Solving this system of equations yields a 3 25 and b

Info icon This preview shows pages 2–4. Sign up to view the full content.

View Full Document Right Arrow Icon
= 0. Solving this system of equations yields A = 3 / 25 and B = 4 / 25. Hence, the general solution of the non-homogeneous equation is y + y p = c 1 e - x + c 2 xe - x + 3 / 25 e x cos x + 4 / 25 e x sin x . c) y 00 + 2 y 0 + y = xe - x . 2
Image of page 2

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

View Full Document Right Arrow Icon
First, we solve the homogeneous equation y 00 +2 y 0 + y = 0. We already found the general solution y = c 1 e - x + c 2 xe - x in part b. Now, we try to find a particular solution of the non-homogeneous equation. Since G ( x ) = xe - x , then a good candidate is y p = ( Ax + B ) e - x . However, this is a solution of the homogenous equation, so we might try y p = ( Ax 2 + Bx ) e - x . But it turns out that this doesn’t work either! Finally, we try y p = ( Ax 3 + Bx 2 ) e - x . We have y 0 p = (3 Ax 2 + 2 Bx ) e - x - ( Ax 3 + Bx 2 ) e - x = (3 Ax 2 + 2 Bx ) e - x - y p and y 00 p = (6 Ax + 2 B ) e - x - (3 Ax 2 +2 Bx ) e - x - y 0 p = (6 Ax +2 B ) e - x - ( y 0 p + y p ) - y 0 p = (6 Ax +2 B ) e - x - 2 y 0 p - y p . So, we have (6 Ax + 2 B ) e - x - 2 y 0 p - y p + 2 y 0 p + y p = xe - x (6 Ax + 2 B ) e - x = xe - x Therefore, 6 A = 1 and 2 B = 0, so A = 1 / 6 and B = 0. Hence, the general solution of the non-homogeneous equation is y + y p = c 1 e - x + c 2 xe - x + 1 / 6 x 3 e - x . 3. Use variation of parameters to find the general solution: a) y 00 - 2 y 0 + y = e x 1 + x 2 . First, we solve the homogeneous equation y 00 - 2 y 0 + y = 0. The characteristic equation is r 2 - 2 r + 1 = 0, which factors as ( r - 1) 2 = 0, so r = 1. The general solution of the homogeneous equation is y = c 1 e x + c 2 xe x . Now, we try to find a particular solution of the non-homogeneous equation of the form y p = u 1 e x + u 2 xe x , where u 1 and u 2 are functions of x . If we require that u 0 1 e x + u 0 2 xe x
Image of page 3
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern