202s04mt1 - B U Department of Mathematics Math 202...

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: B U Department of Mathematics Math 202 Differential Equations Date: April 9, 2004 Full Name : Time: 18:10-19:25 Math 202 Number : Student ID : Spring 2004 First Midterm - Solution Key IMPORTANT 1. Write your name, surname on top of each page. 2. The exam consists of 4 questions some of which have more than one part. 3. Read the questions carefully and write your answers neatly under the corresponding questions. 4. Show all your work. Correct answers without sufficient explanation might not get full credit. 5. Calculators are not allowed. Q1 Q2 Q3 Q4 total 20 pts 25 pts 30 pts 25 pts 100 pts 1.) [20] Solve (sec 2 y ) y + tan y 1 + x = 1 √ 1 + x by using the substitution u = tan y . Solution: Using the given substitution we get: u = tan y ⇒ u = (sec 2 y ) y . The transformed DE for u reads: u + u 1 + x = 1 √ 1 + x , which obviously linear (but not separable). Hence we have to find an integrating factor μ ( x ) by: μ ( x ) = exp Z dx 1 + x = 1 + x. Multiplying the equation by this μ ( x ) factor we get (1 + x ) u + u = √ 1 + x , so that the left hand side becomes a total derivative: [(1 + x ) u ] = √ 1 + x ⇒ (1 + x ) u = 2 3 (1 + x ) 3 / 2 + c. Leaving u alone on the left we get the u function to be: u = 2 3 √ 1 + x + c 1 + x , which then implies after going back to y : y = arctan 2 3 √ 1 + x + c 1 + x . 2.) Consider the differential equation y (4) + 6 y 000 + 9 y 00 = f ( t ). (a) [15] Find the general solution of this differential equation when f ( t ) = 50 e 2 t + 18. Solution: We first need to find the complementary solution. This is a constant coefficient linear DE, so we can use the characteristic equation: r 4 + 6 r 3 + 9 r 2 = 0 which is equivalent to r 2 ( r +3) 2 = 0. We have the roots to be r = 0 ,- 3 both double. Hence the complementary solution is: y c = c 1 + c 2 t + c 3 e- 3 t + c 4 te- 3 t . (Alternatively: you can set y 00 = u and reduce the order by 2, i.e. solve u 00 + 6 u + 9 u = 0 then find y after two integrations.) Now we can look for a particular solution and the most appropriate way to do this is the method of undetermined coefficients. So set y p = Ae 2 t...
View Full Document

This note was uploaded on 11/16/2011 for the course MATH 251 taught by Professor Gurel during the Winter '11 term at Boğaziçi University.

Page1 / 5

202s04mt1 - B U Department of Mathematics Math 202...

This preview shows document pages 1 - 3. Sign up to view the full document.

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