Blog

Systems of Linear Differential Equations

Many systems have several unknown functions that influence each other. Their equations form a system of linear differential equations: \[\left.\begin{aligned} a_{11}(t)x_1'+\cdots+a_{1n}(t)x_n'+b_{11}(t)x_1+\cdots+b_{1n}(t)x_n&=f_1(t)\\ &\vdots\\ a_{n1}(t)x_1'+\cdots+a_{nn}(t)x_n'+b_{n1}(t)x_1+\cdots+b_{nn}(t)x_n&=f_n(t) \end{aligned}\right\}.\] The \(a_{ij}(t)\) and \(b_{ij}(t)\) are known coefficients, and the \(f_i(t)\) are given forcing functions (inputs).

Example An electrical circuit with three loops gives such a system. Applying Kirchhoff’s voltage law to each loop gives three equations for the unknown currents \(i_1,i_2,i_3\):

Theorem 3.9.1

In the standard form, each equation has one derivative on the left, and solutions exist and are unique.

Note that

1. In this standard form the left-hand side contains each derivative on its own. A general system can be brought to this form by Gauss–Jordan elimination on the equations (possible when the matrix \([a_{ij}(t)]\) of derivative coefficients is invertible).

2. The system can be written in vector form, which is how we will work with it from now on: \[\xx'=\mA(t)\xx+\vc f\] where \(\displaystyle\xx=\col{x_1(t)\\\vdots\\x_n(t)}\), \(\mA(t)=[a_{ij}(t)]_{n\times n}\) and \(\displaystyle\vc f=\col{f_1(t)\\\vdots\\f_n(t)}\).

Example Solve a \(2\times2\) system by the Laplace transform. First we bring it to standard form by subtracting the first equation from the second: \[\left.\begin{aligned}x_1'&=x_1+x_2+3t\\x_1'+x_2'&=5x_1+2x_2+5\end{aligned}\right\}\Rightarrow\begin{aligned}x_1'&=x_1+x_2+3t\\x_2'&=4x_1+x_2+5-3t.\end{aligned}\] In vector form, \[\col{x_1'\\x_2'}=\col{1&1\\4&1}\col{x_1\\x_2}+\col{3t\\5-3t}.\] In the Laplace transform domain this becomes an algebraic matrix equation, \[(s\overline{\xx}-\xx_0)=\mA\overline{\xx}+\vc F,\quad\overline{\xx}=\col{\overline X_1(s)\\\overline X_2(s)},\quad\vc F=\col{F_1(s)\\F_2(s)}\] where the transforms of the forcing terms are \begin{align*} F_1(s)&=\int_0^{\infty}3te^{-st}\,dt\\ &=\underbrace{\left.-\frac{3te^{-st}}s\right|_0^{\infty}}_0+\frac3s\underbrace{\int_0^{\infty}e^{-st}\,dt}_{\left.-\frac1s e^{-st}\right|_0^{\infty}=\frac1s}=\frac3{s^2},\\ F_2(s)&=\int_0^{\infty}(5-3t)e^{-st}\,dt=\frac5s-\frac3{s^2}. \end{align*} Collecting the unknowns on the left and solving, \[\Rightarrow(s\mI-\mA)\overline{\xx}=\xx_0+\vc F\qquad\Rightarrow\qquad\overline{\xx}=(s\mI-\mA)^{-1}(\xx_0+\vc F),\] where, for our matrix, the \(2\times2\) inverse formula gives \[(s\mI-\mA)=\col{s-1&-1\\-4&s-1},\qquad(s\mI-\mA)^{-1}=\frac1{s^2-2s-3}\col{s-1&1\\4&s-1}.\]

Note that \(s^2-2s-3=(s-3)(s+1)\). Multiplying out gives two parts, one that depends on the initial conditions and one that depends on the forcing: \[\Rightarrow\overline{\xx}=\frac1{(s-3)(s+1)}\col{s-1&1\\4&s-1}\left(\col{x_1(0)\\x_2(0)}+\col{3/s^2\\5/s-3/s^2}\right)\] \[=\col{\dfrac{(s-1)x_1(0)+x_2(0)}{(s-3)(s+1)}\\[5pt]\dfrac{4x_1(0)+(s-1)x_2(0)}{(s-3)(s+1)}}+\col{\dfrac{(s-1)3}{s^2(s-3)(s+1)}+\dfrac{5s-3}{s^2(s-3)(s+1)}\\[5pt]\dfrac{12}{s^2(s-3)(s+1)}+\dfrac{(5s-3)(s-1)}{s^2(s-3)(s+1)}}\] Each entry is a ratio of polynomials. Expanding each in partial fractions gives \[=\col{\dfrac{2x_1(0)+x_2(0)}{4(s-3)}+\dfrac{2x_1(0)-x_2(0)}{4(s+1)}\\[5pt]\dfrac{x_2(0)+2x_1(0)}{2(s-3)}+\dfrac{x_2(0)-2x_1(0)}{2(s+1)}}+\col{-\dfrac4s+\dfrac2{s^2}+\dfrac{1/2}{s-3}+\dfrac{7/2}{s+1}\\[5pt]\dfrac6s-\dfrac5{s^2}+\dfrac1{s-3}-\dfrac7{s+1}}.\] Inverting term by term, using \(\frac1{s-a}\leftrightarrow e^{at}\), \(\frac1s\leftrightarrow1\) and \(\frac1{s^2}\leftrightarrow t\), gives the solution: \[\xx(t)=\underbrace{\col{\left(\dfrac{2x_1(0)+x_2(0)}4\right)e^{+3t}+\left(\dfrac{2x_1(0)-x_2(0)}4\right)e^{-t}\\[6pt]\dfrac{x_2(0)+2x_1(0)}2e^{3t}+\dfrac{x_2(0)-2x_1(0)}2e^{-t}}}_{\text{determined by the initial conditions }x_1(0),\ x_2(0)}\] \[+\underbrace{\col{-4+2t+\frac12e^{3t}+\frac72e^{-t}\\6-5t+e^{3t}-7e^{-t}}}_{\text{particular solution for zero initial conditions}}.\] There is another approach, based on state-transition matrices. We want to solve \[\xx'=\mA(t)\xx+\vc f(t)\quad\text{with initial value }\xx(0)=\xx_0.\]

Observations

We start with the homogeneous system \(\xx'=\mA(t)\xx\).

(i) The set of solutions of the homogeneous system \[\xx'=\mA(t)\xx\] is an \(n\)-dimensional vector space of functions. It is closed under the vector space operations, because if \(\xx_1(t)\) is a solution then so is \(\alpha\xx_1(t)\), and if \(\xx_2(t)\) is also a solution then so is \(\xx_1(t)+\xx_2(t)\).

(ii) Suppose \(\{\vc\phi_1(t),\ldots,\vc\phi_n(t)\}\) are \(n\) linearly independent solutions of \(\xx'=\mA(t)\xx\). Then every solution has the form \[\xx(t)=\alpha_1\vc\phi_1(t)+\cdots+\alpha_n\vc\phi_n(t)\] for some constants \(\alpha_i\). In matrix form, with the solutions as the columns of a fundamental matrix \(\mPhi(t)\), \[\xx(t)=\mPhi(t)\vc\alpha\] where \(\mPhi(t)=[\vc\phi_1(t)\ \cdots\ \vc\phi_n(t)]\) and \(\displaystyle\vc\alpha=\col{\alpha_1\\\vdots\\\alpha_n}\).

