Find the solution to ivps of the form x ax x t x find

This preview shows page 3 - 7 out of 22 pages.

Find the solution to IVPs of the form: x 0 = Ax , x ( t 0 ) = x 0 . Find the general solution to a 2 × 2 system of linear ODEs with constant matrix A . Must be familiar with: 1. Jordan form method, 2. Undetermined coefficients, 3. Variation of parameters, and 4. Laplace transform. 3
Image of page 3

Subscribe to view the full document.

SUMMARY An Ordinary Differential Equation (ODE) is an equation whose variable is a differen- tiable function and that involves the function and its derivatives: F ( x, y, y 0 , y 00 , . . . , y ( n ) ) = 0 . The order of an ODE is the order of the highest derivative that appears in the equation. For instance, y 0 + y = x is first order while x 3 y 00 - 10 x 7 y 0 = y 4 is second order. A first order ODE of the form y 0 = f ( y ) is called Autonomous . Notice that Autonomous equations are always separable, how- ever the integration involved may be very complicated (if not impossible). One can in any case study solutions by looking at the direction (slope) field for the differential equation. A constant solution to an Autonomous ODE is called an equilibrium solution . These are the zeroes of the function f ( y ). An equilibrium solution is called semistable if as t approaches infinity solutions on one side of the equilibrium solution approach the equilibrium solution while solutions on the other side become further and further away from the equilibrium solution. There are special kinds of ODEs that we know how to solve, these are: 1. Separable: First order ODEs of the form y 0 = F ( x ) G ( y ) These can be solved by dividing both sides by G ( y ) and then integrating with respect to x : Z y 0 ( x ) G ( y ( x )) dx = Z F ( x ) dx, the left hand side is actually the integral with respect to y since if y = y ( x ) then dy = y 0 ( x ) dx , and so the equation becomes Z 1 G ( y ) dy = Z F ( x ) dx. 4
Image of page 4
2. Linear: First order ODEs of the form y 0 + a ( x ) y = b ( x ) . (1) These can be solved by multiplying by the integrating factor μ ( x ) = exp Z a ( x ) dx . Notice that μ 0 ( x ) = a ( x ) μ ( x ) and so multiplying (1) by μ ( x ) we obtain μ ( x ) b ( x ) = μ ( x )( y 0 + a ( x ) y ) = μ ( x ) y 0 + μ 0 ( x ) y = d dx ( μ ( x ) y ) . Thus, y = 1 μ ( x ) Z μ ( x ) b ( x ) dx + C 3. Exact: First order ODEs M ( x, y ) + N ( x, y ) y 0 = 0 (2) that can be written as d dx (Ψ( x, y ( x ))) = 0 (3) for some differentiable function Ψ( x, y ). If it happens that M, N, ∂M ∂y , ∂N ∂x are all continuous in some rectangle then such function Ψ exists if and only if ∂M ∂y = ∂N ∂x If an equation is exact, then the solutions are given by Ψ( x, y ) = c where c is a constant. To find Ψ( x, y ) we notice that for (2) to be equal to (3) we require Ψ ∂x = M, and Ψ ∂y = N. 5
Image of page 5

Subscribe to view the full document.

Therefore Ψ( x, y ) = Z M ( x, y ) dx + C ( y ) for some function C ( y ), depending only on y , and that we determine from the condition Ψ ∂y = N . Some ODEs may not fall directly into any of the groups above, but after a modification or substitution they can be solved by the same methods. 1. Bernoulli Equations: ODEs of the from y 0 + a ( x ) y = b ( x ) y n .
Image of page 6
Image of page 7

{[ 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