This preview has intentionally blurred sections. Sign up to view the full version.
View Full 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
 Spring '07
 Stenstones?

Click to edit the document details