Systems of Linear Ordinary Differential Equations

Systems of Linear Ordinary Differential Equations A Calabia, Universidad de Alcala. (DOI:10.5281/zenodo.17793344) December 2, 2025 1 Introduction Ordinary Differential Equations (ODEs) involve functions of a single variable and their derivatives. They are used to describe the relationship between a function and its derivatives with respect to one independent variable. In contrast, Partial Differential Equations (PDEs) involve functions of multiple variables and their partial derivatives. PDEs are used to describe phenomena where the function depends on several independent variables, such as in heat conduction, wave propagation, and fluid dynamics. This document covers the solution of systems of linear ordinary differential equations (ODEs) using exponential methods in algebra. We will discuss both homogeneous and nonhomogeneous equations. 2 Homogeneous Linear ODEs 2.1 Solving the Homogeneous linear ODE: u′=a·u To solve u′=a·u, we first rewrite the expression as: du dt =a·u Then we separate the variables uand tand integrate both sides: 1 udu =a dt Z1 udu =Za dt After solving both integrals separately using Cas the constant of integration: ln |u|=at +C We can then exponentiate both sides to solve for u: |u|=eat+C 1 Expanding the last expression, and taking eC=C1(where C1as a new constant, which can be positive or negative): |u|=eat ·eC=eat ·C1 If we remove the absolute value (considering C1can absorb the sign): u=C1eat The general solution to u′=a·uis: u(t) = C1eat where C1is an arbitrary constant determined by initial conditions. If we have an initial condition, such as u(0) = u0, we can find C1by substituting t= 0 and u=u0: u0=C1ea·0=⇒u0=C1 Thus, C1=u0, and the specific solution is: u(t) = u0eat 2.2 System of Homogeneous Linear ODEs Now, considering a system of homogeneous linear ODEs in matrix form with where Aas a constant matrix: X′(t) = AX(t) The general solution can be expressed as: X(t) = eAtu0 where eAt is the matrix exponential and u0is a vector of constants determined by initial conditions. Analogous to the Taylor series for the exponential function ex, and knowing eA·0=I, the matrix exponential eAt is defined as: eAt = ∞ X k=0 (At)k k!=I+At +(At)2 2! +(At)3 3! +... and its derivative is: d dteAt =d dt I+At +(At)2 2! +(At)3 3! +...=A+A2t+A3t2 2+... =AeAt 2 Then, the relationship eAt =PeDtP−1comes from the process of diagonalizing the matrix A, i.e., if a matrix Acan be diagonalized, it means there exists an invertible matrix Pand a diagonal matrix Dsuch that: A=PDP−1 where Dis a diagonal matrix whose diagonal elements are the eigenvalues of A, and the columns of Pare the corresponding eigenvectors. Then, we can write: eAt =e(P DP −1)t=PeDtP−1 Since Dis a diagonal matrix, eDt is simply the exponential of each diagonal element: eDt =     eλ1t0· · · 0 0eλ2t· · · 0 . . .. . ..... . . 0 0 · · · eλnt      where λ1, λ2, . . . , λnare the eigenvalues of A. If a matrix Ais not diagonalizable, we can use its Jordan canonical form. For a matrix Athat is not diagonalizable, there exists an invertible matrix P and a Jordan matrix Jsuch that: A=PJP−1 where Jis a block diagonal matrix with Jordan blocks on the diagonal. Each Jordan block corresponds to an eigenvalue of Aand has the form: Jk=       λ1 0 · · · 0 0λ1· · · 0 . . .. . ........ . . 0 0 · · · λ1 0 0 · · · 0λ        The matrix exponential eAt can be computed using the Jordan form: eAt =e(P JP −1)t=P eJtP−1 The exponential of a Jordan block Jkcan be computed as: eJkt=eλt         1tt2 2! · · · tn−1 (n−1)! 0 1 t· · · tn−2 (n−2)! . . .. . ........ . . 0 0 · · · 1t 0 0 · · · 0 1         3 Given the system du dt =Au, the solution can be expressed as: u(t) = eAtu(0) = PeJtP−1u(0) This approach allows you to solve the system even when Ais not diagonalizable by leveraging the Jordan canonical form. Let’s suppose a matrix with eigenvalues {2,2,2,4}. Its Jordan form depends on the geometric multiplicity of each eigenvalue and the structure of the corresponding Jordan blocks. For example, if the geometric multiplicity of eigenvalue 2 is 2 (two linearly independent eigenvectors), the Jordan form is: J=    2100 0200 0020 0004     and the matrix exponential eJt for the entire Jordan form Jis: eJt =  eJ2t0 0 0e2t0 0 0 e4t =    e2t1t 0 10 0 0e2t0 0 0 e4t     =    e2tte2t0 0 0e2t0 0 0 0 e2t0 000e4t     If the geometric multiplicity of eigenvalue 2 is 3 (only one eigenvector), the Jordan form is: J=    2100 0210 0020 0004     and the matrix exponential eJt for the entire Jordan form Jis: eJt =eJ3t0 0e4t=    e2t  1tt2 2! 0 1 t 0 0 1  0 0e4t     =    e2tte2tt2 2! e2t0 0e2tte2t0 0 0 e2t0 0 0 0 e4t     Example 1 Consider the initial value problem: du dt =Au,u(0) = 8 5 where A=4−5 2−3. 4 To solve this system, we use the diagonalization method. First, we find the eigenvalues and eigenvectors of A. The eigenvalues are λ1=−1 and λ2= 2, and the corresponding eigenvectors are v1=1 1and v2=5 2. We can write Aas: A=PDP−1 where P=1 5 1 2, D =−1 0 0 2. The solution to the system is given by: u(t) = eAtu(0) = PeDtP−1u(0) First, compute P−1: P−1=1 det(P)2−5 −1 1 =1 2−52−5 −1 1 =−2 3 5 3 1 3−1 3 Next, compute eDt: eDt =eλ1t0 0eλ2t=e−t0 0e2t Now, combine these results: u(t) = PeDtP−1u(0) = 1 5 1 2e−t0 0e2t−2 3 5 3 1 3−1 38 5 Simplify the expression step-by-step: 1 5 1 2e−t0 0e2t=e−t5e2t e−t2e2t Then, e−t5e2t e−t2e2t−2 3 5 3 1 3−1 3=e−t−2 3+ 5e2t1 3e−t5 3+ 5e2t−1 3 e−t−2 3+ 2e2t1 3e−t5 3+ 2e2t−1 3 Finally, multiply by the initial condition: u(t) = 3e−t+ 5e2t 3e−t+ 2e2t= 3e−t1 1+e2t5 2 Thus, the solution to the initial value problem is: u(t)=3e−t1 1+e2t5 2 5 Example 2 Consider the initial value problem: du dt =Au,u(0) = 1 1 where A=1 1 −1 3. To solve this system, we first find the eigenvalues and eigenvectors of A: det(A−λI) = 0  1−λ1 −1 3 −λ = (1 −λ)(3 −λ)−(−1)(1) = λ2−4λ+ 4 = (λ−2)2= 0 Thus, the eigenvalue is λ= 2 (with algebraic multiplicity 2). Next, we find the eigenvectors corresponding to λ= 2: (A−2I)v= 0 −1 1 −1 1v1 v2=0 0 This gives v1=v2, so an eigenvector is v=1 1. Since Ahas a single eigenvalue with algebraic multiplicity 2 but only one linearly independent eigenvector, the matrix Adoes not diagonalize, and we need to find a generalized eigenvector wsuch that: (A−2I)w=v −1 1 −1 1w1 w2=1 1 Solving this, we get w1= 0 and w2= 1, so a generalized eigenvector is w=0 1. Therefore, the Jordan form Jof Ais: J=2 1 0 2 and the matrix Pof eigenvectors and generalized eigenvectors is: P=1 0 1 1 Which verifies A=P JP −1: A=1 0 1 12 1 0 21 0 1 1−1 =1 1 −1 3 6 Then, the matrix exponential eAt is given by: eAt =PeJtP−1 where eJt =e2t1t 0 1 So that: eAt =1 0 1 1e2t1t 0 11 0 −1 1=e2t1−t t −t1 + t Using the initial condition u(0) = 1 1: u(t) = eAtu(0) = e2t1−t t −t1 + t1 1=e2t1 1 Thus, the solution is: u(t) = e2t1 1 2.3 Solving Homogeneous Linear ODEs of Higher Order To solve a higher-order homogeneous linear ODE using matrix methods, we convert the ODE into a system of first-order linear differential equations. Consider a general n-th order homogeneous linear ODE: an dny dtn+an−1 dn−1y dtn−1+· · · +a1 dy dt +a0y= 0 1. Convert to a System of First-Order Equations. Define a vector usuch that: u=       y y′ y′′ . . . y(n−1)        Then, the system can be written as: u′=Au where Ais an n×nmatrix constructed from the coefficients of the ODE. 2. Find Eigenvalues and Eigenvectors. Solve the characteristic equation det(A−λI) = 0 to find the eigenvalues λand corresponding eigenvectors v. 7 3. The general solution to the system is: u(t) = C1eλ1tv1+C2eλ2tv2+· · · +Cneλntvn where λ1, λ2, . . . , λnare the eigenvalues, and v1,v2,...,vnare the corresponding eigenvectors. 4. Use the initial conditions to solve for the constants C1, C2, . . . , Cn. Example 3 Consider the second-order homogeneous linear ODE: y′′ + 4y′+ 3y= 0 with initial conditions y(0) = 3 and y′(0) = −7. First we create the System of First-Order Equations: u=y y′,u′=y′ y′′=0 1 −3−4y y′ Thus, u′=Au, A =0 1 −3−4 The characteristic equation is: det(A−λI) =  −λ1 −3−4−λ =λ2+ 4λ+ 3 = 0 Solving for λ: λ2+ 4λ+ 3 = (λ+ 1)(λ+ 3) = 0 =⇒λ1=−1, λ2=−3 For λ1=−1: (A+I)v1=1 1 −3−3v11 v12= 0 =⇒v1=1 −1 For λ2=−3: (A+ 3I)v2=3 1 −3−1v21 v22= 0 =⇒v2=1 −3 The General Solution is: u(t) = C1e−t1 −1+C2e−3t1 −3 We apply the initial conditions u(0) = 3 −7: 3 −7=C11 −1+C21 −3 8 Solving the system: (C1+C2= 3 −C1−3C2=−7 We obtain C1= 1 and C2= 2, and the solution is: u(t) = e−t1 −1+ 2e−3t1 −3 Thus: y(t) = e−t+ 2e−3t 3 Nonhomogeneous Linear ODEs 3.1 Solving u′=a·u+f(t) To solve the nonhomogeneous linear ODE u′=a·u+f(t), we use the method of integrating factors as follows. First, rewrite the equation in standard form: u′−a·u=f(t) Next, we find the integrating factor µ(t), a function that, when multiplied by the entire differential equation, allows it to be written in a form where the left-hand side is the derivative of a product of functions: µ(t) = eR−a dt =e−at Multiply both sides of the ODE by the integrating factor: e−atu′−ae−atu=e−atf(t) The left-hand side becomes the derivative of e−atu: d dt e−atu=e−atf(t) Instead of writing an indefinite integral and then solving for the constant of integration, we can apply the initial condition during integration by choosing limits that start at the initial time t= 0 and end at the current time t. This gives: Zt 0 d dτ e−aτ u(τ)dτ =Zt 0 e−aτ f(τ)dτ, so that e−atu(t)−e−a·0u(0) = Zt 0 e−aτ f(τ)dτ. Since e−a·0= 1 and u(0) = u0, we have: e−atu(t)−u0=Zt 0 e−aτ f(τ)dτ. 9