hw2Sols

# hw2Sols - HOMEWORK 2 MATH 265 L Keshet Final version Due in...

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

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: HOMEWORK 2: MATH 265, L Keshet ( Final version ) Due in class on September 29, 2010 NOTE: Most problems on this assignment are straightforward. Problem 4 may take a bit more time and effort. Problem 1: In each case, solve the following second order ODEs for y ( t ): (a) y ′′ + 2 y ′- 3 y = 0 , and y (0) = 1 , y ′ (0) = 2 (b) y ′′- 9 y ′ + 20 y = 0 and y (0) = 1 , y ′ (0) = 0 (c) y ′′- 2 y ′ + 5 y = 0 and y (0) = 1 , y ′ (0) = 1. (d) y ′′- 2 y = 0 and y (0) = 0 , y ′ (0) = 2. Solution to Problem 1: (a) The characteristic equation is r 2 + 2 r- 3 = 0. This factors into ( r + 3)( r- 1) = 0 so has solutions r = 1 ,- 3. The general solution is thus y ( t ) = C 1 e t + C 2 e − 3 t . We find C 1 , C 2 from the initial conditions. We need to find y ′ ( t ) by differentiating y ( t ): we get y ′ ( t ) = C 1 e t- 3 C 2 e − 3 t . Using the initial conditions, we have y (0) = 1 , ⇒ 1 = C 1 + C 2 and y ′ (0) = 2 ⇒ 2 = C 1- 3 C 2 . Solving these equations leads to C 2 =- 1 / 4 and C 1 = 5 / 4 so the solution is y ( t ) = 5 4 e t + 1 4 e − 3 t (b) The characteristic equation is r 2- 9 r + 20 = ( r- 4)( r- 5) = 0, so the roots are r = 4 , 5 and the general solution is y ( t ) = C 1 e 4 t + C 2 e 5 t . Then the initial conditions mean that y (0) = 1 = C 1 + C 2 and y ′ (0) = 0 = 4 C 1 + 5 C 2 . Solving these for C 1 , C 2 leads to y ( t ) = 5 e 4 t- 4 e 5 t . (c) The characteristic equation is r 2- 2 y + 5 = 0. This has the complex roots r = 1 ± 2 i so the general solution is y ( t ) = C 1 e t sin(2 t ) + C 2 e t cos(2 t ). We also need the derivative y ′ ( t ) = e t ( C 1 sin(2 t ) + 2 C 1 cos(2 t ) + C 2 cos(2 t )- 2 C 2 sin(2 t )). Then we use the initial conditions: 1 = y (0) = C 2 and 1 = (2 C 1 + C 2 ) (we used the facts that e = 1 , sin(0) = 0 , cos(0) = 1). This tells us that C 2 = 1 , C 1 = 0 so the solution is y ( t ) = e t cos(2 t ). (d) Characteristic equation: r 2- 2 = 0 so r = ± √ 2 and y ( t ) = C 1 e √ 2 t + C 2 e − √ 2 t . The derivative: y ′ ( t ) = √ 2 C 1 e √ 2 t- √ 2 C 2 e − √ 2 t . Using the I.C’s: y (0) = 0 = C 1 + C 2 and y ′ (0) = 2 = √ 2 C 1- √ 2 C 2 . Solving for constants leads to y ( t ) = √ 2 2 e √ 2 t- √ 2 2 e − √ 2 t . Problem 2: Consider the differential equation ay ′′ + by ′ + cy = 0. Suppose that the two functions y = f 1 ( t ) and y = 1 2 [ f 2 ( t ) + f 1 ( t )] are both solutions to this equation. Show that the function f 2 ( t ) is also a solution....
View Full Document

{[ snackBarMessage ]}

### Page1 / 5

hw2Sols - HOMEWORK 2 MATH 265 L Keshet Final version Due in...

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

View Full Document
Ask a homework question - tutors are online