# 7 - ENM510 Foundations of Engineering Mathematics...

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ENM510 - Foundations of Engineering Mathematics I (Homework #7)Fall Semester, 2015M. Carchidi––––––––––––––––––––––––––––––––––––Problem #1 (25 points)Consider the second-order, linear, homogeneous ordinary differential equa-tion,y00(x) +P(x)y0(x) +Q(x)y(x) = 0.a.)(7 points)Show thaty1(x) =erx(forra constant) is a solution if and onlyif eitherr±, withr±(x) =12¾is constant.
b.)(7 points)Show thaty1(x) =xr(forra constant) is a solution if and onlyif eitherr±, withr±(x) =12¾is constant.
c.)(11 points)Use parts (a) and (b) above to determine a general solution tothe ordinary differential equation
Problem #2 (15 points)Determine the general solution tox(x+ 1)
––––––––––––––––––––––––––––––––––––Problem #3 (20 points)Construct the differential equation having a general solution of the formy(x) =c1xacos(λebx) +c2xasin(λebx)and use your result to determine a general solution to the ordinary differen-tial equation2xy00(x)(x4)y0(x) + (xex1)y(x) = 0for0< x.––––––––––––––––––––––––––––––––––––Problem #4 (15 points)a.)(5 points)Determine a general solution to the ordinary diff
b.)(10 points)Determine the form of the differential equation that has ageneral solution ofy(x) = (c1xn+c2xm)eaxand use your result to determine a general solution to9x2y006x2y0+ (x2+ 2)y= 0for0x.––––––––––––––––––––––––––––––––––––2
––––––––––––––––––––––––––––––––––––Problem #5 (25 points)Use the methods discussed in class to determine a general solution to eachof the following ordinary differential equations:a.)(5 points)xy00(12x2+ 1)y0+ 36x3y= 0for0< x,b.)(5 points)xy00(x+ 1)2y0+ (x2+x+ 1)y= 0for0< x,c.)(5 points)
d.)(5 points)xy00+ 2(x+ 1)y0+ (xλ2+ 2)y= 0for0< x, andλreal,e.)(5 points)y002xy0+ (λ2+x2)y= 0for−∞< x <+and forλreal.
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––––––––––––––––––––––––––––––––––––Solution to Problem #1a.) If we sety1(x) =erx(forrconstant) into the differential equationy00(x) +P(x)y0(x) +Q(x)y(x) = 0,(1.1)we get³r2+rP(x) +Q(x)´erx= 0,orr2+rP(x) +Q(x) = 0.Solving forryieldsr=12½P(x)±qP2(x)4Q(x)¾(1.2)which must therefore be constant. Since the steps above are reversible, wesee that ifrin Equation (1.2) is constant, theny1(x) =erxis a solution toEquation (1.1).b.) If we sety1(x) =xr(forrconstant) into Equation (1.1) we getr(r1)xr2+rxr1P(x) +Q(x)xr= 0,or³r(r
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