sytem of diff eq

This is a given system: $$D\vec{y} = A\vec{y} + \vec{b}$$
With $$A=\left\begin{array}{ccc}1&1&1\\0&2&1\\0&0&3\end{array}\right$$
And $$\vec{b}=\left\begin{array}{c}e^4^t\\0\\0\end{array}\right$$

We find $$\vec{y}_H = Y(t) \cdot \vec{c}$$

With $$Y(t)=\left\begin{array}{ccc}e^t&e^2^t&e^3^t\\0&e^2^t&e^3^t\\0&0&e^3^t\e nd{array}\right$$

Since $$\vec{y} = \vec{y}_H + \vec{y}_P$$ we still need to find is $$\vec{y}_P = Y(t) \cdot \vec{c}(t)$$
$$Y(t)$$ is already known, so whe have to find $$\vec{c}(t)$$
We know that $$D\vec{c}(t)=Y^-^1(t) \cdot \vec{b}$$

Now we are gonna replace $$Y^-^1(t)$$ by $$e^A^(^-^t^)$$

With $$e^A^(^-^t^)$$ being $$e^A^t= Y(t) \cdot Y^-^1(0)$$
So we get $$D\vec{c}(t)= e^A^(^-^t^) \cdot \vec{b}$$

My question actually is why we can replace $$Y^-^1(t)$$ by $$e^A^(^-^t^)$$ since $$e^A^t^= Y(t) \cdot Y^-^1(0)$$?

