Section 116: Vector Functions and Space Curves vector
t 2 is de ned for t 2 The component function ln(8 t) is de ned for t
Method 1: Using the formulas
1 lnt 1 t2 1 t2 lnt+ 1 t2 = 1 t3 lnt+ 1 t3 + 1 t3 lnt = 1 t3: Now, note g(t) is the right-hand side when the equation is in the form y00+ p(t)y0+ q(t)y = g(t): Our equation is t2y00(t) + 3ty0(t) + y(t) = 1 t so we need to divide both sides by t2 Doing so, we obtain the equation y00(t) + 3 t y0(t) + 1 t2 y(t) = 1 t3: Now, we see that g(t) = 1
Properties of the Beurling generalized primes
lnt 1 t2 t (s 1) dt+ Z 1 p 1 ˇ(t) t2 (t 1) dt+ Z 1 p 1 t s lnt dt+ o(1) as s1+:The third integral can be simplified using the change of variables u= (s 1)lnt;followedbyintegrationbyparts Theresultingformulais Z 1 p 1 t s lnt dt= ln 1 s 1 lnlnp 1 + Z 1 0 e ulnudu+ o(1): However,itiswellknown(seeforinstance[4])that R 1 0 e ulnudu= 0(1
1
D Y(t) = 2 5 e t E Y(t) = 2 5 te t 2 The form of a particular solution to the equation 1t2 + C 2t3 + t4 lnt 1 4 A mass weighing 2 lb stretches a spring 6 in
Solutions to Practice Problems II
2 t2 1=2 2 lnt+ 1 t5=2 lnt Simplify the right hand side: 3t =2 lnt 2t3=2 t5=2 lnt = 1 t 2 tlnt Integrate both sides: Z dw w = Z 1 t 2 tlnt dt On the right hand side you will need to use the substitution a = lnt;da = 1 t dt: lnjwj= lnjtj 2lnjlntj Now solve for w But rst we need to combine the log terms on the right hand side: lnjwj= ln 1 t + ln
1(Line integrals{Using parametrization Two types and the ux
We parametrize the curve as ~x(t) = (t;lnt);1 t 2 We can compute that ds= j~x0jdt= p 1 + 1=t2dt Hence, the integral is 2 1 t2 p 1 + 1=t2dt= 2 1 t p t2 + 1dt= 5 2 p u 1 2 du= 1 3 u3=2ju=5 u=2 (c) Let F~ = ( 2y+ 2x;2x 2y) and Cis ~x(t) = (t;t2);0 t 1 Compute the work done by F~along the curve The eld is not conservative and the curve is not
Practice Problems to Midterm Exam 1
t 2 lnt 1 4 t 2 + C +1 pt Math 0230 - Spring 2021 - Tro mov Solutions to Practice Problems to Exam 1 Page 2 of 19 2 (10 points) Evaluate the integral I= Z x p 4 x2 dx
MATH 2202 – Exam 1 Solutions
= t2lnt¡ Z (t+tlnt)dt = t2lnt¡ Z tdt¡ Z tlntdt = t2lnt¡ 1 2 t2¡ Z tlntdt: So far, we have Z tlntdt=t2lnt¡ 1 2 t2¡ Z tlntdt: By adding R tlntdtto both sides of the above equation, we obtain 2 Z tlntdt=t 2lnt¡ 1 2 t: The nal result is Z tlntdt= 1 2 t2lnt¡ 1 4 t2+C: Here is the check: d dt µ1 2 t2lnt¡ 1 4 t2 ¶ = 1 2 t2¢ 1 t +lnt¢t
Section 31 Calculus of Vector-Functions
