\(\text{ }\) Polynomials and their roots. Recall that by the fundamental theorem of algebra, any degree \(k\) polynomial with complex coefficients has exactly \(k\) roots (possibly complex), counted with multiplicity. That is, for any \(a_{0},\ldots,a_{k-1}\in\mathbb{C}\) the polynomial \(p(r)\) from ([eq:ch5_ho_lin_aut_char_poly]) can be written as \[p(r)=(r-r_{1})(r-r_{2})\ldots(r-r_{k}),\tag{5.95}\] where \(r_{1},\ldots,r_{k}\in\mathbb{C}\) are its roots, and where a root \(r\) has multiplicity \(m\) if exactly \(r=r_{i}\) for exactly \(m\) indices \(i.\) Alternatively letting \(r_{1},\ldots,r_{k'}\) where \(k'\le k\) be the distinct roots, and letting \(m_{i}\) be the multiplicity of \(r_{i},\) so that necessarily \(m_{1}+\ldots+m_{k'}=k,\) we can write the polynomial as \[p(r)=\prod_{l=1}^{k'}(r-r_{l})^{m_{l}}.\tag{5.96}\]
This is in contrast to the situation for polynomials over the reals: a degree \(k\) polynomial with real coefficients can have less than \(k\) roots, even counting multiplicity. For instance the polynomial \(x^{2}+1\) has no real roots, and can’t be factored over the reals, but has complex roots \(\pm i\) and complex factorization \[x^{2}+1=(x-i)(x+i).\] The polynomial \(x^{4}+x^{3}+x+1\) has exactly one real root \(x=-1,\) of multiplicity \(2,\) and can be factored over the reals as \[x^{4}+x^{3}+x+1=(x+1)^{2}(x^{2}-x+1),\] and in addition has complex roots \(x=e^{\frac{\pi}{3}i},x=e^{-\frac{\pi}{3}i}\) (the roots of the second quadratic factor), so over \(\mathbb{C}\) it can be factored as \[x^{4}+x^{3}+x+1=(x+1)^{2}(x-e^{\frac{\pi}{3}i})(x-e^{-\frac{\pi}{3}i})=(x+1)(x+1)(x-e^{\frac{\pi}{3}i})(x-e^{-\frac{\pi}{3}i}).\]
\(\text{ }\)Solutions from roots. The first step in solving a higher order linear autonomous ODE is finding the roots of the characteristic polynomial.
The ODE \[y^{(10)}(t)-y(t)=0,\quad\quad t\in\mathbb{R},\tag{5.93}\] is a \(10\)-th order linear autonomous ODE.
Factoring a polynomial with arbitrary coefficients is difficult (for degrees \(3,4\)) or even impossible (for insoluble polynomials of degree \(\ge5\)). In exercises meant to be solved with pen and paper it is likely that there is a “nice” factorization, which can be found for instance by guessing simple roots like \(\pm1,\pm2,\pm3,\ldots.\). Or the problem can be solved without knowing the roots explicitly .
(Complete solution of first order autonomous ODEs with \(a\ne0\)). Assume \(a,b\in\mathbb{R},a\ne0\) and \(I\subset\mathbb{R}\) an interval. Then a function \(y(t)\) is a solution of the ODE \[\dot{y}(t)=ay(t)+b,\quad\quad t\in I,\tag{3.23}\] iff \[y(t)=\alpha e^{at}-\frac{b}{a}\quad\forall t\in I,\text{ for some }\alpha\in\mathbb{R}.\tag{3.24}\]
Assume \(a,b\in\mathbb{R},\) \(a\ne0,\) \(I\subset\mathbb{R}\) an interval \(t_{*}\in I\) and \(y_{0}\in\mathbb{R}.\) Then the ODE \[\dot{y}(t)=ay(t)+b\quad t\in I,\tag{3.31}\] subject to the constraint \[y(t_{*})=y_{0},\tag{3.32}\] has the unique solution \[y(t)=\left(y_{0}+\frac{b}{a}\right)e^{a(t-t_{*})}-\frac{b}{a}\quad\forall t\in I.\tag{3.33}\]
(Uniqueness second order complex linear autonomous ODE) Let \(I\) be a non-empty interval. Let \(a_{1},a_{0}\in\mathbb{C}.\) Consider the complex second order linear autonomous ODE \[x''+a_{1}x'+a_{0}x=0,\quad\quad t\in I.\tag{5.60}\] Let \(r_{1},r_{2}\in\mathbb{C},r_{1}\ne r_{2}\) denote the two roots of its characteristic equation \[r^{2}+a_{1}r+a_{2}r=0.\tag{5.61}\]
a) A function \(x\) solves the ODE ([eq:ch5_so_lin_aut_char_eq_gen]) iff \[x(t)=\alpha_{1}e^{r_{1}t}+\alpha_{2}e^{r_{2}t},\tag{5.62}\] for some \(\alpha_{1},\alpha_{2}\in\mathbb{\mathbb{C}}.\)
b) For any \(t_{0}\in I\) and \(x_{0},x_{1}\in\mathbb{C},\) a function \(y\) solves the constrained homogeneous ODE \[x''+a_{1}x'+a_{0}=0,\quad\quad t\in I,x(t_{0})=x_{0},x'(t_{0})=x_{1},\tag{5.63}\] iff \[x(t)=\frac{-x_{0}r_{2}+x_{1}}{r_{1}-r_{2}}e^{r_{1}(t-t_{0})}+\frac{x_{0}r_{1}-x_{1}}{r_{1}-r_{2}}e^{r_{2}(t-t_{0})}.\tag{5.64}\]
c) If \(b\in\mathbb{R}\) and \(a_{0}\ne0\) a function \(y\) solves the non-homogeneous second order autonomous ODE \[y''+a_{1}y'+a_{0}y=b,\quad\quad t\in I,\tag{5.65}\] iff \[y(t)=\alpha_{1}e^{r_{1}t}+\alpha_{2}e^{r_{2}t}+\frac{b}{a_{0}},\quad\quad t\in I\] and if \(t_{0}\in I,y_{0},y_{1}\in\mathbb{C}\) then a function \(y\) solves the constrained non-homogeneous second order autonomous ODE \[y(t)=\alpha_{1}e^{r_{1}t}+\alpha_{2}e^{r_{2}t}+\frac{b}{a_{0}},\quad\quad t\in I,y(t_{0})=y_{0},y'(t_{0})=y_{1}\] iff \[y(t)=\frac{-\left(x_{0}-\frac{b}{a_{0}}\right)r_{2}+x_{1}}{r_{1}-r_{2}}e^{r_{1}(t-t_{0})}+\frac{\left(x_{0}-\frac{b}{a_{0}}\right)r_{1}-x_{1}}{r_{1}-r_{2}}e^{r_{2}(t-t_{0})}+\frac{b}{a_{0}},\quad\quad t\in I.\]
(Uniqueness second order complex linear autonomous ODE with repeated characteristic polynomial root) Let \(I\) be a non-empty interval. Let \(a_{1},a_{0}\in\mathbb{C}\) be s.t. the two roots of its characteristic equation \[r^{2}+a_{1}r+a_{0}r=0,\tag{5.68}\] has a repeated non-zero root, i.e. \(a_{1}^{2}-4a_{0}=0,a_{0}\ne0.\) Write \(\rho=-\frac{a_{1}}{2}\) for the repeated root. Consider the complex second order linear autonomous ODE \[x''+a_{1}x'+a_{0}x=0,\quad\quad t\in I.\tag{5.69}\]
a) A function \(x\) solves the ODE ([eq:ch5_so_lin_aut_char_eq_gen]) iff \[x(t)=\alpha_{1}e^{\rho t}+\alpha_{2}te^{\rho t},\tag{5.70}\] for some \(\alpha_{1},\alpha_{2}\in\mathbb{\mathbb{C}}.\)
b) For any \(t_{0}\in I\) and \(x_{0},x_{1}\in\mathbb{C},\) a function \(y\) solves the constrained homogeneous ODE \[x''+a_{1}x'+a_{0}=0,\quad\quad t\in I,x(t_{0})=x_{0},x'(t_{0})=x_{1},\tag{5.71}\] iff \[x(t)=x_{0}e^{\rho(t-t_{0})}+\left(x_{1}-x_{0}\right)\frac{t-t_{0}}{\rho}e^{\rho(t-t_{0})}.\tag{5.72}\]
c) If \(b\in\mathbb{R}\) and \(a_{0}\ne0\) a function \(y\) solves the non-homogeneous second order autonomous ODE \[y''+a_{1}y'+a_{0}y=b,\quad\quad t\in I,\tag{5.73}\] iff \[y(t)=\alpha_{1}e^{r_{1}t}+\alpha_{2}e^{r_{2}t}+\frac{b}{a_{0}},\quad\quad t\in I\] and if \(t_{0}\in I,y_{0},y_{1}\in\mathbb{C}\) then a function \(y\) solves the constrained non-homogeneous second order autonomous ODE \[y(t)=\alpha_{1}e^{r_{1}t}+\alpha_{2}e^{r_{2}t}-\frac{b}{a_{0}},\quad\quad t\in I,y(t_{0})=y_{0},y'(t_{0})=y_{1}\] iff \[y(t)=\frac{-\left(x_{0}-\frac{b}{a}\right)r_{2}+x_{1}}{r_{1}-r_{2}}e^{r_{1}(t-t_{0})}+\frac{\left(x_{0}-\frac{b}{a}\right)r_{1}-x_{1}}{r_{1}-r_{2}}e^{r_{2}(t-t_{0})}+\frac{b}{a_{0}},\quad\quad t\in I.\]
The ODE \[y^{(10)}(t)-y(t)=0,\quad\quad t\in\mathbb{R},\tag{5.93}\] is a \(10\)-th order linear autonomous ODE.
(Continuation of Example 5.24). The ODE ([eq:ch5_ho_lin_aut_first_ex]) has characteristic polynomial \[p(r)=r^{10}-1.\] Clearly \(r=1\) is a root. The complete set of roots is the tenth roots of unity: \(r_{l}=e^{\frac{l-1}{10}2\pi i},1\le l\le10.\) Thus for each \(l\) \[y(t)=e^{r_{l}t}\] is a solution of ([eq:ch5_ho_lin_aut_first_ex]), and furthermore since the ODE is homogeneous \[y(t)=\sum_{l=1}^{10}\alpha_{l}e^{r_{l}t}\] is a solution for any \(\alpha_{1},\ldots,\alpha_{10}\in\mathbb{C}.\)
Letting \(r_{n}=e^{2\pi i\frac{n-1}{10}},1\le n\le10\) denote the roots of the characteristic polynomial we know that \[y(t)=\sum_{n=1}^{10}\alpha_{n}e^{tr_{n}}\] is a solution for any \(\alpha_{1},\ldots,\alpha_{10}\in\mathbb{C}.\) The derivatives of this solution are given by \[y^{(l)}(t)=\sum_{n=1}^{10}\alpha_{n}r_{n}^{l}e^{tr_{n}},\quad0\le l\le9,\] and for \(t=0\) \[y^{(l)}(0)=\sum_{n=1}^{10}\alpha_{n}r_{n}^{l},\quad0\le l\le9.\] Thus to find a solution that satisfies the constraints we need to solve the system of ten linear equations \[\sum_{l=0}^{9}\alpha_{n}r_{n}^{l}=y_{l},0\le l\le9,\] where \[y_{l}=\begin{cases} 1 & \text{\,if }l\text{ even},\\ 0 & \text{ if }l\text{\,odd}, \end{cases},\quad\quad0\le l\le9.\] Solving a system of ten linear equations in general requires inverting a \(10\times10\) matrix - a non-trivial task - or using the Gaussian elimination algorithm, which will take a while with pen and paper for \(10\) rows. For these specific \(r_{n}\) and \(y_{l}\) we can exploit a special property to arrive at a solution faster. Write the system as a matrix-vector equation: \[M\alpha=y_{*},\tag{5.99}\] where \(\alpha=(\alpha_{1},\ldots,\alpha_{n})\in\mathbb{C}^{10},y_{*}=(1,0,1,0,1,0,1,0,1,0)\in\mathbb{C}^{10}\) and \[M=\left(\begin{matrix}1 & 1 & \ldots & 1 & \ldots & 1 & 1\\ r_{1} & r_{2} & \ldots & r_{6} & \ldots & r_{9} & r_{10}\\ r_{1}^{2} & r_{2}^{2} & \ldots & r_{6}^{2} & \ldots & r_{9}^{2} & r_{10}^{2}\\ & & \ldots & \ldots & \ldots\\ & & \ldots & \ldots & \ldots\\ r_{1}^{8} & r_{2}^{8} & \ldots & r_{6}^{8} & \ldots & r_{9}^{8} & r_{10}^{8}\\ r_{1}^{9} & r_{2}^{9} & \ldots & r_{6}^{9} & \ldots & r_{9}^{9} & r_{10}^{9} \end{matrix}\right)=\left(\begin{matrix}1 & 1 & \ldots & 1 & \ldots & 1 & 1\\ 1 & r_{2} & \ldots & -1 & \ldots & r_{9} & r_{10}\\ 1 & r_{2}^{2} & \ldots & 1 & \ldots & r_{9}^{2} & r_{10}^{2}\\ & & \ldots & \ldots & \ldots\\ & & \ldots & \ldots & \ldots\\ 1 & r_{2}^{8} & \ldots & 1 & \ldots & r_{9}^{8} & r_{10}^{8}\\ 1 & r_{2}^{9} & \ldots & -1 & \ldots & r_{9}^{9} & r_{10}^{9} \end{matrix}\right),\] whereby the first column is \(M_{1\cdot}=(1,\ldots,1)\) since \(r_{1}=e^{\frac{1-1}{10}2\pi i}=e^{0}=1\) and the \(6\)-th column is \(M_{6\cdot}=(1,-1,1,\ldots,-1)\) since \(r_{6}=e^{\frac{6-1}{10}2\pi i}=-1.\) Our target vector \(y_{*}\) is a linear combination of \(M_{1\cdot},M_{6\cdot},\) more precisely \[y_{*}=\frac{M_{1\cdot}+M_{6\cdot}}{2}.\] Therefore \(\alpha_{1}=\alpha_{6}=\frac{1}{2}\) and \(\alpha_{i}=0\) for other \(i\) solves ([eq:ch5_ho_lin_aut_example_IVP_matrix_eq]). Thus \[y(t)=\alpha_{1}e^{r_{1}t}+\alpha_{6}e^{r_{5}t}=\frac{e^{t}+e^{-t}}{2}\tag{5.100}\] solves the constrained ODE ([eq:ch5_ho_lin_aut_first_example_IVP]).
