Q4_1130 - Can y = sin ( t 2 ) be a solution on an interval...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
MATH 415.01 QUIZ 4A Feb 3, 2010 SHOW ALL YOUR WORK! GOOD LUCK! NAME: 1. (6.5 pts.) In the following differential equation, determine the values of α , if any, for which all solutions tend to zero as t → ∞ . y 00 + (4 - α ) y 0 - 3( α - 1) y = 0 r 2 + (4 - α ) r - 3( α - 1) = 0 .i.e. ( r + 3)( r - ( α - 1)) = 0. Thus, r 1 = - 3 , r 2 = ( α - 1). General Solution is y = c 1 e - 3 t + c 2 e ( α - 1) t Obviously, e - 3 t 0 as t → ∞ . y will tend to zero if we also have e ( α - 1) t 0. This is true if ( α - 1) < 0. i.e. α < 1 2. (6 pts.) Find the Wronskian of 2 x and xe x . W = ± ± ± ± 2 x xe x 2 e x + xe x ± ± ± ± = 2 x ( e x + xe x ) - 2 xe x = 2 x 2 e x
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
MATH 415.01 QUIZ 4B Feb 3, 2010 SHOW ALL YOUR WORK! GOOD LUCK! NAME: 1. (6.5 pts.) Find the solution of the initial value problem y 00 - y = 0 , y (0) = 5 2 , y 0 (0) = 3 2 r 2 - 1 = 0 i.e. ( r - 1)( r + 1) = 0. So, r 1 = 1 ,r 2 = - 1. Thus, general solution is y = c 1 e t + c 2 e - t Then, y 0 = c 1 e t - c 2 e - t . Use initial conditions, c 1 + c 2 = 5 / 2 c 1 - c 2 = 3 / 2 Thus, c 1 = 2 and c 2 = 1 / 2 Finally solution becomes y = 2 e t + 1 2 e - t 2. (6 pts.)
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Can y = sin ( t 2 ) be a solution on an interval containing t = 0 of an equation y 00 + p ( t ) y + q ( t ) y = 0 with continuous coecients? Explain your answer. NO. We have y ( t ) = sin ( t 2 ), then y ( t ) =-2 t cos ( t 2 ). So, y (0) = 0 and y (0) = 0. Now, assume that, y ( t ) = sin ( t 2 ) is a solution of an equation y 00 + p ( t ) y + q ( t ) y = 0. Then, it should also satisfy initial conditions y (0) = 0 and y (0) = 0. But, this initial value problem has a UNIQUE solution y = 0. (check your notes). There is only one solution y = 0 and obviously sin ( t 2 ) 6 = 0 . Hence, y ( t ) = sin ( t 2 ) is NOT a solution of this dierential equation....
View Full Document

Page1 / 2

Q4_1130 - Can y = sin ( t 2 ) be a solution on an interval...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online