Section 3 1 Calculus of Vector-Functions De &nition A vector-valued function is a rule that assigns a vector to each member in a subset of R1: In other words, a vector-valued function is
Math 128A Spring 2003 Week 12 Solutions
t2y00 −2ty0 +2y = t3 lnt, 1 ≤ t ≤ 2, y(1) = 1, y0(1) = 0, with h = 0 1; actual solution y(t) = 7 4 t+ 1 2 t3 lnt
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Solutions to Practice Problems II
Written by Victoria Kala
vtkala@math.ucsb.eduSH 6432u Oce Hours: T 12:451:45pm
Last updated 7/12/2016
Answers
This page contains answers only. Detailed solutions are on the following pages.1. (a)y=e2x(c1cosp3x+c2sinp3x)
(b)y=c1+c2ex+c3xex (c)u=c1e12 t+c2e3t (d)r=c1es+c2es+c3coss+c4sins2.W= 2e6x
3. (a)y2=x2lnx
(b)W=x3 (c)y=c1x2+c2x2lnx4.y2=t1=2
5. (a)yp=At2+Bt+C
(b) Answers will vary (c)yp=At2+Bt+C+Ee2t (d)yp=Ae3tsin4t+Be3tcos4t (e) Answers will vary (f)yp=A+ (Bt2+Ct+E)et (g)yp=Acos2t+Bsin2t (h) Answers will vary (i)yp= (At+B) + (Ct+E)sint+ (Ft+G)cost6.y=c1+c2e2t+32
t12 sin2t12 cos2t7.y= 2e2t+ 9te2t+16
t3+32 t2e2t8.y=c1cost+c2sintcostlnjsect+ tantj:
9.y=c1cos(lnt) +c2sin(lnt) + lnjcos(lnt)j(cos(lnt)) + lnjtj(sin(lnt))
1Detailed Solutions
1. Find the general solution of the following homogeneous higher-order dierential equations:
(a)y00+ 4y0+ 7y= 0 (usexas the independent variable) Solution.The characteristic equation isr2+4r+7 = 0. Using the quadratic equation, we see the roots are r=4p424(1)(7)2
)r=4i2p3 2 =2ip3: We have two complex roots, hence the solution to the dierential equation is y=e2x(c1cosp3x+c2sinp3x)(b)y(3)+ 2y00+y0= 0 (usexas the independent variable) Solution.The characteristic equation isr3+ 2r2+r= 0 which factors as r(r2+ 2r+ 1) = 0)r(r+ 1)2= 0: Therefore the roots arer= 0 multiplicity 1,r=1 multiplicity 2. The solution to the dierential equation is therefore y=c1e0x+c2e1x+c3xe1xory=c1+c2ex+c3xex:(c) 2 d2udt25dudt
3u= 0 Solution.The characteristic equation is 2r25r3 = 0, which factors as (2r+ 1)(r3) = 0:The roots arer=12
;r= 3 each with multiplicity 1. The solution to the dierential equation is therefore u=c1e12 t+c2e3t:(d) d4rds 4r= 0 Solution.The characteristic equation ism41 = 0 (we cannot useras our characteristic equation variable since it is being used in the dierential equation). This factors as (m21)(m2+ 1) = 0)(m1)(m+ 1)(mi)(m+i) = 0: The roots arem= 1;1;i. The solution to the dierential equation isr=c1es+c2es+e0s(c3coss+c4sins))r=c1es+c2es+c3coss+c4sins:2. CalculateW(y1;y2;y3) wherey1=ex;y2=e2x;y3=e3x.
2Solution.