We double-check our calculation by computing from ([eq:ch5_ho_lin_aut_first_example_IVP_final_sol]) \[y^{(l)}(t)=\frac{e^{t}+(-1)^{l}e^{-t}}{2},\quad l\ge0,\tag{5.101}\] and plugging in to the ODE in ([eq:ch5_ho_lin_aut_first_example_IVP]), obtaining \[y^{(10)}(t)-y(t)=\frac{e^{t}+(-1)^{10}e^{-t}}{2}-\frac{e^{t}+e^{-t}}{2}=\frac{e^{t}+e^{-t}}{2}-\frac{e^{t}+e^{-t}}{2}=0\quad\forall t\checkmark,\] and checking the constraints from ([eq:ch5_ho_lin_aut_first_example_IVP]) by noting from ([eq:ch5_ho_lin_aut_first_example_IVP_final_sol_derivs]) that \[y^{(l)}(0)=\frac{1+(-1)^{l}}{2}=\begin{cases} 1 & \text{ if }l\text{ even,}\\ 0 & \text{ if }l\text{\,odd}. \end{cases}\checkmark\]
Even though we went through complex valued computations, we ended up with a real valued answer, a phenomenon we already saw in the second order case with complex roots, and which makes perfect sense since the ODE has real coefficients and the r.h.s. of each constraint is real. Below we will prove that in this situation there is always a unique solution, and this solution is real valued.
Like for the case \(k=2,\) it turns out that when the characteristic polynomial has no repeated roots the linear combinations of the \(k\) solutions \(t\to e^{rt},\) for each root \(r,\) is the complete set of solutions (the proof will be given below)
\(\text{ }\) Repeated roots. Let us investigate the case of a repeated root. Then the functions \(t\to e^{rt}\) yield less than \(k\) solutions. Assume that \(r_{*}\) is a repeated root. By analogy to the second order case it makes sense to test if \[y(t)=te^{r_{*}t}\tag{5.103}\] is a solution.
To compute the derivatives we use a general formula for the \(l\)-th derivative of a product \(f(t)g(t)\) of differentiable functions \(f(t),g(t).\) The case \(l=1\) is just the usual product rule: \[\frac{d}{dt}\left\{ f(t)g(t)\right\} =f'(t)g(t)+f(t)g'(t).\tag{5.104}\] We can compute the second \(l=2\) and third \(l=3\) derivatives by iterating the \(l=1\) formula: \[\begin{array}{ccl} \frac{d^{2}}{dt^{2}}\left\{ f(t)g(t)\right\} & = & f''(t)g(t)+2f'(t)g'(t)+f(t)g''(t)\\ \frac{d^{3}}{dt^{3}}\left\{ f(t)g(t)\right\} & = & f'''(t)g(t)+3f''(t)g'(t)+3f'(t)g''(t)+f(t)g'''(t). \end{array}\] The general formula (which can be proved e.g. by induction) for the \(l\)-th derivative is \[\frac{d^{l}}{dt^{l}}\left\{ f(t)g(t)\right\} =\sum_{i=0}^{l}{l \choose i}f^{(i)}(t)g^{(l-i)}(t),\tag{5.105}\] (note the analogy to the binomial formula). All the formulas ([eq:ch5_ho_lin_aut_product_rule])-([eq:ch5_ho_lin_aut_iterated_product_rule]) hold both for real-valued functions \(f,g:I\to\mathbb{R}\) and more generally for complex-valued functions \(f,g:\mathbb{R}\to\mathbb{C}.\)
Using ([eq:ch5_ho_lin_aut_iterated_product_rule]) with \(f(t)=t\) (so \(f'(t)=1,f''(t)=0,f'''(t)=0,\ldots\)) and \(g(t)=e^{r_{*}t}\) (so \(g^{(l)}(t)=r_{*}^{l}e^{r_{*}t}\)) we can compute the \(l\)-th derivative of \(y(t)\) in ([eq:ch5_ho_lin_aut_t_exp_t_ansatz]) as \[y^{(l)}(t)=tr_{*}^{l}e^{r_{*}t}+lr_{*}^{l-1}e^{r_{*}t},\quad l\ge0.\] Thus ([eq:ch5_ho_lin_aut_t_exp_t_ansatz]) satisfies \[y^{(k)}+\sum_{l=0}^{k-1}a_{l}y^{(l)}=tr_{*}^{k}e^{r_{*}t}+kr_{*}^{k-1}e^{r_{*}t}+\sum_{l=0}^{k-1}a_{l}\left\{ tr_{*}^{l}e^{r_{*}t}+lr_{*}^{l-1}e^{r_{*}t}\right\} .\tag{5.106}\] For convenience we let \(a_{k}=1,\) so we can write the general homogeneous ODE ([eq:ch5_ho_lin_aut_ODE_homo_def]) as \[\sum_{l=0}^{k}a_{l}y^{(l)}=0,\tag{5.107}\] the characteristic polynomial ([eq:ch5_ho_lin_aut_char_poly]) as \[p(r)=\sum_{l=0}^{k}a_{k}r^{l}.\tag{5.108}\] Then ([eq:ch5_ho_lin_aut_t_exp_t_ansatz_plugged_in]) equals \[\sum_{l=0}^{k}a_{l}y^{(l)}=\sum_{l=0}^{k}a_{l}\left\{ tr_{*}^{l}e^{r_{*}t}+lr_{*}^{l-1}e^{r_{*}t}\right\} .\] We can rearrange the r.h.s. as \[te^{r_{*}t}\sum_{l=0}^{k}a_{l}r_{*}^{l}+e^{rt}\sum_{l=0}^{k}a_{l}lr_{*}^{l-1}.\tag{5.109}\] Note that the first of sum on the r.h.s. is simply \(p(r_{*}),\) cf. ([eq:ch5_ho_lin_aut_char_poly_with_ak]). Inspecting the second sum, we see that it coincides with the derivative \(p'(r_{*}),\) since \[p'(r)=\sum_{l=0}^{k}a_{k}\frac{d}{dr}\left\{ r^{l}\right\} =\sum_{l=0}^{k}a_{k}lr^{l-1}!\] Thus for \(y\) as in ([eq:ch5_ho_lin_aut_t_exp_t_ansatz]) \[\sum_{l=0}^{k}a_{l}y^{(l)}=te^{r_{*}t}p(r_{*})+e^{r_{*}t}p'(r_{*}).\tag{5.110}\] Since \(r\) is a solution of the characteristic equation it holds that \(p(r_{*})=0,\) so the first term vanishes. What about \(p'(r_{*})\)? Recall from ([eq:ch5_ho_lin_aut_FTOA_with_mults]) from that if \(r_{*}\) has multiplicity \(m\) then \[p(r)=(r-r_{*})^{m}q(r)\tag{5.111}\] for the polynomial \(q(r)=\prod_{s\text{ root}:s\ne r_{*}}(r-s).\) Thus \[p'(r)=m(r-r_{*})^{m-1}q(r)+(r-r_{*})^{m}q'(r),\] and if \(m\ge2\) then both terms on the r.h.s. vanish if \(r=r_{*},\) showing that \[p'(r_{*})=0\text{ if }r_{*}\text{\,is a repeated root}.\tag{5.112}\] Thus it follows from ([eq:ch5_ho_lin_aut_repeated_root_rhs_in_terms_of_p_pprime]) that \[\sum_{l=0}^{k}a_{l}y^{(l)}=te^{r_{*}t}\cdot0+e^{r_{*}t}\cdot0=0,\] so \(y(t)=te^{r_{*}t}\) is indeed a solution of ([eq:ch5_ho_lin_aut_ODE_homo_def_with_ak]) if \(r_{*}\) is a repeated root, just as in the second order case.
