ID # : October 14, 2008

Tutorial # : Instructors: Dr. J.-P. Gabardo

Dr. Z. V. Kovarik Test duration: 1 hour

If you don’t remember your tutorial number, note that they are scheduled as follows: Tutorial 01: Thursday 9:30 am -10:30 am

Tutorial 02: Tuesday 5:30 pm- 6:20 pm

Instructions: You must use permanent ink. Tests submitted in pencil will not be consid-ered later for remarking. This exam consists of 8 problems on 10 pages (make sure you have all 10 pages). The last page is for scratch or overflow work. The total number of points is 50. Do not add or remove pages from your test. No books, notes, or “cheat sheets” allowed. The only calculator permitted is the McMaster Standard Calculator, the Casio fx 991. There is a formula that might be useful on the last page. GOOD LUCK!

### SOLUTIONS

# Mark

PART I: Multiple choice. Indicate your choice very clearly. There is only one correct answer in each multiple-choice problem. Circle the letter (a,b,c,d or e) corresponding to your choice. Ambiguous answers will be marked as wrong.

1. *(4 pts.)* Let*y(x) be the unique solution to the initial value problem*

(

*y0* _{= (2}* _{−}_{y)}*2

_{(y}

_{+ 1) (e}

*y*

_{−}_{1)}

_{,}*y(0) =a.*

For which two values of *a* below is it true that lim

*x→*+*∞* *y(x) = 2.*

(Hint: Do not try to solve the ODE)
(a) *a*= 0 and *a*= 1

*→* (b) *a*= 2 and *a*= 0.1
(c) *a*= 1.9 and*a* = 2.1
(d) *a* =*−1 and* *a*= 0.5
(e) *a* = 1.7 and *a*=*−0.7*

Solution. The critical points are *−1, 0 and 2. The right-hand side of the DE is positive*
for *y >*2, 0*< y <*2 and *y <−1 and negative for* *−1< y <* 0. Thus we have

lim

*x→*+*∞* *y(x) = 2*

2. *(4 pts.)* Find the general solution of the differential equation
*y*(5)_{−}_{4}* _{y}*(3)

_{+}

*(2)*

_{y}

_{−}_{4}

_{y}_{= 0.}The answer (where

*C*1

*, . . . , C*5 denote arbitrary constants) is: (a)

*y(x) =*

*C*1

*ex*+

*C*2 cos(x) +

*C*3 sin(x) +

*C*4

*e−x*

*√*

3*/*2 _{cos(x/2) +}* _{C}*
5

*e−x*

*√*

3*/*2_{sin(x/2)}
(b) *y(x) =* *C1* cos(2*x) +C2* sin(2*x) +C3ex*_{+}* _{C4}_{e}−x/*2

_{cos(x}

*√*

_{3) +}

*2*

_{C5}_{e}−x/_{sin(x}

*√*

_{3)}(c)

*y(x) =*

*C*1

*e−x*+

*C*2

*x e−x*+

*C*3

*e*2

*x*+

*C*4 cos(x

*√*

3/2) +*C*5 sin(x
*√*

3/2)
(d) *y(x) =* *C*1*ex*+*C*2*e−x*+*C*3*e−*2*x*+*C*4*e−x* cos(x

*√*

3) +*C*5*e−x*sin(x
*√*

3)

*→* (e) *y(x) =* *C1e−x*_{+}* _{C2}_{e}*2

*x*

_{+}

*2*

_{C3}_{e}−*x*

_{+}

*2*

_{C4}_{e}x/_{cos(x}

*√*

_{3/2) +}

*2*

_{C5}_{e}x/_{sin(x}

*√*

_{3/2)}

Solution. The auxiliary equation is

*m*5*−4m*4+m2*−4 = 0 or (m*3+1) (m2*−4) = 0 or (m+1) (m*2*−m+1) (m+2) (m−2) = 0*
The roots of *m*2_{−}_{m}_{+ 1 = 0 are} _{m}_{=} 1

2 *±*

*√*

3

2 *i. We find thus 3 real roots−1, 2,* *−2 and two*
complex roots 1

2 *±*

*√*

3

2 *i, all simple. The general solution is thus*

*y(x) =C*1*e−x*+*C*2*e*2*x*+*C*3*e−*2*x*+*C*4*ex/*2 cos(x
*√*

3/2) +*C*5*ex/*2sin(x
*√*

3. *(4 pts.)* Let*a* be a real number and let*y(x) be the unique solution of the initial value*

problem: _{(}

2* _{dx}dy* =

*−y*3

_{cos}

_{x}*y(0) =a.*

Find a value of*a*such that the largest interval containing*x*= 0 where the solution is defined
and continuous is the interval (−π/2, 3*π/2).*

*→* (a) *a*= 1
(b) *a*= 1/2
(c) *a* =*√*2
(d) *a*= 1/3
(e) *a*= 0

Solution. Using the method of separable equations, we can solve the DE:
*−2y−*3 *dy*

*dx* = cos*x*

or Z

*−2y−*3*dy*=

Z

cos*x dx.*
Therefore,

*y−*2 = sin*x*+*C,* where *C* =*a−*2*.*
and

*y*=*±√* 1
sin*x*+*a−*2
For*a*= 1, the solution is*y* = * _{√}* 1

sin*x*+1. For*a*= 1/2, the solution is *y*=
1

*√*

sin*x*+4. For *a*=
*√*

2,
the solution is *y* = * _{√}* 1

sin*x*+1*/*2. For *a* = 1/3, the solution is *y* =
1

*√*

4. *(4 pts.)* A model for the velocity *v* of a sprinter is given by the differential equation
*dv*

*dt* =*F*0*−k*0*v,*

where *F*0 is the force (assumed constant) exerted by the sprinter and *k*0 is the coefficient of
friction due to air resistance. Suppose that*k*0 = ln 2. If the race begins with a standing start
(i. e. *v(0) = 0) and the velocity of the sprinter is 4 m/s after 1 second, find the approximate*
velocity at the end of the race

*v∞*= lim

*t→∞* *v(t).*

(a) *v∞*= 6.5

(b) *v∞*= 7

(c) *v∞*= 7.5

*→* (d) *v∞*= 8

(e) *v∞*= 8.5

Solution. The DE is linear:

*dv*

*dt* + ln 2*v* =*F*0
The integrating factor is

*u(t) =* *e*ln 2R 1*dt* =*e*ln 2*t*= 2*t*
We have thus

*dv*
*dt* *e*

ln 2*t*_{+ ln 2}* _{v e}*ln 2

*t*

_{=}

_{F}0*e*ln 2*t* or
*d*
*dt*

©

*v e*ln 2*t*ª_{=}_{F}

0*e*ln 2*t.*

Integration yields

*v e*ln 2*t*_{=} *F*0
ln 2 *e*

ln 2*t*_{+}_{C}

and since *v(0) = 0,* *C* =*−F*0

ln 2. Thus,

5. *(4 pts.)* Let*y(x) be the unique solution to the initial value problem*

½

*y0* _{= 2 + sin(1/y)}

*y(0) = 1/π.*

Then, the value of the second derivative of *y(x) evaluated atx*= 0 is:
(Hint: Do not try to solve the D.E.)

(a) *y00*_{(0) = 2}_{−}_{4}_{π}

(b) *y00*_{(0) =}* _{−1}_{−}_{π}*2

*(c)*

_{/2}*y00*(0) = 3 +

*π*

*→*(d)

*y00*

_{(0) = 2}

*2 (e)*

_{π}*y00*

_{(0) = 3}

_{π}Solution. Note first that *y0*_{(0) = 2 + sin(1/y(0)) = 2 + sin(π) = 2. Using the chain rule}

and the original DE,

*y00* _{= cos(1/y)}
µ

*−y0*
*y*2

¶

and thus

*y00*(0) =*−*cos(π) 2

Part II: Provide all details and fully justify your answer in order to receive credit.

6. *(10 pts.)* Use the method of undetermined coefficients to compute in explicit form
the solution of the initial value problem

*y00*_{+ 4}_{y}_{= cos(x)}

with initial conditions *y(0) =y0*_{(0) = 0.}

Solution. The auxiliary equation is *m*2 _{+ 4 = 0 with solutions} _{m}_{=} _{±2}_{i}_{(both simple).}
The complementary solution is thus

*yc*(x) = *c*1 cos(2*x) +c*2 sin(2*x),* *c*1*, c*2 arbitrary.