P.S.: I'll post the whole excersise later if it's necessary, but I had to much trouble with the Latex code for now ;)
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 Would like to take a shot at this; will do so in parts. First, a simple answer is that since the column vectors of $Y(t)$ each represent a solution to the ODE and since the product $Y(t)\cdot Y^-^1(0)$ produces column vectors that are linear combinations of the column vectors of $Y(t)$ then they are again solutions .To see this, perform the the matrix inner product multiplications indicated by $Y(t)\cdot Y^-^1(0)$ and confirm that in the resultant , column vector1 = column vector 1; column vector 2 = column vector 2 minus column vector 1 and column vector 3 = column vector 3 minus column vector 2. The Principle of Superposition Theorem (linearity of differentiation) assures that linearly combining constant multiplies of known solutions (all of which equal zero in a homogeneous system) will again produce a solution. So the point is that you can make the replacement because you are simply substituting one fundamental matrix of solutions for another. See future posts for expansion on this terminology. OK, your question circles around the basic concepts of the use of the Matrix Exponential and one thing I found helpful is to realize that this methodology generalizes the notation of systems of ODE's. By this I mean that the system of equations can be viewed as a "Matrix valued equation" and much of the notation of single equations and vector valued equations can be used to express the concepts. If this doesn't communicate, never mind. One way to find $\vec{y}_P$ is to use the concept of undetermined coefficients. Much of the time this will be more difficult to evaluate than the concept of variation of parameters which also has application here ( will deal with this in a future post). This problem is simple enough however that undetermined coefficients works to quickly establish a $\vec{y}_P$. Since the non-homogeneous term , the forcing function $\;\vec{b}(t)\;\mbox{is linear in the\;} exp^{4t} \mbox{\,function, postulate a particular solution,}$ $$\vec{y}_p=\left[\begin{array}{c}a_1e^4^t\\a_2 e^4^t\\a_3 e^4^t\end{array }\right]$$ and substitue this into the original DE and then evaluate the unknown coefficients. $$D\vec{y}_p=\left[\begin{array}{c}4a_1e^4^t\\4a_2 e^4^t\\4a_3 e^4^t\end{array }\right]=A \cdot \vec {y}_p +\vec{b}=\left[\begin{array}{ccc}1&1&1\\0&2&1\\0&0&3\end{array}\right]\;\vec{y}_p\;+\;\left[\begin{array}{c}{exp(4t)\\ 0\\ 0\end{array}\right]$$ This will quickly produce a system of 3 equations in the three unknown coefficients: $$4a_1 = a_1 + a_2 + a_3 +1$$ $$4a_2 = 2a_2 +a_3$$ $$4a_3 = 3a_3$$ This solves by inspection to $$\vec{a}=\{a_1,a_2,a_3\}^T=\{\frac{1}{3},0,0\}^T$$ Therefore the desired particular solution is: $$\vec{y}_p =\vec{a}\cdot exp(4t) = \left[\begin{array}{c}1/3\\0\\ 0\end{array}\right]exp(4t) = \left[\begin{array}{c}\frac{exp(4t)}{3}\\0\\ 0\end{array}\right]$$ a solution vector to the Matrix valued equation, $$D\vec{y}_p=A \cdot \vec {y}_p +\vec{b}$$. To verify this: Left hand side $$D\vec{y}_p = D\left[\begin{array}{c}\frac{exp(4t)}{3}\\ 0\\0\end{array}\right]}= \left[\begin{array}{c}\frac{4exp(4t)}{3} \\0\\ 0\end{array}\right]$$ Right hand side $$\left[\begin{array}{ccc}1&1&1\\0&2&1\\0&0&3\end{array}\right]\cdot\left[\begin{array}{c}\frac{exp(4t)}{3}\\ 0\\ 0\end{array}\right] + \left[\begin{array}{c}{exp(4t)\\ 0\\ 0\end{array}\right] = \left[\begin{array}{c}\frac{exp(4t)}{3}\\0\\0\end{array}\right] + \left[\begin{array}{c}exp(4t)\\0\\0\end{array}\right]$$$$=\left[\begin{array}{c}\frac{4exp(4t)}{3}\\ 0\\ 0\end{array}\right]$$
 Continued: Here is the next sequential short installment. Referencing Reason's query setup: "We find $Y_h = Y(t)*\vec{c}$ " and then he gives a matrix of solutions, Y(t), arranged in column form, also known as a fundamental matrix for the associated homgeneous system. % $Y_h\cdot\vec{c}$ a matrix representation of the fundamental matrix times an arbitrary constant vector which gives the general solution to the associated homogeneous equation. This is to say that $Y' = A\cdot Y$ is a matrix differential equation. To see if the relation is true, it suffices to multiply each column $vector_i$ in Y(t) by A and compare the result to differentiating each element in said vector wrt to t. I will reference this again below and rename $$Y_h = Y(t)$$ to $$\Phi(t)$$ for the sake of generality. %The fundamental matrix can be arrived at by simply reading the eigenvalues, $\lambda_i$ , {1,2,3} of the coefficient matrix from the diagonal (in this case of a simple upper triangular form) and using them to compute 3 independent eigenvectors using the Eigenvalue Method of solving the system $[A -\lambda_i\cdot I_3]\cdot \{a,b,c\}= \{0,0,0\}$ in turn for each eigenvalue. Multiplying the resulting eigenvectors,$\{a_i,b_i,c_i\}$ by exp( $\lambda_i \;t$) respectively and arranging them in column form produces Y(t). You can read the eigenvectors so produced from the fundamental matrix Y(t). Next time I will try a short derivation/explantion of extending variation of parameters to the matrix valued equation to solve again for $\vec{y}_p$ , a particular solution.

sytem of diff eq

Find out more about using LaTeX on this website in this thread:

 Thanks to TALewis for the link to Tex; have perused and researched since last post and have decided to complete this explanation first using raw text and then gradually edit with Tex. UNDER CONSTRUCTION. So with the previous replies as background, here continues the attempt to answer the original query. Referencing an understanding of the variation of parameters technique for solving for a particular solution of a single linear ODE given the general solution of the associated homogeneous equation, recall that variable coefficients are permitted (thus this technique is more general). This method can be extended to systems of linear ODE's. Reflecting the more general nature of variation of parameters one can write $\vec{y}\;' = P(t)\;\vec{y} + \vec{b}(t)$ with $\vec{y}=\{y_1(t),y_2(t),y_3(t)\}^T$ representing a vector of unknown functions for which we hope to solve by evaluating the system of differential equations represented. The P(t) replaces the matrix A of constant coefficients although constants are permitted; P(t) = A is not excluded. We already have the solution of the complementary equation and will write that generally as $\vec{y}_c (t) = \mathbf\Phi(t) \cdot \vec{c}$ to reflect the general approach with c being merely a vector of arbitrary constants associated with a general solution. The concept is to replace the vector parameter c with a variable vector u(t) and to find one particular solution of the form: $\vec{y}_p (t) = \mathbf \Phi (t) \cdot \vec{u}(t)$ , with u(t) such that it satisfies $D_t\vec{y}_p = P(t)\cdot\vec{y}_p + \vec{b}(t)$ Taking the derivative $\vec {y} \ ' _p (t) = \mathbf\Phi' (t) \cdot \vec{u} (t) +\mathbf \Phi (t) \cdot\vec {u}\ ' (t)$ using the product rule. Substituting $\vec{y}_p$ and $\vec{y}_p\; '$ to $\vec{y}\;' = P(t) \vec{y} +\vec{b}(t)$ gives $$\mathbf\Phi'(t) \ u(t) + \mathbf\Phi(t) \; u'(t) = P(t) \; \mathbf \Phi(t) \ u(t) + \vec{b}$$ Since each column vector of $\mathbf\Phi(t)$, the fundamental matrix of solutions, satisfies $y' = P(t) \; y$, then $\mathbf\Phi'(t) = P(t) \;\mathbf \Phi(t)[/tex] also and rewriting [itex] P(t) \;\mathbf \Phi(t)\; u(t) +\mathbf\Phi(t) \;u'(t) = P(t)\mathbf\Phi(t) \;u(t) +\vec{b}$ which reduces to $\mathbf\Phi(t) \ u'(t) = \vec{b} [/tex] Therefore [itex]u'(t) =\mathbf \Phi^-^1(t) \ \vec{b}$ the inverse of $\mathbf\Phi(t)$ existing because it consists of linearly independent solutions and is thereby nonsingular (its determinant is not equal to zero). The particular solution desired is then $y_p(t) = \mathbf\Phi(t) \ u(t) =\mathbf \Phi(t) \ \int\mathbf\Phi^-^1 \; \vec{b}\dt$ This completes the raw bones answer and I will use it to produce again $\vec{y}_p$, a single particular solution which already was calculated using unknown coefficients and try to answer the original question better. (next post) The explanation above belongs entirely to Profs. C.H. Edwards and David E. Penney and is outlined in Chap. 5 Sect.8 of their book Elementary Differential Equations 4th Ed. Any errors in presentation and typos belong entirely to me, of course.
 $y_p(t) = \mathbf\Phi(t) \ \int\mathbf\Phi^-^1 \ \vec{b} \ \dt$ was derived and so now let's apply it to the present example by first displaying the required inverse of the fundamental matrix without going in to its calculation $$\mathbf\Phi^-^1(t) =\left[\begin{array}{ccc}e^-^t&e^-^t&0\\0&e^-^2^t&-e^-^2^t\\0&0&e^-^3^t\end{array}\right]$$ so that $$y_p = \left[\begin{array}{ccc}e^t&e^2^t&e^3^t\\0&e^2^t&e^3^t\\0&0&e^3^t\end{array}\ right]\cdot\int\left[\begin{array}{ccc}e^-^t&e^-^t&0\\0&e^-^2^t&-e^-^2^t\\0&0&e^-^3^t\end{array}\right] \cdot\:\left[\begin{array}{c}e^4^t\\0\\0\end{array }\right] dt$$ performing the matrix multiplication inside the integral $$\left[\begin{array}{ccc}e^t&e^2^t&e^3^t\\0&e^2^t&e^3^t\\0&0&e^3^t\end{array}\ right]\cdot\int\left[\begin{array}{c}e^3^t\\0\\0\end{array}\right]dt$$ the integration is element by element and so this reduces to the matrix multiplication $$\left[\begin{array}{ccc}e^t&e^2^t&e^3^t\\0&e^2^t&e^3^t\\0&0&e^3^t\end{array}\ right]\cdot\left[\begin{array} {c} \frac {e^3^t} {3}\\0\\0\end{array}\right] =\left[\begin{array}{c}\frac{e^4^t}{3}\\0\\0\end{array}\right]= \vec{y}_p$$ , the desired particular solution.