dstefan
07-02-2008, 06:46 PM
Chapter 6, Problem 8
Show that the ODE y''(x) - 2 y'(x) + x^2 y(x) ~=~ 0 can be written as Y'(x) ~=~ f(x,Y(x)), where
Y(x) ~=~ \left( \begin{array}{c} y(x) \\ y'(x) \end{array} \right); f(x,Y(x)) ~=~ \left(\begin{array}{cc}0 & 1 \\-x^2 & 2 \end{array} \right) Y(x)
Solution:
Note that y''(x) = 2 y'(x) - x^2 y(x). Then,
Y'(x) = \left( \begin{array}{c} y'(x) \\ y''(x) \end{array} \right) ~=~ \left( \begin{array}{c} y'(x) \\ 2 y'(x) - x^2 y(x) \end{array} \right) = \left( \begin{array}{cc}0 & 1 \\-x^2 & 2 \end{array} \right) \left( \begin{array}{c} y(x) \\ y'(x) \end{array} \right) = \left( \begin{array}{cc} 0 & 1 \\ -x^2 & 2 \end{array} \right) Y(x)
Show that the ODE y''(x) - 2 y'(x) + x^2 y(x) ~=~ 0 can be written as Y'(x) ~=~ f(x,Y(x)), where
Y(x) ~=~ \left( \begin{array}{c} y(x) \\ y'(x) \end{array} \right); f(x,Y(x)) ~=~ \left(\begin{array}{cc}0 & 1 \\-x^2 & 2 \end{array} \right) Y(x)
Solution:
Note that y''(x) = 2 y'(x) - x^2 y(x). Then,
Y'(x) = \left( \begin{array}{c} y'(x) \\ y''(x) \end{array} \right) ~=~ \left( \begin{array}{c} y'(x) \\ 2 y'(x) - x^2 y(x) \end{array} \right) = \left( \begin{array}{cc}0 & 1 \\-x^2 & 2 \end{array} \right) \left( \begin{array}{c} y(x) \\ y'(x) \end{array} \right) = \left( \begin{array}{cc} 0 & 1 \\ -x^2 & 2 \end{array} \right) Y(x)