0 then this amounts to solving the following system

This preview shows 3 out of 4 pages.

= 0, then this amounts to solving the following system of equations: u 0 1 e x + u 0 2 ( e x + xe x ) = e x 1 + x 2 u 0 1 e x + u 0 2 xe x = 0 If we subtract the second equation from the first, we get u 0 2 e x = e x / (1+ x 2 ), or u 0 2 = 1 / (1+ x 2 ). Integrating this shows that u 2 = arctan x . On the other hand, substituting in u 0 2 = 1 / (1+ x 2 ) into the second equation gives u 0 1 e x + xe x / (1 + x 2 ) = 0, or u 0 1 = - x/ (1 + x 2 ). Integrating this shows that u 1 = - 1 / 2 ln(1 + x 2 ). Putting it all together, the general solution of the non-homogeneous equation is given by y + y p = c 1 e x + c 2 xe x - 1 / 2 ln(1+ x 2 ) e x + xe x arctan x . c) y 00 + 4 y 0 + 4 y = e - 2 x x 3 . First, we solve the homogeneous equation y 00 + 4 y 0 + 4 y = 0. The characteristic equation is r 2 + 4 r + 4 = 0, which factors as ( r + 2) 2 = 0, so r = - 2. The general solution of the homogeneous equation is y = c 1 e - 2 x + c 2 xe - 2 x . Now, we try to find a particular solution of the non-homogeneous equation of the form y p = u 1 e - 2 x + u 2 xe - 2 x , where u 1 and u 2 are functions of x . If we require that u 0 1 e - 2 x + u 0 2 xe - 2 x = 0, then this amounts to solving the 3
Image of page 3

Subscribe to view the full document.

following system of equations: - 2 u 0 1 e - 2 x + u 0 2 ( e - 2 x - 2 xe - 2 x ) = e - 2 x x 3 u 0 1 e - 2 x + u 0 2 xe - 2 x = 0 If we multiply the second equation by 2 and add the result to the first equation, we get u 0 2 e - 2 x = e - 2 x /x 3 , or u 0 2 = 1 /x 3 . Integrating this shows that u 2 = - 1 / (2 x 2 ). On the other hand, substituting in u 0 2 = 1 /x 3 into the second equation gives u 0 1 e - 2 x + xe - 2 x /x 3 = 0, or u 0 1 = - 1 /x 2 . Integrating this shows that u 1 = 1 /x . Putting it all together, the general
Image of page 4
You've reached the end of this preview.
  • Fall '08
  • Reshetiken
  • u2, Heterogeneity, Homogeneity, general solution, 1 2 g

{[ 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