list_of_concepts_test2-mat_275 - MAT 275 TEST 2 LIST OF...

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MAT 275 TEST 2 - LIST OF CONCEPTSExistence, Uniqueness, Fundamental sets for second order linear differential equations(Section 3.2)Know how to determine whether two functions of one variable are linearly independent or dependent on a given interval by determining whether one of them is a constant multiple of the other or not.A second order linear differential equation in standard form has the form: y ' 'pty 'qty=gt.If gt=0,it is called homogeneous (HODE); otherwise non-homogeneous.Know Theorem 3.2.1: Existence and Uniqueness. Consider the Initial Value Problem y ' 'pty 'qty=gt,yt0=y0,y 't0=y'0where p, q and g are continuous on an open intervalI that contains t0.Then there is exactly one solution of this problem, and the solution exists throughout the interval I.Know Theorem 3.2.2: Principle of Superposition. Ify1andy2are solutions of the homogeneousdifferential equation y ' 'pty 'qty=0,so is c1y1c2y2for any constantc1andc2.Know Theorem 3.2.4: Wronskian of Solutions: Given two solutionsy1andy2of a homogeneous ODE, always check whether the Wronskian of the two solutions is noteverywhere zero. In this is case the two solutions are linearly independent and we say they form a Fudamental Set of Solutions for the ODE. The general solution is given byc1y1c2y2withc1andc2arbitrary constants.Linear Homogeneous DEs with constant coefficients (Sections 3.1, 3.3, 3.4)Second order, linear, homogeneous, constant coefficient ODE:Given the homogeneous ODE a y ' 'b y 'c y=0,with a, band cconstant:Letr1andr2be the roots of the characteristic equation, ar2brc=0If r1andr2are real and distinct, then the general solution of the homogeneous ODE is y=c1er1tc2 er2tIfr1=r2=r ,then the general solution isy=c1c2tert.If the roots of the characteristic equation,r1andr2are complex conjugate a±bi ,then the general solution is y=c1ea tcosbtc2eatsinbt

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