MATH 4B Final Exam Study Guide

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

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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

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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
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

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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 .
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