(iii) Notation. We write \[\mPhi(t;t_0,\mX_0)=[\vc\phi(t;t_0,\xx_1)\ \cdots\ \vc\phi(t;t_0,\xx_n)],\] where \(\vc\phi(t;t_0,\xx_i)\) is the solution of \[\xx'=\mA\xx,\quad\text{with }\xx(t_0)=\xx_i.\] (iv) The special case \(\mX_0=\mI\), that is, starting each solution from a standard basis vector, is called the state-transition matrix: \[\mPhi(t;t_0)=[\vc\phi(t;t_0,\ev_1)\ \cdots\ \vc\phi(t;t_0,\ev_n)].\] (v) The unique solution of \(\xx'=\mA(t)\xx\) with \(\xx(t_0)=\xx_0\) can be written with either matrix. Write the initial vector in the starting vectors \(\xx_i\), with coefficients \(\vc\alpha\), and use the fact that a combination of solutions is a solution: \[\vc\phi(t;t_0,\xx_0)=\sum_{i=1}^n\vc\phi(t;t_0,\xx_i)\alpha_i=\mPhi(t;t_0,\mX_0)\vc\alpha,\] where \[\xx_0=\alpha_1\xx_1+\cdots+\alpha_n\xx_n=\underbrace{[\xx_1\ \cdots\ \xx_n]}_{\mX_0}\vc\alpha\Rightarrow\vc\alpha=\mX_0^{-1}\xx_0.\] So \[\Rightarrow\vc\phi(t;t_0,\xx_0)=\mPhi(t;t_0,\mX_0)\mX_0^{-1}\xx_0.\] The choice \(\mX_0=\mI\) gives the state-transition matrix:

In words, the state-transition matrix carries the state at time \(t_0\) to the state at time \(t\).

Properties of the State-Transition Matrix

Three properties of \(\mPhi(t;t_0)\) make it easy to work with. Throughout, \(I\) is the interval on which the solutions exist.

(i) Non-singularity. \(\mPhi(t;t_0,\mX_0)\) is non-singular for all \(t\) and \(t_0\) in \(I\), whenever \(\mX_0\) is non-singular.

Proof We argue by contradiction. Assume that \(\mPhi(t_1;t_0,\mX_0)\) is singular for some \(t_1\) and \(t_0\) in \(I\). Then there is a vector \(\vc c\ne\nul\) with \[\mPhi(t_1;t_0,\mX_0)\vc c=\nul.\] Define \(\vc\psi(t)=\mPhi(t;t_0,\mX_0)\vc c\). This is a combination of solutions, hence a solution of \(\xx'=\mA(t)\xx\), and it satisfies \(\vc\psi(t_1)=\nul\). The zero function is also a solution with this initial value, and the solution is unique (Theorem 3.9.1), so \(\vc\psi(t)=\nul\) for all \(t\), the zero function. In particular \[\vc\psi(t_0)=\mX_0\vc c=\nul,\] which says \(\mX_0\) is singular, contradicting our assumption. \(\blacksquare\)

(ii) The composition rule. For all \(t,t_1,t_0\) in \(I\), \[\mPhi(t;t_1)\mPhi(t_1;t_0)=\mPhi(t;t_0).\] In words: going from \(t_0\) to \(t_1\) and then on to \(t\) is the same as going from \(t_0\) to \(t\) directly.

Proof Let \(\Psi(t)=\mPhi(t;t_1)\mPhi(t_1;t_0)\). Differentiating, and using that \(\mPhi(t;t_1)\) solves the matrix equation, \[\Psi'=\mPhi'(t;t_1)\mPhi(t_1;t_0)=\mA(t)\underbrace{\mPhi(t;t_1)\mPhi(t_1;t_0)}_{\Psi(t)}.\] \[\Rightarrow\Psi'=\mA(t)\Psi(t),\] and at \(t=t_1\) the initial value is \[\Psi(t_1)=\underbrace{\mPhi(t_1;t_1)}_{\mI}\mPhi(t_1;t_0)=\mPhi(t_1;t_0).\] Thus \(\Psi(t)\) is the unique solution of \(\mX'=\mA(t)\mX\) with \(\mX(t_1)=\mPhi(t_1;t_0)\). But \(\mPhi(t;t_0)\) solves the same problem, so by uniqueness \[\Psi(t)=\mPhi(t;t_0)\ \Rightarrow\ \mPhi(t;t_0)=\mPhi(t;t_1)\mPhi(t_1;t_0).\quad\blacksquare\] (iii) The inverse. Taking \(t=t_0\) in rule (ii) gives \[\mPhi(t_0;t_0)=\mI=\mPhi(t_0;t_1)\mPhi(t_1;t_0)\qquad\forall\,t_0,t_1\in I,\] so \(\mPhi(t_0;t_1)\) is the inverse of \(\mPhi(t_1;t_0)\). Renaming the variables:

Running the system backward in time is the same as inverting the forward map.

