{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Fall 2006 - Hall's Class - Practice Exam 2

# Fall 2006 - Hall's Class - Practice Exam 2 - Math 20D...

This preview shows pages 1–3. Sign up to view the full content.

Math 20D Practice Exam II Solutions 1. (a) (5 points) Find the general solution to the homogeneous differ- ential equation y 00 + 4 y 0 + 4 y = 0 . y = c 1 e - 2 t + c 2 te - 2 t (b) (8 points) Use the method of Undetermined Coefficients to find a particular solution to the non-homogeneous differential equation y 00 + 4 y 0 + 4 y = 2 t + 1 . Guess Y = at + b . Plug in to get 4 a + 4 at + 4 b = 2 t + 1, i.e., a = 1 / 2 , b = - 1 / 4 Y = (1 / 2) t - 1 / 4 (c) (8 points) Use the method of Variation of Parameters to find a particular solution to the non-homogeneous differential equation y 00 + 4 y 0 + 4 y = e - 2 t 1 - t 2 . (Hint: R 1 1 - x 2 dx = arcsin( x ) + C ) W ( y 1 , y 2 ) = e - 2 t te - 2 t - 2 e - 2 t e - 2 t - 2 te - 2 t = e - 4 t u 0 1 = - y 2 g ( t ) W ( y 1 ,y 2 ) = - t 1 - t 2 u 1 = 1 - t 2 u 0 2 = y 1 g ( t ) W ( y 1 ,y 2 ) = 1 1 - t 2 u 2 = sin - 1 t Y = 1 - t 2 e - 2 t + (sin - 1 t ) te - 2 t (d) (4 points) Write the general solution to the the non-homogeneous differential equation y 00 + 4 y 0 + 4 y = 2 t + 1 + e - 2 t 1 - t 2 . y = c 1 e - 2 t + c 2 te - 2 t + (1 / 2) t - 1 / 4 + 1 - t 2 e - 2 t + (sin - 1 t ) te - 2 t 2. Consider the initial value problem 2 t 2 y 00 + ty 0 - 3 y = 0 , y (1) = - 2 , y 0 (1) = 1 . 1

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

View Full Document
(a) (5 points) Without solving the initial value problem, explain why its solution will only be defined for t > 0. Writing the equation in standard form y 00 + 1 2 t y 0 - 3 2 t 2 y = 0, we see that the coefficient functions are not defined at t = 0. Since the initial value of t is t 0
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}