingroup$
          Thank you for your answer, but I am having difficulty in understanding what this ' fundamental matrix ' is, as well as its notation. What is that semicolon's purpose?
          $endgroup$
          – Gradient Entropy
          Mar 17 at 21:53




          $begingroup$
          Thank you for your answer, but I am having difficulty in understanding what this ' fundamental matrix ' is, as well as its notation. What is that semicolon's purpose?
          $endgroup$
          – Gradient Entropy
          Mar 17 at 21:53












          $begingroup$
          I replaced the semicolons with commas, to keep the notation in line with the Wikipedia entry.
          $endgroup$
          – user539887
          Mar 17 at 22:00




          $begingroup$
          I replaced the semicolons with commas, to keep the notation in line with the Wikipedia entry.
          $endgroup$
          – user539887
          Mar 17 at 22:00












          $begingroup$
          One last question. I'm quite unfamiliar with control theory, and I don't know how this state transition matrix is composed of. I assume it's a 3x3 matrix (according to my example) and it's invertible, because the wronskian determinant is non zero. This means we can write: $vec{c}'=Phi^{-1}(x,x_0)vec{b}$, then integrate to find the vector c. Is this correct? If so, how is that done? That is why I'm asking about the elements of the state transition matrix.
          $endgroup$
          – Gradient Entropy
          Mar 17 at 23:01






          $begingroup$
          One last question. I'm quite unfamiliar with control theory, and I don't know how this state transition matrix is composed of. I assume it's a 3x3 matrix (according to my example) and it's invertible, because the wronskian determinant is non zero. This means we can write: $vec{c}'=Phi^{-1}(x,x_0)vec{b}$, then integrate to find the vector c. Is this correct? If so, how is that done? That is why I'm asking about the elements of the state transition matrix.
          $endgroup$
          – Gradient Entropy
          Mar 17 at 23:01














          $begingroup$
          It is only the name "state-transition matrix" that is used in Control Theory. It is one of the most fundamental concepts in the theory of linear systems of ODEs: it is the fundamental matrix that takes value $I$ at $x_0$. In other words, if $Psi(x)$ is any fundamental matrix of $(1)$ then the state-transition matrix of $(1)$ equals $Phi(x,x_0)=Psi(x)Psi^{-1}(x_0)$. Another name (when $x_0$ is fixed): principal fundamental matrix, see, e.g., Linear Differential System.
          $endgroup$
          – user539887
          Mar 18 at 7:18




          $begingroup$
          It is only the name "state-transition matrix" that is used in Control Theory. It is one of the most fundamental concepts in the theory of linear systems of ODEs: it is the fundamental matrix that takes value $I$ at $x_0$. In other words, if $Psi(x)$ is any fundamental matrix of $(1)$ then the state-transition matrix of $(1)$ equals $Phi(x,x_0)=Psi(x)Psi^{-1}(x_0)$. Another name (when $x_0$ is fixed): principal fundamental matrix, see, e.g., Linear Differential System.
          $endgroup$
          – user539887
          Mar 18 at 7:18


















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Mathematics Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3151773%2fhow-exactly-is-the-wronskian-tied-with-cramers-rule-in-a-system-of-odes%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          Popular posts from this blog

          Nidaros erkebispedøme

          Birsay

          Was Woodrow Wilson really a Liberal?Was World War I a war of liberals against authoritarians?Founding Fathers...