The Non-Homogeneous Case

Now we add a forcing term and solve \[\xx'=\mA(t)\xx+\vc f(t).\] Let \(\mPhi(t;t_0,\mX_0)\) be any fundamental matrix of the homogeneous problem. Then \(\xx=\mPhi(t;t_0,\mX_0)\vc c\) solves the homogeneous problem for every constant vector \(\vc c\).

Idea. Use the variation of parameters technique: replace the constant vector \(\vc c\) by an unknown function \(\vv(t)\), so we look for a solution of the form \[\xx=\mPhi(t;t_0,\mX_0)\vv(t).\] Substituting this guess into the non-homogeneous equation and using the product rule, \begin{align*} \xx'&=\mPhi'(t;t_0,\mX_0)\vv(t)+\mPhi(t;t_0,\mX_0)\vv'(t)\\ &=\mA(t)\underbrace{\mPhi(t;t_0,\mX_0)\vv(t)}_{\xx}+\mPhi(t;t_0,\mX_0)\vv'(t)\\ &=\mA(t)\xx+\mPhi(t;t_0,\mX_0)\vv'(t). \end{align*} Comparing with the original equation, the extra term must equal the forcing: \[\Rightarrow\mPhi(t;t_0,\mX_0)\vv'(t)=\vc f(t)\] \[\Rightarrow\vv'(t)=\mPhi^{-1}(t;t_0,\mX_0)\vc f(t)\] \[\Rightarrow\vv(t)=\underbrace{\int\mPhi^{-1}(t;t_0,\mX_0)\vc f(t)\,dt}_{\vc V(t)}+\vc c=\vc V(t)+\vc c.\] Hence the general solution is the sum of a particular part and a complementary part: \[\xx=\mPhi(t;t_0,\mX_0)\cdot(\vc V(t)+\vc c)=\underbrace{\vc\phi_p(t)}_{\text{particular solution}}+\underbrace{\vc\phi_c(t)}_{\text{complementary solution}}.\]

If an initial condition \(\xx(t_0)=\xx_0\) is given, we choose \(\mPhi(t;t_0,\mX_0)\) to be the state-transition matrix and integrate from \(t_0\): \[\vc V(t)=\int_{t_0}^t\mPhi^{-1}(\tau;t_0)\vc f(\tau)\,d\tau=\int_{t_0}^t\mPhi(t_0;\tau)\vc f(\tau)\,d\tau.\] Then \(\vc V(t_0)=\nul\), so \(\vc c\) must be \(\xx_0\). Using \(\mPhi(t;t_0)\mPhi(t_0;\tau)=\mPhi(t;\tau)\) from rule (ii), the solution becomes:

The first term is the response of the system to its initial state alone. The second adds up the response to the input: the input applied at time \(\tau\) is carried to time \(t\) by \(\mPhi(t;\tau)\).

