[PDF] Math 54 Selected Solutions for Week 11 Section - Berkeley Math PDF Homework%2011%20-%20Solutions.pdf

Determine whether the given functions are linearly dependent or linearly independent on the specified interval Justify your decision 1sinx,cosx,tanxl on (- π/2,

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[PDF] Math 54 Selected Solutions for Week 11 Section - Berkeley Math

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Math 54. Selected Solutions for Week 11

Section 4.6 (Page 435)

7. Find a general solution to the dierential equation

00+ 4y0+ 4y=e2tlnt

using the method of variation of parameters. The characteristic polynomial isr2+4r+4 = (r+2)2, so a fundamental solution set isy1=e2t,y2=te2t. One therefore needs to solve the system e

2tv01+te2tv02= 0 ;

2e2tv01+ (12t)e2tv02=e2tlnt :

Adding two times the rst equation to the second gives e

2tv02=e2tlnt ;

sov02= lnt, and (from the rst equation)v01=tlnt. Integrating gives v 1=t22 lnt+t24 andv2=tlntt :

Therefore

y p= t22 lnt+t24 e

2t+ (tlntt)te2t

lnt2 34
t 2e2t:

The general solution is therefore

y=lnt2 34
t

2e2t+c1e2t+c2te2t:

Section 6.1 (Page 482)

2. Determine the largest interval (a;b) for which Theorem 1 guarantees the existence of

a unique solution on (a;b) to the given initial value problem: y

000pxy= sinx;y() = 0; y0() = 11; y00() = 3:

Since pxis only dened forx0, the largest possible interval is (0;1). 1 2

4. Determine the largest interval (a;b) for which Theorem 1 guarantees the existence of

a unique solution on (a;b) to the given initial value problem: x(x+ 1)y0003xy0+y= 0 ;y(1=2) = 1; y0(1=2) =y00(1=2) = 0: We have to divide the equation byx(x+ 1) to get the coecient ofy000to equal

1 (which the theorem requires). This function is zero atx= 0 andx=1 (and

nowhere else), so the largest interval is (1;0).

10. Determine whether the given functions are linearly dependent or linearly independent

on the specied interval. Justify your decision. fsinx;cosx;tanxgon (=2;=2): They are linearly independent. To show this, we use the method of Example 3.

Suppose that

1sinx+c2cosx+c3tanx= 0

for allx2(=2;=2). Plugging inx= 0 givesc2= 0, leavingc1sinx+c3tanx= 0.

Plugging in two other values gives

c 12 +c3p3 = 0 (x==3) ; c 1p2 +c3= 0 (x==4):

Since the matrix

1=2 1=p3

1=p2 1

is invertible, this forcesc1=c3= 0, so the functions are linearly independent.

12. Determine whether the given functions are linearly dependent or linearly independent

on the specied interval. Justify your decision. fcos2x;cos2x;sin2xgon (1;1): We have cos2x= cos2xsin2xon (1;1), so the functions are linearly de- pendent.

34.Constructing Dierential Equations.Given three functionsf1(x),f2(x),f3(x) that

are each three times dierentiable and whose Wronskian is never zero on (a;b), show that the equationf

1(x)f2(x)f3(x)y

01(x)f02(x)f03(x)y0

001(x)f002(x)f003(x)y00

0001(x)f0002(x)f0003(x)y000

= 0 3 is a third-order linear dierential equation for whichff1;f2;f3gis a fundamental so- lution set. What is the coecient ofy000in this equation? The functiony=f1satises the dierential equation because the rst and fourth columns of the matrix are equal, so the determinant is zero. Similarly,y=f2and y=f3are also solutions. This equation is a linear dierential equation because you can expand about the last column to get an expression C

44y000+C34y00+C24y0+C14y= 0;

where the cofactorsCi4are functions ofxnot involvingy. The coecient ofy000is just the Wronskian off1;f2;f3, and we are given that it is never zero, so the equation must be of third order (the termC44y000does not disappear), and we can divide by the Wronskian and apply Theorem 3 to nd thatf1,f2, andf3form a fundamental set of solutions of the dierential equation.

Section 6.2 (Page 488)

14. Find a general solution for the dierential equation

y (4)+ 2y000+ 10y00+ 18y0+ 9y= 0 withxas the independent variable. [Hint:y(x) = sin3xis a solution.] The auxiliary polynomial isr4+2r3+10r2+18r+9. We are given that sin3xis a solution, which suggests that3iare roots. In fact, dividing byr2+ 9 works out, and we have the factorization (r2+ 9)(r2+ 2r+ 1) = (r2+ 9)(r+ 1)2. Therefore the general solution isy=c1sin3x+c2cos3x+c3ex+c4xex.

Section 9.1 (Page 503)

4. Express the system of dierential equations in matrix notation:

01=x1x2+x3x4;

02=x1+x4;

x 03=px 1x3; x

04= 0:

2 6 4x 1 x 2 x 3 x 43
7 50
=2 6

411 11

1 0 0 1p01 0

0 0 0 03

7 52
6 4x 1 x 2 x 3 x 43
7 5. 4

12. Express the given system of higher-order dierential equations as a matrix system in

normal form: x

00+ 3x0y0+ 2y= 0;

00+x0+ 3y0+y= 0:

Lettingx1=x,x2=x0,x3=y, andx4=y0gives

2 6 4x 1 x 2 x 3 x 43
7 50
=2 6

40 1 0 0

032 1

0 0 0 1

01133
7 52
6 4x 1 x 2 x 3 x 43
7 5:quotesdbs_dbs17.pdfusesText_23

[PDF] [PDF] Math 54 Selected Solutions for Week 11 Section - Berkeley Math

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