Title Test 19 - Solving Differential Equations, Newton's Law solutions_vF.pdf ✅

Test 19 - Differential Equations and Newton’s Law Question 1 2 marks Find the value(s) of k ∈ R for which y = kex 2 is a solution of d 2 y d x2 + d y d x = xex 2 . A. 0 B. 1 2 C. 2 D. 0 or 1 2 E. Such a real constant does not exist Solution: First, find the derivative with a higher order. d 2 y d x2 = d d x μ d d x 3 kex 2 ́ ¶ = d d x 3 2kxex 2 ́ = 2kex 2 +4kx2 e x 2 Substitute d 2 y d x2 = 2kex 2 +4kx2 e x 2 and d y d x = 2kxex 2 into the differential equation. d 2 y d x2 + d y d x = xex 2 2kex 2 +4kx2 e x 2 +2kxex 2 = xex 2 2k +4kx2 +2kx = x This equation must be true for all x, so the coefficients must be equal. By compar- ing the coefficients of each power of x, a system of equations is formed:    2k = 0 4k = 0 2k = 1 This is an inconsistent system of equations, so it has no solutions. This means that there is no constant k that will make y a solution of the differential equation. The correct answer is E . Specialist Maths 3/4 © EdAtlas | Page 1 © EdAtlas