Example Find the solution of the initial value problem \[\xx'=\col{-1&1\\-2&1}\xx+\col{1\\2}u(t),\quad\xx(0)=\nul,\quad u(t)=\begin{cases}1&t>0\\0&t\le0.\end{cases}\] First we need a fundamental matrix for the homogeneous problem. The two functions \[\vc\phi_1=\col{\cos t\\\cos t-\sin t},\qquad\vc\phi_2=\col{\sin t\\\cos t+\sin t}\] are linearly independent solutions, so a fundamental matrix is \[\mPhi(t)=[\vc\phi_1(t)\ \vc\phi_2(t)]=\col{\cos t&\sin t\\\cos t-\sin t&\cos t+\sin t}.\] The state-transition matrix is then \(\mPhi(t;t_0)=\mPhi(t)\mPhi^{-1}(t_0)\), which evaluates to \begin{align*} \mPhi(t;t_0)&=\col{\cos t&\sin t\\\cos t-\sin t&\cos t+\sin t}\col{\cos t_0+\sin t_0&-\sin t_0\\-\cos t_0+\sin t_0&\cos t_0}\\ &=\col{\cos(t-t_0)-\sin(t-t_0)&\sin(t-t_0)\\-2\sin(t-t_0)&\cos(t-t_0)+\sin(t-t_0)}. \end{align*} As expected for a time-invariant system, it depends only on \(t-t_0\).

For \(t>0\) we have \(u(t)=1\), so \(\vc f(t)=(1,2)^T\). We solve by variation of parameters, with the fundamental matrix \(\mPhi(t)\) found above (any fundamental matrix works): \[\xx=\mPhi(t)\vv(t).\] \begin{align*} \vv'(t)&=\mPhi^{-1}(t)\vc f(t)=\col{\cos t+\sin t&-\sin t\\-\cos t+\sin t&\cos t}\col{1\\2},\quad t>0\\ &=\col{\cos t-\sin t\\\cos t+\sin t}. \end{align*} \[\Rightarrow\vv(t)=\col{\sin t+\cos t+c_1\\\sin t-\cos t+c_2}.\] Multiplying by \(\mPhi(t)\) gives the general solution: \[\xx=\mPhi(t)\vv(t)=\col{1+c_1\cos t+c_2\sin t\\(c_1+c_2)\cos t+(c_2-c_1)\sin t}.\] We impose the initial condition to find \(c_1\) and \(c_2\): \[\xx(0)=\nul\Rightarrow\col{1+c_1\\c_1+c_2}=\nul\Rightarrow c_1=-1,\quad c_2=1.\] Hence the solution is

A Second Way: the Formula Directly

Since \(\xx(0)=\nul\), the initial-condition term vanishes and the input term gives the answer in one step, with no unknown constants to fix: \begin{align*} \xx(t)&=\int_0^t\mPhi(t;\tau)\vc f(\tau)\,d\tau\\ &=\int_0^t\col{\cos(t-\tau)-\sin(t-\tau)&\sin(t-\tau)\\-2\sin(t-\tau)&\cos(t-\tau)+\sin(t-\tau)}\col{1\\2}\,d\tau\\ &=\int_0^t\col{\cos(t-\tau)+\sin(t-\tau)\\2\cos(t-\tau)}\,d\tau=\col{1-\cos t+\sin t\\2\sin t}, \end{align*} which agrees with the result above.

Supplement supplied separately; wording and calculations retained as written.

Example A time-varying system with fixed eigenvectors. Consider the homogeneous, time-varying linear system of differential equations \[\xx'=\mA(t)\xx\qquad(\mathrm{Eq.1})\] where \(\mA(t)\) is \(n\times n\) and has the form \(\mA(t)=\mV\underbrace{\mLambda(t)}_{\text{diagonal}}\mV^{-1}\). That is, for every \(t\) the matrix \(\mA(t)\) has the same eigenvectors, but the corresponding eigenvalues are functions of time.

In this special case we can decouple the system. Define the new unknown \(\vc z=\mV^{-1}\xx\), and substitute \(\xx=\mV\vc z\) into (Eq.1). Because \(\mV\) is constant, \(\xx'=\mV\vc z'\), so \[\mV\vc z'=\mV\mLambda(t)\mV^{-1}\mV\vc z\] \[\Rightarrow\vc z'=\mLambda(t)\vc z\qquad(\mathrm{Eqn.2}).\] Since \(\mLambda(t)\) is diagonal, (Eqn.2) is a set of \(n\) separate first-order equations: \[z_k'=\lambda_k(t)z_k,\quad1\le k\le n.\] Each one is solved by separating variables: \[\frac{dz_k}{dt}=\lambda_k(t)z_k\Rightarrow\int_{z_k(0)}^{z_k(t)}\frac{dz_k}{z_k}=\int_0^t\lambda_k(\tau)\,d\tau\] \[\Rightarrow\ln|z_k|=\int_0^t\lambda_k(\tau)\,d\tau+d_k.\] Exponentiating gives

Returning to the original unknown, with \(\vv_k\) the columns of \(\mV\), \[\xx(t)=\mV\vc z(t)=\sum_{k=1}^n\vv_kz_k(t)=\sum_{k=1}^n\vv_kC_ke^{\int_0^t\lambda_k(\tau)\,d\tau}.\] Note that if the system starts at an eigenvector, \(\xx(0)=\vv_m\), then \(C_k=0\) for \(k\ne m\) and \(C_m=1\), so the solution stays on the line through \(\vv_m\). Collecting these \(n\) solutions as columns gives a fundamental matrix: \[\underbrace{\mPhi(t;0,\mV)}_{\text{fundamental matrix}}=[\vv_1e^{\int_0^t\lambda_1(\tau)\,d\tau}\ \cdots\ \vv_ne^{\int_0^t\lambda_n(\tau)\,d\tau}].\]

Supplement supplied separately; wording and calculations retained as written.

The same fundamental matrix can be written as a product with a diagonal matrix: \[\mPhi(t;0,\mV)=\mV\col{e^{\int_0^t\lambda_1(\tau)\,d\tau}&&\\&\ddots&\\&&e^{\int_0^t\lambda_n(\tau)\,d\tau}}.\] The state-transition matrix is obtained by normalizing with \(\mV^{-1}\): \[\mPhi(t;0)=\mPhi(t;0,\mV)\mV^{-1}=\mV\col{e^{\int_0^t\lambda_1(\tau)\,d\tau}&&\\&\ddots&\\&&e^{\int_0^t\lambda_n(\tau)\,d\tau}}\mV^{-1}.\] To start at a general time \(t_0\) we use the composition and inverse rules of the previous section, \(\mPhi(t;t_0)=\mPhi(t;0)\mPhi^{-1}(t_0;0)\). The middle factors \(\mV^{-1}\mV\) cancel: \begin{align*} \mPhi(t;t_0)&=\mPhi(t;0)\mPhi^{-1}(t_0;0)\\ &=\mV\col{e^{\int_0^t\lambda_1(\tau)\,d\tau}&&\\&\ddots&\\&&e^{\int_0^t\lambda_n(\tau)\,d\tau}}\mV^{-1}\mV\col{e^{-\int_0^{t_0}\lambda_1(\tau)\,d\tau}&&\\&\ddots&\\&&e^{-\int_0^{t_0}\lambda_n(\tau)\,d\tau}}\mV^{-1}. \end{align*} Combining the exponentials, the two integrals merge into one from \(t_0\) to \(t\):

Example Find the state-transition matrix of \[\xx'=\underbrace{\col{1-t&t/2-1\\0&-t/2}}_{\mA(t)}\xx.\] Since \(\mA(t)\) is upper triangular, its eigenvalues are the diagonal entries, \(\lambda_1=1-t\) and \(\lambda_2=-\dfrac t2\). We find the eigenvectors from \((\mA-\lambda\mI)\vv=\nul\): \[(\mA-\lambda_1\mI)\vv_1=\nul\Rightarrow\col{0&t/2-1\\0&t/2-1}\underbrace{\col{v_1\\v_2}}_{\vv_1}=\nul\ \Rightarrow\ \vv_1=\col{1\\0},\] \[(\mA-\lambda_2\mI)\vv_2=\nul\Rightarrow\col{1-t/2&t/2-1\\0&0}\vv_2=\nul\ \Rightarrow\ \vv_2=\col{1\\1}.\] Both eigenvectors are constants, which is what (Eqn.3) requires.

Supplement supplied separately; wording and calculations retained as written.

