{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Assignment5-solution-public

# Assignment5-solution-public - Math 201 Assignment#5 Due...

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

Math 201 Assignment #5 Due: November 1 in tutorial TAs: grade problems 2 and 3. 15 marks each. Give effort marks for 1, 4, 5, and 6, 5 marks each fi shown effort to finish, or 0 if left empty. 1. Solve the initial value problem 4 y 00 - 4 y 0 + y = 0 , y (0) = 1 , y 0 (0) = 1 . Auxiliary equation: 4 λ 2 - 4 λ + 1 = (2 λ - 1) 2 = 0 , λ = 1 / 2 , repeated root y 1 = e x/ 2 , y 2 = xy 1 = xe x/ 2 General solution: y = c 1 y 1 + c 2 y 2 = c 1 e x/ 2 + c 2 xe x/ 2 Determine c 1 and c 2 using initial conditions: y (0) = c 1 = 1 , y 0 ( x ) = c 1 2 e x/ 2 + c 2 e x/ 2 + c 2 x 2 e x/ 2 , y 0 (0) = c 1 2 + c 2 = 1 , c 2 = 1 / 2 y = e x/ 2 + 1 2 xe x/ 2 2. Find the general solution of y 000 + y 0 = xe x . Homogeneous part: y 000 + y 0 = 0 Auxiliary equation: λ 3 + λ = λ ( λ 2 + 1) = 0 , λ = 0 , λ = ± i . Three independent solutions: * From λ = 0: y 1 = e λ 1 t = e 0 t = 1 , * From λ = ± i , y 2 = e 0 t cos t = cos t , y 3 = e 0 t sin t = sin t . y c = c 1 y 1 + c 2 y 2 + c 3 y 3 = c 1 + c 2 cos t + c 3 sin t Particular solution: Guess for y p = ( ax + b ) e x y 0 p = ae x + ( ax + b ) e x = ( ax + a + b ) e x , y 00 p = ae x + ( ax + a + b ) e x = ( ax + 2 a + b ) e x , y 000 p = ( ax + 3 a + b ) e x , y 000 p + y 0 p = ( ax + 3 a + b ) e x + ( ax + a + b ) e x = (2 ax + 4 a + 2 b ) e x = xe x Thus, 2 ax + 4 a + 2 b = x , comparing the coefficients, 2 a = 1 , 4 a + 2 b = 0 , a = 1 / 2 , b = - 1 .

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.

{[ snackBarMessage ]}

### Page1 / 4

Assignment5-solution-public - Math 201 Assignment#5 Due...

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

View Full Document
Ask a homework question - tutors are online