W(ex;e2x;e3x) =
e xe2xe3x (ex)0(e2x)0(e3x)0 (ex)00(e2x)00(e3x)00 e xe2xe3x e x2e2x3e3x e x4e2x9e3x =ex2e2x3e3x4e2x9e3x
exe2xe3x4e2x9e3x
+exe2xe3x2e2x3e3x
=ex(18e5x12e5x)ex(9e5x4e5x) +ex(3e5x2e3x) =ex(6e5x5e5x+e5x) = 2e6x3. (a) The functiony1=x2is a solution ofx2y003xy0+4y= 0. Use the method of reduction of order to nd a second solutiony2to the dierential equation on the interval (0;1). Proof.We need to write the dierential equation in standard form by dividing through byx2: y 003x y0+4x2y= 0 (1)
Lety2=vy1wherevis a function ofx. Sincey1=x2, theny2=x2v. We wish to plug this into the equation above, so we need to ndy02;y002using the product rule: y 2=x2v y02= (x2)0v+x2v0= 2xv+x2v0
y002= (2xv)0+ (x2v0)0= (2x)0v+ 2xv0+ (x2)0v0+x2v00= 2v+ 2xv0+ 2xv0+x2v00= 2v+ 4xv0+x2v00
Substitute these into the standard form equation (5) above:2v+ 4xv0+x2v003x
(2xv+x2v0) +4x2(x2v) = 0
)2v+ 4xv0+x2v006xv3xv0+ 4v= 0 )x2v00+xv0= 0 (2)Now letw=v0, thenw=v00and (6) becomes
x2w0+xw= 0)x2dwdx
=xw: This is a separable equation. Separate the variables: dww =1x dxIntegrate both sides:
Zdww =Z1x dx)lnw=lnx)lnw= ln1xSolve forw:
w=1xBut,w=v0, and so we have
v 0=1x )v=Z1x dx= lnxTherefore
y2=x2v)y2=x2lnx:3
(b) Show thaty1andy2form a fundamental set of solutions to the dierential equation. Solution.We need to show thatW(y1;y2)6= 0 in the interval (0;1).W(y1;y2) =x2x2lnx
(x2)0(x2lnx)0 =x2x2lnx2x2xlnx+x
=x2(2xlnx+x)2x(x2lnx) =x3W=x3which is nonzero since we are in the interval (0;1).(c) Write the general solution of the dierential equation usingy1andy2.
Solution.The general solution to the dierential equation isy=c1y1+c2y2, which is y=c1x2+c2x2lnx:4. Use reduction of order to nd a second solutiony2to the dierential equation4t2y00+y= 0
given thaty1=t1=2lntis a solution. Solution.Lety2=vy1=v(t1=2lnt). We need to ndy02;y002: y02=v0(t1=2lnt) +v(t1=2lnt))0=v0(t1=2lnt) +v12
t1=2lnt+t1=2 =v0(t1=2lnt) +v t 1=212 lnt+ 1 y002=v00(t1=2lnt) +v0(t1=2lnt)0+v0
t 1=212 lnt+ 1 +v t 1=212 lnt+ 1 0 =v00(t1=2lnt) + 2v0 t 1=212 lnt+ 1 +v 12 t3=212 lnt+ 1 +t1=212t =v00(t1=2lnt) + 2v0 t 1=212 lnt+ 1 +v 14 t3=2lntNow plug these into the original equation:
4t2y00+y= 4t2
v00(t1=2lnt) + 2v0
t 1=212 lnt+ 1 +v 14 t3=2lnt +v(t1=2lnt) = 4v00t5=2lnt+ 8t2v0 t 1=212 lnt+ 1 t2v t3=2lnt +v(t1=2lnt) = 4v00t5=2lnt+ 8t2v0 t 1=212 lnt+ 1Now set this equal to 0:
4v00t5=2lnt+ 8t2v0
t 1=212 lnt+ 1 = 0Letw=v0, thenw0=v00:
4w0t5=2lnt+ 8t2w
t 1=212 lnt+ 1 = 0 4 This is a separable equation. We can rewritew0asdwdx and getw's on one side,t's on the other: 4 dwdt t5=2lnt=8t2w t 1=212 lnt+ 1 dww =2t2t1=212 lnt+ 1t5=2lnt
Simplify the right hand side:
t3=2lnt2t3=2t5=2lnt=1t
2tlntIntegrate both sides:
Zdww =Z 1t 2tlnt dt On the right hand side you will need to use the substitutiona= lnt;da=1t dt: lnjwj=lnjtj 2lnjlntj Now solve forw. But rst we need to combine the log terms on the right hand side: lnjwj= ln1t + ln1(lnt)2 lnjwj= ln1t(lnt)2Take the exponential of both sides:
w=1t(lnt)2But,w=v0, so
v=Z1t(lnt)2dt and using the substitutiona= lnt;da=1t dt, v=Z1a2da=1a
=1lnt:Therefore
y2=v2t1=2lnt=1lntt1=2lnt=t1=2:5. Suppose you are solving the equationy00+P(t)y0+Q(t)y=g(t), whereP(t) andQ(t) are constants,
using the method of undetermined coecients. Complete the table below. Assumeg(t) has no function in common with the homogeneous solutionyh. List your solutions for (a)(i). g(t)Form ofypg(t)Form ofyp3t22(a)1t2et(f) (b)Ae5t3cos2t(g)
6t2+ 212e2t(c)(h)At+Be
3xsin4x(d)4t(1 + 3sint)(i)
(e)(At+B)e3tSolution.(a)g(t) is a second degree polynomial, so we must guess a second degree polynomial:
y p=At2+Bt+C 5 (b)ypis an exponential function, sogmust have also been an exponential function:g= 100e5t (answers will vary) (c)g(t) is a second degree polynomial added to an exponential function, so we must guess a second degree polynomial added to an exponential function:yp=At2+Bt+C+Ee2t.(d)g(t) is an exponential function multiplied by a sine function, so we must also guess an exponential
function multiplied by a sine function (we will also have to include a cosine function since sine and cosine come in pairs):yp=Ae3tsin4t+Be3tcos4t (e)ypis a rst degree polynomial multiplied by an exponential function, sogmust have been a rst degree polynomial multiplied by an exponential function:g= 18te3t(answers will vary)(f)g(t) is a constant term added to a second degree polynomial multiplied by an exponential function,
so we guess the same thing:yp=A+ (Bt2+Ct+E)et(g)g(t) is a cosine function, so we must also guess a cosine function (we will also have to include a
sine function since sine and cosine come in pairs):yp=Acos2t+Bsin2t (h)ypis a rst degree polynomial, sogmust have also been a rst degree polynomial:g=t+ 3 (answers will vary) (i) Distribute rst: g= 4t+ 12tsint gis a rst degree polynomial added to a rst degree polynomial multiplied by a sine function, so we must also guess the same thing (we will also have to include a cosine function since sine andcosine come in pairs):yp=At+B+ (Ct+E)sint+ (Ft+G)cost6. Solve the given dierential equation using the method of undetermined coecients.
y00+ 2y0= 3 + 4sin2t:
Solution.We rst nd the homogeneous solution:
y00+ 2y0= 0
has the characteristic equation r2+ 2r= 0)r(r+ 2) = 0r= 0;r=2
therefore the homogeneous solution isyh=c1+c2e2t. Now we guess a particular solution.g(t) = 3+4sin2t, which is a constant term added to a sine term. We must also guess the same thing, but we will need to include a cosine term as well since sine and cosine come in pairs: y p=A+Bsin2t+Ccos2t: Now we compare withyhto check if there are any solutions in common. Notice thatyphas a constant term in common withyh(c1), so we must multiply our constantAbyt(\bump it up") so there is no more overlap: y p=At+Bsin2t+Ccos2t: We compare again withyh, and we see there are no more solutions in common, so this is our nal guess.Findy0p;y00p:
y0p=A+ 2Bcos2t2Csin2t
y00p=4Bsin2t4Ccos2t
6Substitute these values into our equation above:
y00+ 2y0=4Bsin2t4Ccos2t+ 2(A+ 2Bcos2t2Csin2t)
=4Bsin2t4Ccos2t+ 2A+ 4Bcos2t4Csin2tSet this equal tog(t):
4Bsin2t4Ccos2t+ 2A+ 4Bcos2t4Csin2t= 3 + 4sin2t
Set like terms equal to each other:
2A= 34B4C= 4
4C+ 4B= 0
The solution to this system isA=32
;B=12 ;C=12 , therefore the particular solution isyp= 32t12 sin2t12 cos2t. The general solution isy=yh+yp: y=c1+c2e2t+32 t12 sin2t12 cos2t:7. Solve the given initial-value problem. y