\(\text{ }\) Repeated roots with higher multiplicity. What if there are repeated roots of higher multiplicity? It turns out that then \(t^{2}e^{rt},t^{3}e^{rt},\ldots\) can be solutions. Indeed consider for \(n\in\{0,1,2,\ldots\}\) \[y(t)=t^{n}el^{rt}.\tag{5.113}\] Let us plug this in to ([eq:ch5_ho_lin_aut_ODE_homo_def_with_ak]). Since the computation is fairly long we encapsulate it in the following lemma, which generalizes ([eq:ch5_ho_lin_aut_repeated_root_rhs_in_terms_of_p_pprime]) to higher powers of \(t.\)
Let \(I\subset\mathbb{R}.\) Let \(k\ge1,a_{0},\ldots,a_{k}\in\mathbb{C}\) and let \(p(r)=\sum_{l=0}^{k}a_{l}r^{l}\) be a polynomial. Furthermore let \(r\in\mathbb{C},n\in\{0,1,2,\ldots\},\) and \(y(t)\) be as in ([eq:ch5_ho_lin_aut_repeated_root_sol_higher_t_power]). Then \[\sum_{l=0}^{k}a_{l}y^{(l)}(t)=\sum_{i=0}^{n}{n \choose i}t^{n-i}p^{(i)}(r),\quad\quad t\in I.\tag{5.114}\]
Proof. Using the iterated product rule ([eq:ch5_ho_lin_aut_iterated_product_rule]) with \(f(t)=t^{n}\) (so that \(f^{(i)}=0\) if \(i>n\) and \(f^{(i)}=n(n-1)\ldots(n-i+1)t^{n-i}\) otherwise) and \(g(t)=e^{rt}\) we obtain the formula \[y^{(l)}(t)=\sum_{i=0}^{\max(l,n)}{l \choose i}n(n-1)\ldots(n-i+1)t^{n-i}r^{l-i}e^{rt},\quad\quad l\ge0,\tag{5.115}\] for the \(l\)-th derivative. Define \(C_{l,i}=l(l-1)\ldots(l-k+1).\) With this notation the constant can be rewritten as \[{l \choose i}n(n-1)\ldots(n-i+1)=\frac{l(l-1)\ldots(l-i+1)}{i!}n(n-1)\ldots(n-i+1)={n \choose i}C_{l,i}.\] Then ([eq:ch5_ho_lin_aut_repeated_root_sol_higher_t_power_derivs_first_step]) becomes \[y^{(l)}(t)=e^{rt}\sum_{i=0}^{\max(l,n)}{n \choose i}C_{l,i}t^{n-i}r^{l-i},\quad\quad l\ge0,\] Plugging this into the r.h.s. of ([eq:ch5_ho_lin_aut_repeated_root_sol_higher_t_power_derivs]) gives \[\sum_{l=0}^{k}a_{l}y^{(l)}(t)=e^{rt}\sum_{l=0}^{k}a_{l}\sum_{i=0}^{\max(l,n)}{n \choose i}C_{l,i}t^{n-i}r^{l-i}.\tag{5.116}\] To facilitate switching the order of summation, write the sums as \[\sum_{l=0}^{k}\sum_{i=0}^{n}1_{\left\{ i\le l\right\} }a_{l}{n \choose i}C_{l,i}t^{n-i}r^{l-i}.\] Now the order of the sums can be switched to rewrite the last expression as \[\begin{array}{ccl} \sum_{i=0}^{n}\sum_{l=0}^{k}1_{\left\{ i\le l\right\} }a_{l}{n \choose i}C_{l,i}t^{n-i}r^{l-i} & = & \sum_{i=0}^{n}\sum_{l=i}^{k}a_{l}{n \choose i}C_{l,i}t^{n-i}r^{l-i}\\ & = & \sum_{i=0}^{n}{n \choose i}t^{n-i}\sum_{l=i}^{k}a_{l}C_{l,i}r^{l-i}. \end{array}\] We have arrived at \[\sum_{l=0}^{k}a_{l}y^{(l)}(t)=e^{rt}\sum_{i=0}^{n}t^{n-i}\sum_{l=i}^{k}a_{l}{n \choose i}C_{l,i}r^{l-i}.\tag{5.117}\] The claim now follows since \[p^{(i)}(r)=\sum_{l=i}^{k}a_{l}l(l-1)(l-2)\ldots(l-i+1)r^{l-i}=\sum_{l=i}^{k}a_{l}C_{l,i}r^{l-i}.\]
To check if \(y\) as in ([eq:ch5_ho_lin_aut_repeated_root_sol_higher_t_power]) is a solution to ([eq:ch5_ho_lin_aut_ODE_homo_def_with_ak]) we need to evaluate the r.h.s. of this expression, i.e. we need to compute the derivatives \(p^{(i)}(r).\) But it is a basic property of polynomials that if a root \(r\) has multiplicity \(m,\) then \(p^{(i)}(r)=0\) for all \(i<m,\) as the following generalization of the formula ([eq:ch5_ho_lin_aut_ODE_homo_p_prime_of_rep_r]) proves.
Let \(h(r)\) be a non-zero complex-valued polynomial of degree \(k\) in \(r,\) and \(s\) a root of multiplicity \(m\in\{1,2,\ldots,k\}.\) Then \(h^{(i)}(s)=0\) for all \(i<m.\)
Proof. Recall from ([eq:ch5_ho_lin_aut_repeated_root_rhs_in_terms_of_p_factored_out_root_of_mult_m]), if \(s\) is repeated root with multiplicity \(m\) then \[h(r)=(r-s)^{m}g(r)\] for a polynomial \(g(r).\) Using the iterated product rule ([eq:ch5_ho_lin_aut_iterated_product_rule]) with \(f(r)=(r-s)^{m}\) we obtain that the \(i\)-th derivative of \(h(r)\) is given by \[h^{(i)}(r)=\sum_{j=0}^{i}{i \choose j}f^{(j)}(r)g^{(i-j)}(r).\tag{5.118}\] Since \[f^{(j)}(r)=m(m-1)\ldots(m-j+1)(r-s)^{m-j}\text{\,for }j\le m,\] in particular \[f^{(j)}(s)=0\text{ for }j<m.\] Thus the claim follows since all the terms in ([eq:ch5_ho_lin_aut_homo_deriv_repeated_root_step_1]) vanish.
Combining this lemma and the previous one, we see that if \(r\) is a root of the characteristic polynomial ([eq:ch5_ho_lin_aut_char_poly]) of multiplicity \(m\in\{1,2,\ldots,\}\) then \[e^{rt},te^{rt},\ldots,t^{m-1}e^{rt}\] are \(m\) distinct solutions of the corresponding homogeneous ODE ([eq:ch5_ho_lin_aut_ODE_homo_def]). Note that this means the for any characteristic polynomial of degree \(k,\) regardless of the multiplicity of its roots, there are exactly \(k\) distinct solutions of the form \(t^{l}e^{rt}.\) Furthermore, by taking linear combinations of these one sees that for any complex polynomial \(q(t)\) of degree at most \(m-1\) \[y(t)=q(t)e^{rt}\] is a solution of ([eq:ch5_ho_lin_aut_ODE_homo_def]).
Let us solve the IVP \[y''''-2y'''+2y'-y=0,\quad\quad y(0)=y'(0)=y''(0)=0,y'''(0)=1.\tag{5.120}\] Its characteristic polynomial is \[p(r)=r^{4}-2r^{2}+2r-1.\] We note that \(p(1)=0\) so \(r=1\) is a root, which we can factor out to obtain \[p(r)=(r-1)(r^{3}-r^{2}-r+1).\] Again \(r=1\) is a root of the second factor \(r^{3}-r^{2}-r+1,\) so we can factor out another \((r-1)\) to obtain \[p(r)=(r-1)^{2}(r^{2}-1).\] The last factor is easily factorized as \(r^{2}-1=(r-1)(r+1),\) so the final factorization giving all the roots is \[p(r)=(r-1)^{3}(r+1).\] The root \(r=1\) has multiplicity \(3,\) so for any \(\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4}\in\mathbb{C}\) the function \[y(t)=(\alpha_{1}+t\alpha_{2}+t^{2}\alpha_{3})e^{t}+\alpha_{4}e^{-t}\] is a solution of the ODE. This function has derivatives \[\begin{array}{ccl} y'(t) & = & (\alpha_{2}+2t\alpha_{3})e^{t}+(\alpha_{1}+t\alpha_{2}+t^{2}\alpha_{3})e^{t}-\alpha_{4}e^{-t},\\ y''(t) & = & 2\alpha_{3}e^{t}+2(\alpha_{2}+2t\alpha_{3})e^{t}+(\alpha_{1}+t\alpha_{2}+t^{2}\alpha_{3})e^{t}+\alpha_{4}e^{-t},\\ y'''(t) & = & 6\alpha_{3}e^{t}+3(\alpha_{2}+2t\alpha_{3})e^{t}+(\alpha_{1}+t\alpha_{2}+t^{2}\alpha_{3})e^{t}-\alpha_{4}e^{-t}, \end{array}\] so that after plugging in \(t=0\) \[\begin{array}{ccl} y(0) & = & \alpha_{1}+\alpha_{4},\\ y'(0) & = & \alpha_{2}+\alpha_{1}-\alpha_{4},\\ y''(0) & = & 2\alpha_{3}+2\alpha_{2}+\alpha_{1}+\alpha_{4},\\ y'''(0) & = & 6\alpha_{3}+3\alpha_{2}+\alpha_{1}-\alpha_{4}. \end{array}\] To satisfy the constraints we need that \[\begin{array}{ccl} 0 & = & \alpha_{1}+\alpha_{4},\\ 0 & = & \alpha_{2}+\alpha_{1}-\alpha_{4},\\ 0 & = & 2\alpha_{3}+2\alpha_{2}+\alpha_{1}+\alpha_{4},\\ 1 & = & 6\alpha_{3}+3\alpha_{2}+\alpha_{1}-\alpha_{4}. \end{array}\] The first line implies \(\alpha_{4}=-\alpha_{1},\) the second \(\alpha_{2}=-2\alpha_{1}\) and then the third \(\alpha_{3}=2\alpha_{1}.\) Plugging these into the last line yields \(1=8\alpha_{1},\) so a solution is given by \[\alpha_{1}=\frac{1}{8},\alpha_{2}=-\frac{1}{4},\alpha_{3}=\frac{1}{4},\alpha_{4}=-\frac{1}{8}.\] Thus \[y(t)=\left(\frac{1}{8}-\frac{t}{4}+\frac{t^{2}}{4}\right)e^{t}-\frac{1}{8}e^{-t}\] solves ([eq:ch5_ho_lin_aut_homo_rep_root]) (when solving a problem yourself this is where you would double-check by differentiating and plugging in).
Let \(a_{0},\ldots,a_{k-1},b\in\mathbb{C}\) and \(I\) be a non-empty interval.
a) Assume \(a_{0}\ne0.\) Then a function \(y\) is a solution to ([eq:ch5_ho_lin_aut_ODE_def]) iff \(x=y-\frac{b}{a_{0}}\) is a solution to ([eq:ch5_ho_lin_aut_ODE_homo_def]).
b) \(y(t)=0\) is a solution to ([eq:ch5_ho_lin_aut_ODE_homo_def]).
c) If \(y\) is a solution to ([eq:ch5_ho_lin_aut_ODE_homo_def]) then \(\alpha y\) is also a solution for any \(\alpha\in\mathbb{R}.\)
d) If \(y_{1}\) and \(y_{2}\) are solutions to ([eq:ch5_ho_lin_aut_ODE_homo_def]) then \(y_{1}+y_{2}\) is also a solution to ([eq:ch5_ho_lin_aut_ODE_homo_def]).
(All solutions of higher order complex linear autonomous ODEs) Let \(I\subset\mathbb{R}\) be a non-empty interval. Let \(k\ge1,\) \(a_{0},a_{1},\ldots,a_{k-1}\in\mathbb{C}.\) Consider the ODE \[x^{(k)}(t)+\sum_{l=0}^{k-1}a_{l}x^{(l)}(t)=0.\tag{5.124}\] Assume that the characteristic polynomial \(p(r)=r^{k}+\sum_{l=0}^{k-1}a_{l}r^{l}\) has \(k'\) distinct roots \(r_{1},\ldots,r_{k'}\in\mathbb{C},\) each of multiplicity \(m_{1},m_{2},\ldots,m_{k'}\in\{1,2,\ldots\},\) where \(m_{1}+\ldots+m_{k'}=k.\) Let \[x_{1}(t),\ldots,x_{k}(t)\text{ denote the functions }e^{r_{1}t},te^{r_{1}t},\ldots,t^{m_{1}-1}e^{r_{1}t},e^{r_{2}t},te^{r_{2}t},\ldots,t^{m_{k'}-1}e^{r_{k'}t}.\]
a) A function \(x\) solves the ODE ([eq:ch5_ho_lin_aut_complex_all_sols_hom_ODE]) iff \[x(t)=\sum_{i=1}^{k}\alpha_{i}x_{i}(t),\tag{5.125}\] for some \(\alpha\in\mathbb{C}^{k}.\)
b) For any \(t_{*}\in I,\) the possibly complex \(k\times k\) matrix \(D\) with entries \[D_{i,l}=x_{i}^{(l)}(t_{*}),\] is non-degenerate.
c) Let \(t_{*}\in I\) and \(x_{*}\in\mathbb{C}^{k}.\) A function \(x\) solves the constrained homogeneous ODE \[x^{(k)}(t)+\sum_{l=0}^{k-1}a_{l}x^{(l)}(t)=0,\quad\quad t\in I,\quad\quad x^{(l)}(t_{*})=x_{*,l+1},0\le l\le k-1,\tag{5.126}\] iff \[x(t)=\sum_{i=1}^{k}\alpha_{i}x_{i}(t),\] where \(\alpha\in\mathbb{C}^{k}\) is the unique solution to \[D\alpha=x_{*},\quad\quad\text{i.e. }\quad\quad\alpha=D^{-1}x_{*}.\]
d) If \(b\in\mathbb{C}\) and \(a_{0}\ne0\) a function \(y\) solves the non-homogeneous autonomous ODE \[y^{(k)}(t)+\sum_{l=0}^{k-1}a_{l}y^{(l)}(t)=b,\quad\quad t\in I,\tag{5.127}\] iff \[y(t)=x(t)+\frac{b}{a_{0}},\quad\quad t\in I,\] for an \(x(t)\) as in part a). If \(t_{*}\in I,y_{*}\in\mathbb{C}\) then a function \(y\) solves the constrained non-homogeneous second order autonomous ODE \[y^{(k)}(t)+\sum_{l=0}^{k-1}a_{l}y^{(l)}(t)=b,\quad\quad t\in I,y^{(l)}(t_{*})=y_{*,l+1},0\le l\le k-1,\] iff \[y(t)=x(t)+\frac{b}{a_{0}},\quad\quad t\in I,\] where \(x(t)\) is the unique function from part c) with \[\alpha=D^{-1}z_{*}\quad\quad\text{ for }\quad\quad z_{*}=y_{*}-\left(\begin{matrix}\frac{b}{a_{0}}\\
0\\
\ldots\\
0
\end{matrix}\right).\]
Let \(a_{0},\ldots,a_{k-1},b\in\mathbb{C}\) and \(I\) be a non-empty interval.
a) Assume \(a_{0}\ne0.\) Then a function \(y\) is a solution to ([eq:ch5_ho_lin_aut_ODE_def]) iff \(x=y-\frac{b}{a_{0}}\) is a solution to ([eq:ch5_ho_lin_aut_ODE_homo_def]).
b) \(y(t)=0\) is a solution to ([eq:ch5_ho_lin_aut_ODE_homo_def]).
c) If \(y\) is a solution to ([eq:ch5_ho_lin_aut_ODE_homo_def]) then \(\alpha y\) is also a solution for any \(\alpha\in\mathbb{R}.\)
d) If \(y_{1}\) and \(y_{2}\) are solutions to ([eq:ch5_ho_lin_aut_ODE_homo_def]) then \(y_{1}+y_{2}\) is also a solution to ([eq:ch5_ho_lin_aut_ODE_homo_def]).
Assume \(a,b\in\mathbb{C},\) \(a\ne0,\) \(I\subset\mathbb{R}\) an interval. Then a complex differentiable function \(y:I\to\mathbb{C}\) is a solution to \[\dot{y}(t)=ay(t)+b\quad t\in I,\tag{5.57}\] iff \(y(t)=\alpha e^{at}-\frac{b}{a}\) for some \(\alpha\in\mathbb{C}.\) For any \(t_{*}\in I\) and \(y_{0}\in\mathbb{C}\) the same ODE subject to the constraint \[y(t_{*})=y_{0},\tag{5.58}\] has the unique solution \[y(t)=\left(y_{0}+\frac{b}{a}\right)e^{a(t-t_{*})}-\frac{b}{a}\quad\forall t\in I.\tag{5.59}\]
(Uniqueness second order complex linear autonomous ODE) Let \(I\) be a non-empty interval. Let \(a_{1},a_{0}\in\mathbb{C}.\) Consider the complex second order linear autonomous ODE \[x''+a_{1}x'+a_{0}x=0,\quad\quad t\in I.\tag{5.60}\] Let \(r_{1},r_{2}\in\mathbb{C},r_{1}\ne r_{2}\) denote the two roots of its characteristic equation \[r^{2}+a_{1}r+a_{2}r=0.\tag{5.61}\]
a) A function \(x\) solves the ODE ([eq:ch5_so_lin_aut_char_eq_gen]) iff \[x(t)=\alpha_{1}e^{r_{1}t}+\alpha_{2}e^{r_{2}t},\tag{5.62}\] for some \(\alpha_{1},\alpha_{2}\in\mathbb{\mathbb{C}}.\)
b) For any \(t_{0}\in I\) and \(x_{0},x_{1}\in\mathbb{C},\) a function \(y\) solves the constrained homogeneous ODE \[x''+a_{1}x'+a_{0}=0,\quad\quad t\in I,x(t_{0})=x_{0},x'(t_{0})=x_{1},\tag{5.63}\] iff \[x(t)=\frac{-x_{0}r_{2}+x_{1}}{r_{1}-r_{2}}e^{r_{1}(t-t_{0})}+\frac{x_{0}r_{1}-x_{1}}{r_{1}-r_{2}}e^{r_{2}(t-t_{0})}.\tag{5.64}\]
c) If \(b\in\mathbb{R}\) and \(a_{0}\ne0\) a function \(y\) solves the non-homogeneous second order autonomous ODE \[y''+a_{1}y'+a_{0}y=b,\quad\quad t\in I,\tag{5.65}\] iff \[y(t)=\alpha_{1}e^{r_{1}t}+\alpha_{2}e^{r_{2}t}+\frac{b}{a_{0}},\quad\quad t\in I\] and if \(t_{0}\in I,y_{0},y_{1}\in\mathbb{C}\) then a function \(y\) solves the constrained non-homogeneous second order autonomous ODE \[y(t)=\alpha_{1}e^{r_{1}t}+\alpha_{2}e^{r_{2}t}+\frac{b}{a_{0}},\quad\quad t\in I,y(t_{0})=y_{0},y'(t_{0})=y_{1}\] iff \[y(t)=\frac{-\left(x_{0}-\frac{b}{a_{0}}\right)r_{2}+x_{1}}{r_{1}-r_{2}}e^{r_{1}(t-t_{0})}+\frac{\left(x_{0}-\frac{b}{a_{0}}\right)r_{1}-x_{1}}{r_{1}-r_{2}}e^{r_{2}(t-t_{0})}+\frac{b}{a_{0}},\quad\quad t\in I.\]
Let \(I\subset\mathbb{R}.\) Let \(k\ge1,a_{0},\ldots,a_{k}\in\mathbb{C}\) and let \(p(r)=\sum_{l=0}^{k}a_{l}r^{l}\) be a polynomial. Furthermore let \(r\in\mathbb{C},n\in\{0,1,2,\ldots\},\) and \(y(t)\) be as in ([eq:ch5_ho_lin_aut_repeated_root_sol_higher_t_power]). Then \[\sum_{l=0}^{k}a_{l}y^{(l)}(t)=\sum_{i=0}^{n}{n \choose i}t^{n-i}p^{(i)}(r),\quad\quad t\in I.\tag{5.114}\]
Let \(I\subset\mathbb{R}.\) Let \(k\ge1,a_{0},\ldots,a_{k}\in\mathbb{C}\) and let \(p(r)=\sum_{l=0}^{k}a_{l}r^{l}\) be a polynomial. Furthermore let \(r\in\mathbb{C},n\in\{0,1,2,\ldots\},\) and \(y(t)\) be as in ([eq:ch5_ho_lin_aut_repeated_root_sol_higher_t_power]). Then \[\sum_{l=0}^{k}a_{l}y^{(l)}(t)=\sum_{i=0}^{n}{n \choose i}t^{n-i}p^{(i)}(r),\quad\quad t\in I.\tag{5.114}\]
c) This is a direct consequence of a), b).
