Download Exam 1 Solution | Differential Equations and Partial Differential Equations | MA 30300 and more Exams Differential Equations in PDF only on Docsity! Math 303-Exam 1-Solution Name: Student I.D.: 1. (10 pts) Determine a lower bound for the radius of convergence of series solutions (x2 − 4x+ 5)y′′ + (x+ 3)y′ + 4(x2 − 4x+ 5)y = 0 about x0 = 1. Solution) Note that x0 = 1 is an ordinary point and p(x) = x+ 3 x2 − 4x+ 5 , q(x) = 4. The roots of x2 − 4x + 5 = 0 are 2 + i and 2 − i. The distance from x0 = 1 to the nearest root 2 + i is √ 2 (you may take 2− i as well), and so the radius of convergence of p(x) is ρp = √ 2. The radius of convergence of q(x) is ρq =∞. Therefore, min{ √ 2,∞} = √ 2 and so we find that the radius of convergence ρ of the series solution is at least √ 2, which is a lower bound. 2. Consider the differential equation (x− 2)2(x+ 1)y′′ + 3(x2 + x− 6)y′ + (4x+ 1)y = 0. (a) (8 pts) Show that x0 = 2 is a regular singular point. Solution) Clearly, x0 = 2 is a singular point because (x− 2)2(x+ 1) = 0 =⇒ x = 2,−1. Use x2 + x− 6 = (x− 2)(x+ 3) and compute p0 = lim x→2 (x− 2)p(x) = lim x→2 (x− 2)3(x− 2)(x+ 3) (x− 2)2(x+ 1) = lim x→2 3(x+ 3) (x+ 1) = 5 and q0 = lim x→2 (x− 2)2q(x) = lim x→2 (x− 2)2 4x+ 1 (x− 2)2(x+ 1) = lim x→2 4x+ 1 x+ 1 = 3. Therefore x0 = 1 is a regular singular point. (b) (7 pts) Find the indicial equation of a series solution of the form y = φ(r, x) = ∞∑ n=0 an(r)(x− 2)r+n and also find the exponents at the singular point x0 = 2. Solution) The indicial equation is 0 = r(r − 1) + p0r + q0 = r(r − 1) + 5r + 3 =⇒ r2 + 4r + 3 = 0 =⇒ r = −1,−3 and the exponents of singularity at x0 = 2 are r = −1,−3. 3. Consider a series solution y = ∑∞ n=0 anx n about x0 = 0 of y′′ − xy′ − 2y = 0. (a) (10 pts) Find the recurrence relation for an. (b) (5 pts) Find a general formula for an. (c) (5 pts) Find two linearly independent series solutions. Solution) (a) Compute y′ = ∞∑ n=1 nanx n−1, y′′ = ∞∑ n=2 n(n− 1)anxn−2 1 2 and put y, y′, y′′ into the differential equation to find that 0 = y′′ − xy′ − 2y = ∞∑ n=2 n(n− 1)anxn−2 − x ∞∑ n=1 nanx n−1 − 2 ∞∑ n=0 anx n = ∞∑ n=2 n(n− 1)anxn−2 − ∞∑ n=1 nanx n − 2 ∞∑ n=0 anx n Use the shifting formula n→ n+ 2 ∞∑ n=2 n(n− 1)anxn−2 = ∞∑ n=0 (n+ 2)(n+ 1)an+2xn to get 0 = ∞∑ n=0 (n+ 2)(n+ 1)an+2xn − ∞∑ n=1 nanx n − 2 ∞∑ n=0 anx n = (2a2 − 2a0) + ∞∑ n=1 [ (n+ 2)(n+ 1)an+2 − (n+ 2)an ] and { a2 = a0 (n+ 2)(n+ 1)an+2 − (n+ 2)an = 0, n ≥ 1 (b) The recurrence relation can be simplified to an+2 = 1 n+ 1 an, n ≥ 1. Considering even and odd cases we see that a2m = a0 1 · 3 · 5 · · · (2m− 3) · (2m− 1) , m ≥ 1 and a2m+1 = a1 2 · 4 · 6 · · · (2m− 2) · (2m) , m ≥ 0. (c) The general solution is y = ∞∑ n=0 anx n = a0 + ∞∑ m=1 a2mx 2m + ∞∑ m=0 a2m+1x 2m+1 = a0 + ∞∑ m=1 a0 1 · 3 · 5 · · · (2m− 3) · (2m− 1) x2m + ∞∑ m=0 a1 2 · 4 · 6 · · · (2m− 2) · (2m) x2m+1 and so y1(x) = 1 + ∞∑ m=1 1 1 · 3 · 5 · · · (2m− 3) · (2m− 1) x2m and y2(x) = ∞∑ m=0 1 2 · 4 · 6 · · · (2m− 2) · (2m) x2m+1 are two linearly independent solutions. 4. (15 pts) Use the Laplace transform to solve the initial value problem y′′ − 7y′ + 12y = 0, y(0) = 1, y′(0) = 0.
