Exercises 4.415.For this problem, the corresponding homogeneous equation isy0−5y0+6y= 0, which hasthe associated auxiliary equationr2−5r+ 6 = 0. The roots of this equation arer=3andr= 2. Therefore, neithery=exnory=xexsatis±es the homogeneous equation, and in theexpressionyp(x)=xs(Ax+B)exfor a particular solution we can takes=0. Soyp(x)=(Ax+B)ex⇒y0p(xAx+B+A)ex⇒y0p(xAx+B+2A)ex⇒(Ax+BA)ex−5(Ax+B+A)ex+6(Ax+B)ex=xex⇒(2Ax−3AB)ex=xex⇒2A=1,−3AB=0⇒A/2,B=3/4,andyp(xx/2+3/4)ex.16.The corresponding homogeneous equation has the auxiliary equationr2−1 = 0, whose rootsarer=±1. Thus, in the expressionθp(tA1t+A0)cost+(B1t+B0)sintnone of theterms is a solution to the homogeneous equation. We ±ndθp(tA1t+A0tB1t+B0t⇒θ0p(tA1cost−(A1t+A0t+B1sintB1t+B0t=(B1t+A1+B0t−A1t−A0+B1t⇒θ0p(tB1cost−(B1t+B0+A1t−A1sint−
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at Berkeley.