\(\text{ }\) ODEs with real coefficients. Let us now specialize to the situation where the ODE has real coefficients. A degree \(k\) polynomial \(p(r)\) with real coefficients has \(k\) roots counted with multiplicity, but they may be complex. Complex roots must appear in complex conjugate pairs: \(r,r'\) s.t. \(r'=\overline{r}\) (where \(\overline{r}\) denotes complex conjugation, i.e. \(\overline{a+ib}=a-ib\) for \(a+ib\in\mathbb{C}\)): this is because if \(p(r)\) has real coefficients, then \(\overline{p(r)}=p(\overline{r}),\) so if \(r\) is a root then so is \(\overline{r}.\) Each non-real complex conjugate pair \(r=a+ib,\overline{r}=a-ib,\) can give rise to the functions \(\cos,\sin,\) as we saw in the second order case: if \(\alpha\in\mathbb{R},0\le l\le m-1,a,b\in\mathbb{R}\) and \(r=a+ib\) is root of multiplicity \(m,\) then \[\frac{\alpha}{2}t^{l}e^{rt}+\frac{\alpha}{2}t^{l}e^{\overline{r}}=\alpha t^{l}e^{at}\cos(bt)\quad\quad\frac{\alpha}{2i}t^{l}e^{rt}-\frac{\alpha}{2i}t^{l}e^{-rt}=\alpha t^{l}e^{at}\sin(bt),\] are solutions.
(All solutions of higher order complex linear autonomous ODEs) Let \(I\subset\mathbb{R}\) be a non-empty interval. Let \(k\ge1,\) \(a_{0},a_{1},\ldots,a_{k-1}\in\mathbb{C}.\) Consider the ODE \[x^{(k)}(t)+\sum_{l=0}^{k-1}a_{l}x^{(l)}(t)=0.\tag{5.124}\] Assume that the characteristic polynomial \(p(r)=r^{k}+\sum_{l=0}^{k-1}a_{l}r^{l}\) has \(k'\) distinct roots \(r_{1},\ldots,r_{k'}\in\mathbb{C},\) each of multiplicity \(m_{1},m_{2},\ldots,m_{k'}\in\{1,2,\ldots\},\) where \(m_{1}+\ldots+m_{k'}=k.\) Let \[x_{1}(t),\ldots,x_{k}(t)\text{ denote the functions }e^{r_{1}t},te^{r_{1}t},\ldots,t^{m_{1}-1}e^{r_{1}t},e^{r_{2}t},te^{r_{2}t},\ldots,t^{m_{k'}-1}e^{r_{k'}t}.\]
a) A function \(x\) solves the ODE ([eq:ch5_ho_lin_aut_complex_all_sols_hom_ODE]) iff \[x(t)=\sum_{i=1}^{k}\alpha_{i}x_{i}(t),\tag{5.125}\] for some \(\alpha\in\mathbb{C}^{k}.\)
b) For any \(t_{*}\in I,\) the possibly complex \(k\times k\) matrix \(D\) with entries \[D_{i,l}=x_{i}^{(l)}(t_{*}),\] is non-degenerate.
c) Let \(t_{*}\in I\) and \(x_{*}\in\mathbb{C}^{k}.\) A function \(x\) solves the constrained homogeneous ODE \[x^{(k)}(t)+\sum_{l=0}^{k-1}a_{l}x^{(l)}(t)=0,\quad\quad t\in I,\quad\quad x^{(l)}(t_{*})=x_{*,l+1},0\le l\le k-1,\tag{5.126}\] iff \[x(t)=\sum_{i=1}^{k}\alpha_{i}x_{i}(t),\] where \(\alpha\in\mathbb{C}^{k}\) is the unique solution to \[D\alpha=x_{*},\quad\quad\text{i.e. }\quad\quad\alpha=D^{-1}x_{*}.\]
d) If \(b\in\mathbb{C}\) and \(a_{0}\ne0\) a function \(y\) solves the non-homogeneous autonomous ODE \[y^{(k)}(t)+\sum_{l=0}^{k-1}a_{l}y^{(l)}(t)=b,\quad\quad t\in I,\tag{5.127}\] iff \[y(t)=x(t)+\frac{b}{a_{0}},\quad\quad t\in I,\] for an \(x(t)\) as in part a). If \(t_{*}\in I,y_{*}\in\mathbb{C}\) then a function \(y\) solves the constrained non-homogeneous second order autonomous ODE \[y^{(k)}(t)+\sum_{l=0}^{k-1}a_{l}y^{(l)}(t)=b,\quad\quad t\in I,y^{(l)}(t_{*})=y_{*,l+1},0\le l\le k-1,\] iff \[y(t)=x(t)+\frac{b}{a_{0}},\quad\quad t\in I,\] where \(x(t)\) is the unique function from part c) with \[\alpha=D^{-1}z_{*}\quad\quad\text{ for }\quad\quad z_{*}=y_{*}-\left(\begin{matrix}\frac{b}{a_{0}}\\
0\\
\ldots\\
0
\end{matrix}\right).\]
(All solutions of higher order real linear autonomous ODEs) Let \(I\subset\mathbb{R}\) be a non-empty interval. Let \(k\ge1,\) \(a_{0},a_{1},\ldots,a_{k-1}\in\mathbb{R}.\) Consider the ODE \[x^{(k)}(t)+\sum_{l=0}^{k-1}a_{l}x^{(l)}(t)=0.\tag{5.134}\] Assume that the characteristic polynomial \(p(r)=r^{k}+\sum_{l=0}^{k-1}a_{l}r^{l}\) has \(k'\) distinct real roots \(r_{1},\ldots,r_{k'}\in\mathbb{R},\) each of multiplicity \(m_{1},m_{2},\ldots,m_{k'}\in\{1,2,\ldots\},\) and \(k''\) complex roots \(r_{k'+1},\ldots,r_{k'+k''}\in\mathbb{C}\) with positive imaginary part with multiplicities \(m_{k'+1},\ldots,m_{k'+k''}\in\{1,2,\ldots\},\) where \(m_{1}+\ldots+m_{k'}+2m_{k'+1}+\ldots+2m_{k'+k''}=k.\) Let \[x_{1}(t),\ldots,x_{k'}(t)\text{ denote the functions }e^{r_{1}t},te^{r_{1}t},\ldots,t^{m_{1}-1}e^{r_{1}t},e^{r_{2}t},te^{r_{2}t},\ldots,t^{m_{k'}-1}e^{r_{k'}t},\] and \[\begin{array}{c} x_{k'+1}(t),\ldots,x_{k+k'}(t)\text{ denote the functions }\\ e^{\mathrm{Re}r_{1}t}\cos(\mathrm{Im}r_{1}t),e^{\mathrm{Re}r_{1}t}\sin(\mathrm{Im}r_{1}t),te^{\mathrm{Re}r_{1}t}\cos(\mathrm{Im}r_{1}t),\ldots,t^{m_{k'+l''}-1}e^{\mathrm{Re}r_{k'+k''}t}\sin(\mathrm{Im}r_{k'+k''}t). \end{array}\]
a) A function \(x\) solves the ODE ([eq:ch5_ho_lin_aut_real_all_sols_hom_ODE]) iff \[x(t)=\sum_{i=1}^{k}\alpha_{i}x_{i}(t),\tag{5.135}\] for some \(\alpha\in\mathbb{R}^{k}.\)
b) For any \(t_{*}\in I,\) the \(k\times k\) real matrix \(D\) with entries \[D_{i,l}=x_{i}^{(l)}(t_{*}),\] is non-degenerate.
c) Let \(t_{*}\in I\) and \(x_{*}\in\mathbb{R}^{k}.\) A function \(x\) solves the constrained homogeneous ODE \[x^{(k)}(t)+\sum_{l=0}^{k-1}a_{l}x^{(l)}(t)=0,\quad\quad t\in I,\quad\quad x^{(l)}(t_{*})=x_{*,l+1},0\le l\le k-1,\tag{5.136}\] iff \[x(t)=\sum_{i=1}^{k}\alpha_{i}x_{i}(t),\tag{5.137}\] where \(\alpha\in\mathbb{R}^{k}\) is the unique solution to \[D\alpha=y_{*},\quad\quad\text{i.e. }\quad\quad\alpha=D^{-1}x_{*}.\]
d) If \(b\in\mathbb{R}\) and \(a_{0}\ne0\) a function \(y\) solves the non-homogeneous autonomous ODE \[y^{(k)}(t)+\sum_{l=0}^{k-1}a_{l}y^{(l)}(t)=b,\quad\quad t\in I,\tag{5.138}\] iff \[y(t)=x(t)+\frac{b}{a_{0}},\quad\quad t\in I,\] for an \(x(t)\) as in part a). If \(t_{*}\in I,y_{*}\in\mathbb{R}^{k}\) then a function \(y\) solves the constrained non-homogeneous second order autonomous ODE \[y^{(k)}(t)+\sum_{l=0}^{k-1}a_{l}y^{(l)}(t)=b,\quad\quad t\in I,y^{(l)}(t_{*})=y_{*,l+1},0\le l\le k-1,\] iff \[y(t)=x(t)+\frac{b}{a_{0}},\quad\quad t\in I,\] where \(x(t)\) is the unique function from part c) with \[\alpha=D^{-1}z_{*}\quad\quad\text{ for }\quad\quad z_{*}=y_{*}-\left(\begin{matrix}\frac{b}{a_{0}}\\ 0\\ \ldots\\ 0 \end{matrix}\right).\]
(All solutions of higher order complex linear autonomous ODEs) Let \(I\subset\mathbb{R}\) be a non-empty interval. Let \(k\ge1,\) \(a_{0},a_{1},\ldots,a_{k-1}\in\mathbb{C}.\) Consider the ODE \[x^{(k)}(t)+\sum_{l=0}^{k-1}a_{l}x^{(l)}(t)=0.\tag{5.124}\] Assume that the characteristic polynomial \(p(r)=r^{k}+\sum_{l=0}^{k-1}a_{l}r^{l}\) has \(k'\) distinct roots \(r_{1},\ldots,r_{k'}\in\mathbb{C},\) each of multiplicity \(m_{1},m_{2},\ldots,m_{k'}\in\{1,2,\ldots\},\) where \(m_{1}+\ldots+m_{k'}=k.\) Let \[x_{1}(t),\ldots,x_{k}(t)\text{ denote the functions }e^{r_{1}t},te^{r_{1}t},\ldots,t^{m_{1}-1}e^{r_{1}t},e^{r_{2}t},te^{r_{2}t},\ldots,t^{m_{k'}-1}e^{r_{k'}t}.\]
a) A function \(x\) solves the ODE ([eq:ch5_ho_lin_aut_complex_all_sols_hom_ODE]) iff \[x(t)=\sum_{i=1}^{k}\alpha_{i}x_{i}(t),\tag{5.125}\] for some \(\alpha\in\mathbb{C}^{k}.\)
b) For any \(t_{*}\in I,\) the possibly complex \(k\times k\) matrix \(D\) with entries \[D_{i,l}=x_{i}^{(l)}(t_{*}),\] is non-degenerate.
c) Let \(t_{*}\in I\) and \(x_{*}\in\mathbb{C}^{k}.\) A function \(x\) solves the constrained homogeneous ODE \[x^{(k)}(t)+\sum_{l=0}^{k-1}a_{l}x^{(l)}(t)=0,\quad\quad t\in I,\quad\quad x^{(l)}(t_{*})=x_{*,l+1},0\le l\le k-1,\tag{5.126}\] iff \[x(t)=\sum_{i=1}^{k}\alpha_{i}x_{i}(t),\] where \(\alpha\in\mathbb{C}^{k}\) is the unique solution to \[D\alpha=x_{*},\quad\quad\text{i.e. }\quad\quad\alpha=D^{-1}x_{*}.\]
d) If \(b\in\mathbb{C}\) and \(a_{0}\ne0\) a function \(y\) solves the non-homogeneous autonomous ODE \[y^{(k)}(t)+\sum_{l=0}^{k-1}a_{l}y^{(l)}(t)=b,\quad\quad t\in I,\tag{5.127}\] iff \[y(t)=x(t)+\frac{b}{a_{0}},\quad\quad t\in I,\] for an \(x(t)\) as in part a). If \(t_{*}\in I,y_{*}\in\mathbb{C}\) then a function \(y\) solves the constrained non-homogeneous second order autonomous ODE \[y^{(k)}(t)+\sum_{l=0}^{k-1}a_{l}y^{(l)}(t)=b,\quad\quad t\in I,y^{(l)}(t_{*})=y_{*,l+1},0\le l\le k-1,\] iff \[y(t)=x(t)+\frac{b}{a_{0}},\quad\quad t\in I,\] where \(x(t)\) is the unique function from part c) with \[\alpha=D^{-1}z_{*}\quad\quad\text{ for }\quad\quad z_{*}=y_{*}-\left(\begin{matrix}\frac{b}{a_{0}}\\
0\\
\ldots\\
0
\end{matrix}\right).\]
Let \(I\subset\mathbb{R}.\) Let \(k\ge1,a_{0},\ldots,a_{k}\in\mathbb{C}\) and let \(p(r)=\sum_{l=0}^{k}a_{l}r^{l}\) be a polynomial. Furthermore let \(r\in\mathbb{C},n\in\{0,1,2,\ldots\},\) and \(y(t)\) be as in ([eq:ch5_ho_lin_aut_repeated_root_sol_higher_t_power]). Then \[\sum_{l=0}^{k}a_{l}y^{(l)}(t)=\sum_{i=0}^{n}{n \choose i}t^{n-i}p^{(i)}(r),\quad\quad t\in I.\tag{5.114}\]
Let \(I\subset\mathbb{R}.\) Let \(k\ge1,a_{0},\ldots,a_{k}\in\mathbb{C}\) and let \(p(r)=\sum_{l=0}^{k}a_{l}r^{l}\) be a polynomial. Furthermore let \(r\in\mathbb{C},n\in\{0,1,2,\ldots\},\) and \(y(t)\) be as in ([eq:ch5_ho_lin_aut_repeated_root_sol_higher_t_power]). Then \[\sum_{l=0}^{k}a_{l}y^{(l)}(t)=\sum_{i=0}^{n}{n \choose i}t^{n-i}p^{(i)}(r),\quad\quad t\in I.\tag{5.114}\]