We will also show how to sketch phase portraits associated with real repeated eigenvalues (improper nodes). Repeated Roots and Reduction of Order. differential equations. In this section we will solve systems of two linear differential equations in which the eigenvalues are real repeated (double in this case) numbers. The first one is easy enough, since it's not part of the homogeneous solution. Rearranging and factoring gives the indicial equation + (−) + = We then solve for m. There are three particular cases of interest: Case #1 of two distinct roots, m 1 and m 2; Case #2 of one real repeated root, m; Case #3 of complex roots, α ± βi. The example below demonstrates the method. After solving the characteristic equation the form of the complex roots of r1 and r2 should be: λ ± μi. We have already addressed how to solve a second order linear homogeneous differential equation with constant coefficients where the roots of the characteristic equation are real and distinct. Here is a set of practice problems to accompany the Repeated Roots section of the Second Order Differential Equations chapter of the notes for Paul Dawkins Differential Equations course at Lamar University. Now that we know how to solve second order linear homogeneous differential equations with constant coefficients such that the characteristic equation has distinct roots (either real or complex), the next task will be to deal with those which have repeated roots.We proceed with an example. used in solving second order constant coefficient differential equations There are three types of roots, Distinct, Repeated and Complex, which determine which of the three types of general solutions is used in solving a problem. We will also derive from the complex roots the standard solution that is typically used in this case that will not involve complex numbers. We will now explain how to handle these differential equations when the roots are complex. Phase portrait for repeated eigenvalues Subsection 3.5.2 Solving Systems with Repeated Eigenvalues ¶ If the characteristic equation has only a single repeated root, there is a single eigenvalue. It's simply $$ y_1(x) = a_1 x^3 + a_2 x^2 + a_3 x + a_4 $$ The second one is a polynomial multiplied with a solution of the homogeneous equation, so we should guess a solution of the same form. In case #1, the solution is Repeated Eignevalues Again, we start with the real 2 × 2 system. x = Ax. (1) We say an eigenvalue λ 1 of A is repeated if it is a multiple root of the char acteristic equation of A; in our case, as this is a quadratic equation, the only possible case is when λ 1 is a double real root. Distinct Real Roots If the roots have opposite sign, the graph will be have a saddle point where only two asymptotic We refer back to the characteristic equation, we then assume that all the solution to the differential equation will be: y(t) = e^(rt) By plugging in our two roots into the general formula of the solution, we get: y1(t) = e^(λ + μi)t Repeated Eigenvalues 1. This will include deriving a second linearly independent solution that we will need to form the general solution to the system. If roots are real and repeated ( ), natural solution becomes 4 2 0 2 ... a1=-3 →s1 =s2 a1=2. The form of the general solution varies depending on whether the characteristic equation has distinct, real roots; a single, repeated real root; or complex conjugate roots.