Solving Fourth-Order Linear Differential Equation with Given Initial Conditions, Quizzes of Linear Algebra

Download Solving Fourth-Order Linear Differential Equation with Given Initial Conditions and more Quizzes Linear Algebra in PDF only on Docsity! SOLVING LINEAR EQUATIONS 𝑦(4) − 6𝑦′′′ + 7𝑦′′ + 6𝑦′ − 8𝑦 = 0 𝑦(0) = −2, 𝑦′(0) = −8, 𝑦′′(0) = −14, 𝑦′′′(0) = −62 The characteristic polynomial is: 𝑟4 − 6𝑟3 + 7𝑟2 + 6𝑟 − 8 = 0 The constant term ir 𝒂𝟎 = −𝟖. From the rational root theorem, the possible roots are ±1, ±2, ±4, ±8. Then 𝑓1(1) = 1 − 6 + 7 + 6 − 8 = 0 so by the remainder theorem, 𝑟 = 1 is a lot. Dividing 𝑓1(1) by 𝑟 − 1 using synthetic division, we obtain: 1 -6 7 6 -8 1 1 -5 2 8 1 -5 2 8 0 Let 𝑓2(𝑟) = 𝑟3 − 5𝑟2 + 2𝑟 + 8. checking for 𝑟 = 1, 𝑓2(1) = 1 − 5 + 2 + 8 ≠ 0, so we now discard 𝑟 = 1, Moving on to the next wot, 𝑟 = −1, we obtain 1 -5 2 -8 -1 -1 6 8 -1 -6 8 0 Let 𝑓3(𝑟) = 𝑟2 − 6𝑟 + 8 = 0. Checking for 𝑟 = −1, 𝑓3(−1) = 1 − 6 + 8 + 0.50 we finally discard 𝑟 = −1. Since the function is quadratic, it will be inefficient to use the Rational root. By factoring, we obtain 𝑓4(𝑟) = (𝑟 − 4)(𝑟 − 2) Equating both factors to zero, we get the last two roots 𝑟 = 2 and 𝑟 = 4. Therefore, the roots are: −1,1,2,4. Hence, the general solution is given by: 𝑦 = 𝑐1𝑒−𝑥 + 𝑐2𝑒𝑥 + 𝑐3𝑒2𝑥 + 𝑐4𝑒4𝑥 The first, second and third derivatives of 𝑦 are: