midterm3sol

# midterm3sol - Solutions to the Second Midterm Exam Problem...

This preview shows pages 1–3. 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: Solutions to the Second Midterm Exam Problem 1. a) We look for a solution in the form y = ae 2 x for some a . Substituting y into the equation, we get 4 ae 2 x + 2 ae 2 x + ae 2 x = e 2 x , from which 7 a = 1 and a = 1 / 7. Answer: A particular solution is y = e 2 x 7 . b) First, we look for a particular solution in the form y = ae 2 x for some a . Thus we get y ′′ = 4 ae 2 x and the equation becomes 4 ae 2 x- 4 ae 2 x = e 2 x , which is a signature equation of duplication (resonance). Hence we try y = axe 2 x instead, so y ′ = e 2 x ( a + 2 ax ) and y ′′ = e 2 x (2 a + 4 ax ) + 2 ae 2 x = e 2 x (4 a + 4 ax ). Substituting that into the equation, we get e 2 x (4 a + 4 ax )- 4 axe 2 x = 4 ae 2 x = e 2 x , from which a = 1 / 4. Answer: A particular solution is y = xe 2 x 4 . c) First, we solve the homogeneous equation y ′′- y = 0 . Looking for the solution in the form y = e rx we get r 2- 1 = 0, from which the general solution is y = C 1 e x + C 2 e − x for some constants C 1 and C 2 . Now, we look for the solution of the original equation in the form y = C 1 ( x ) e x + C 2 ( x ) e − x . Differentiating, we get y ′ = C ′ 1 e x + C 1 e x + C ′ 2 e − x- C 2 e − x , 1 which, after the imposing the condition (1.1) C ′ 1 e x + C ′ 2 e − x = 0 becomes just y ′ = C 1 e x- C 2 e − x . Differentiating further, we get y ′′ = C ′ 1 e x + C 1 e x- C ′ 2 e − x + C 2 e − x . Hence (1.2) y ′′- y = C ′ 1 e x- C ′ 2 e − x = xe 2 x . Now we solve (1.1) and (1.2) for C ′ 1 and C ′ 2 . If we add (1.1) and (1.2), we get 2 C ′ 1 e x = xe 2 x , so C ′ 1 = xe x / 2. Similarly, if we subtract (1.2) from (1.1), we get 2 C ′ 2 e − x =- xe 2 x , so C ′ 2 =- xe 3 x / 2. Integrating by parts, we get C 1 ( x ) = 1 2 integraldisplay xe x dx = xe x 2- e x 2 + constant and C 2 ( x ) =- 1 2 integraldisplay xe 3 x dx =- xe 3 x 6 + e 3 x 18 + some other constant. Since we are interested in particular solutions, we can choose the constants equal to 0 and get y = C 1 ( x ) e x + C 2 ( x ) e − x = xe 2 x 2- e 2 x 2- xe 2 x 6 + e 2 x 18 = xe 2 x 3- 4 e 2 x 9 . Answer: A particular solution is y ( x ) = e 2 x parenleftbigg x 3- 4 9 parenrightbigg . d) From the general equation we have mx ′′ + kx = 0, where m = 10 and k is determined from the equation F = kx with F = 5 and x = 1, so k = 5. Thus the equation reads 10 x ′′ + 5 x = 0 or, equivalently, x ′′ + 1 2 x = 0, from which the general solution is x ( t ) = A sin t √ 2 + B cos t √ 2 . Hence the resonance occurs at ω = 1 / √ 2 and ω =- 1 / √ 2. Answer: The resonance occurs at ω = 1 / √ 2, and, if we assume that ω can be negative, at ω =- 1 / √ 2....
View Full Document

### Page1 / 9

midterm3sol - Solutions to the Second Midterm Exam Problem...

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

View Full Document
Ask a homework question - tutors are online