Therefore \(\mA(t)\) factors as \begin{align*} \mA(t)&=[\vv_1\ \vv_2]\col{\lambda_1(t)&\\&\lambda_2(t)}[\vv_1\ \vv_2]^{-1}\\ &=\underbrace{\col{1&1\\0&1}}_{\mV}\underbrace{\col{1-t&0\\0&-t/2}}_{\mLambda(t)}\underbrace{\col{1&-1\\0&1}}_{\mV^{-1}}. \end{align*} By (Eqn.3) of the previous page, the state-transition matrix is \[\mPhi(t;t_0)=\mV\col{e^{\int_{t_0}^t\lambda_1(\tau)\,d\tau}&0\\0&e^{\int_{t_0}^t\lambda_2(\tau)\,d\tau}}\mV^{-1}.\] The two integrals are \[\int_{t_0}^t\lambda_1(\tau)\,d\tau=\int_{t_0}^t(1-\tau)\,d\tau=\left.\tau-\frac12\tau^2\right|_{t_0}^t=(t-t_0)-\frac12(t^2-t_0^2),\] \[\int_{t_0}^t\lambda_2(\tau)\,d\tau=\int_{t_0}^t-\frac\tau2\,d\tau=\left.-\frac{\tau^2}4\right|_{t_0}^t=-\frac14(t^2-t_0^2).\] Thus \begin{align*} \mPhi(t;t_0)&=\col{1&1\\0&1}\col{e^{(t-t_0)-\frac12(t^2-t_0^2)}&0\\0&e^{-\frac14(t^2-t_0^2)}}\col{1&-1\\0&1}\\ &=\col{1&1\\0&1}\col{e^{(t-t_0)-\frac12(t^2-t_0^2)}&-e^{(t-t_0)-\frac12(t^2-t_0^2)}\\0&e^{-\frac14(t^2-t_0^2)}}. \end{align*} Carrying out the last product gives the final answer:

Supplement supplied separately; wording and calculations retained as written.

How to Find the State-Transition Matrix

This page collects the standard ways to find the state-transition matrix (STM) \(\mPhi(t;t_0)\), first for time-invariant systems \(\xx'=\mA\xx\), then for a time-varying case.

Time-Invariant Homogeneous Equations

Method 1: diagonalization (or the Cayley–Hamilton method). If \(\mA=\mQ\mLambda\mQ^{-1}\), the STM is the matrix exponential, which is computed on the eigenvalues:

Method 2: inverse Laplace transform. Transforming \(\xx'=\mA\xx\) gives \(\overline{\xx}=(s\mI-\mA)^{-1}\xx_0\), so

Method 3: from a fundamental matrix. Find \[\mPhi(t;t_0,\mX_0)=[\vc\phi(t;t_0,\xx_1)\ \cdots\ \vc\phi(t;t_0,\xx_n)]\] by building the solutions from the eigenpairs \((\lambda_k,\uu_k)\); for \(n=2\) these are combinations \(C_1\uu_1e^{\lambda_1t}+C_2\uu_2e^{\lambda_2t}\). Then normalize with the inverse of the starting matrix: \[\mPhi(t;t_0)=\mPhi(t;t_0,\mX_0)\cdot\mX_0^{-1}.\]

Time-Varying Homogeneous Equations

Diagonalization with constant eigenvectors. If \(\mA(t)=\mQ\mLambda(t)\mQ^{-1}\) with a constant eigenvector matrix \(\mQ\), then each eigenvalue is integrated separately: \[\mPhi(t;t_0)=\mQ\col{e^{\int_{t_0}^t\lambda_1(\tau)\,d\tau}&&\\&\ddots&\\&&e^{\int_{t_0}^t\lambda_n(\tau)\,d\tau}}\mQ^{-1}.\]

Time-Invariant Non-Homogeneous Equations

For \(\xx'=\mA\xx+\mB\vc u(t)\) with \(\xx(t_0)=\xx_0\), the solution \(\vc\phi(t;t_0,\xx_0)\) combines the response to the initial state with the response to the input: