Ma 221 - Differential Equations Homework 4 Name (Printed): Due: Oct 6, 2016 Recitation: Collaborators: I pledge my honor that I have abided by the Stevens Honor System. Sign: General Instructions: Write up solutions to the following set of questions and submit in recitation on the date indicated. Please staple this cover sheet to your solution pages. Legibility, organization of the solution, and clearly stated reasoning where appropriate are all important. Points will be deducted for disorganized work or insufficient explanations. Collaboration with classmates is acceptable and encouraged but all students must write up the solutions on their own. Collaborators (up to groups of three) should be identified on the top of the front page. All submitted work must be pledged and signed. 1. Find a homogeneous linear ODE with constant coefficients whose general solution is, y = C1 cos 2x + C2 sin 2x + C3 e2x + C4 e2x . 2. Find the general solution to y 000 + 6y 0 20y = 0 if one of the solutions is known to be y1 (x) = ex sin(3x). 3. Use Variation of Parameters to solve the initial value problem, y 00 4y = te2t , y(0) = 1/32, y 0 (0) = 0 . 4. Find the general solution to the differential equation y 00 + 4y 0 + 4y = e2t ln t. 1) The given general solution is Now And And implies the roots of auxiliary equation are implies the roots of auxiliary equation is implies the roots of auxiliary equation is Therefore, auxiliary equation is Therefore, the corresponding differential equation is 2) The given differential equation is Or we can write Here the auxiliary equation is ---- (1) Given that one of the solutions is So, the corresponding value of Since we know that complex root exits in pair. Therefore, another value of So, we have To find the third factor of , Divide , we get Hence, (1) can be written as Therefore, the general solution is Where are arbitrary constants 3) The given differential equation is --- (1) Compare (1) from we get Or we can write Here the auxiliary equation is So, the complementary solution is Where are arbitrary constants Let Then of Then particular solution is Where And Therefore, is Hence, the general solution is ---- (2) Where are arbitrary constants And So, ---- (3) Given that So, put in (2), we get --- (4) Given that So, put in (3), we get ---- (5) Adding (4) and (5), we get And Hence, required solution is 4) The given differential equation is --- (1) Compare (1) from we get Or we can write Here the auxiliary equation is So, the complementary solution is Where are arbitrary constants Let Then of Then particular solution is Where is And Therefore, Hence, the general solution is Where are arbitrary constants 1) The given general solution is Now And And implies the roots of auxiliary equation are implies the roots of auxiliary equation is implies the roots of auxiliary equation is Therefore, auxiliary equation is Therefore, the corresponding differential equation is 2) The given differential equation is Or we can write Here the auxiliary equation is ---- (1) Given that one of the solutions is So, the corresponding value of Since we know that complex root exits in pair. Therefore, another value of So, we have To find the third factor of , Divide , we get Hence, (1) can be written as Therefore, the general solution is Where are arbitrary constants 3) The given differential equation is --- (1) Compare (1) from we get Or we can write Here the auxiliary equation is So, the complementary solution is Where are arbitrary constants Let Then of Then particular solution is Where And Therefore, is Hence, the general solution is ---- (2) Where are arbitrary constants And So, ---- (3) Given that So, put in (2), we get --- (4) Given that So, put in (3), we get ---- (5) Adding (4) and (5), we get And Hence, required solution is 4) The given differential equation is --- (1) Compare (1) from we get Or we can write Here the auxiliary equation is So, the complementary solution is Where are arbitrary constants Let Then of Then particular solution is Where is And Therefore, Hence, the general solution is Where are arbitrary constants