Using the method of undetermined coefficients, a particular solution has the form
*yp*(x) =*A* cos*x*+*B* sin*x.*

Since

*y00 _{p}* + 4

*yp*=

*−A*cos

*x−B*sin

*x*+ 4 (A cos

*x*+

*B*sin

*x) = 3A*cos

*x*+ 3

*B*sin

*x,*

it follows that *A*= 1/3 and*B* = 0. The general solution is thus

*y*=*yp* +*yc*=

1

3 cos*x*+*c*1 cos(2*x) +c*2 sin(2*x)*
and

*y0* =*−*1

3 sin*x−*2*c1* sin(2*x) + 2c2* cos(2*x).*
The condition *y(0) = 0 yields* 1

3 +*c*1 = 0 or *c*1 = *−*13, while the condition *y0*(0) = 0 yields
2*c*2 = 0 or *c*2 = 0. The solution of the IVP is thus

*y*= 1

3 cos*x−*
1

7. (a) *(4 pts.)* Consider the initial value problem

(

(x2_{−}_{9)}_{y}00_{+ (x}_{+ 1)}_{y}0_{+}_{y}_{= sin}_{x}

*y(0) = 0, y0*_{(0) = 0.}

Find the largest open interval *I* containing *x* = 0 where you can say for sure that the
solution is unique, well-defined and twice differentiable, without having to actually compute
it. Explain.

Solution. In standard form the DE is written as
*y00*_{+} *x*+ 1

*x*2_{−}_{9}*y*

*0*_{+} 1

*x*2 _{−}_{9}*y*=
sin*x*
*x*2_{−}_{9}

and the coefficients of *y* and *y0* _{as well as the right-hand side are discontinuous at} _{x}_{=}_{±3.}

According to the existence and uniqueness theorem for linear differential equations, we are
guaranteed the existence of a solution on the largest open interval containing 0 but not*±3.*
This is the interval

*I* = (−3,3) *.*

(b) *(6 pts.)* If the method of undetermined coefficients is used to find a particular solution
*yp*(x) of the differential equation

*y*(5)*−*2*y*(4)+ 8*y*(3)*−*16*y*(2)+ 16*y0−*32*y*=*x ex*+ 2*e*2*x*+*x* cos(2*x),*

what would be the general form of a particular solution *yp*(x) obtained in this way? Note

that the corresponding auxiliary equation can be factorized as (m*−*2) (m2_{+ 4)}2 _{= 0.}
(DO NOT SOLVE FOR THE COEFFICIENTS!)

Solution.

8. *(10 pts.)* Use the variation of parameters method to find the general solution of the
differential equation

*x y00 _{−}_{y}0*

_{+ 4}

*3*

_{x}

_{y}_{=}

*3*

_{x}

_{,}_{for}

_{x >}_{0,}

given that the pair*y*1(x) = cos(x2) and*y*2(x) = sin(x2) forms a fundamental set of solutions
for the associated homogeneous equation.

Solution. In standard form the DE is written as

*y00 _{−}_{x}−*1

_{y}0_{+ 4}

*2*

_{x}

_{y}_{=}

*2*

_{x}

_{,}_{for}

_{x >}_{0,}and thus

*f(x) =*

*x*2

_{. The Wronskian associated with}

_{y}1 and *y*2 is
*W*(y1*, y*2)(x) =

¯
¯
¯
¯*y _{y}*1

*0*

*y*2

1 *y0*2

¯ ¯ ¯ ¯= ¯ ¯ ¯ ¯ cos(x

2_{)} _{sin(x}2_{)}
*−2x* sin(x2_{) 2}_{x}_{cos(x}2_{)}

¯
¯
¯
¯= 2*x*

¡

cos2_{(x}2_{) + sin}2_{(x}2_{)}¢_{= 2}_{x.}

A particular solution is

*yp*(x) = *−*

·Z

*y*2(x)*f*(x)
*W*(y1*, y*2)

*dx*

¸

*y*1(x) +

·Z

*y*1(x)*f*(x)
*W*(y1*, y*2)

*dx*

¸

*y*2(x)

=*−*

·Z

sin(x2_{)}* _{x}*2
2

*x*

*dx*

¸

cos(x2_{) +}

·Z

cos(x2_{)}* _{x}*2
2

*x*

*dx*

¸

sin(x2_{)}

=*−*1
2

·Z

sin(x2_{)}_{x dx}

¸

cos(x2_{) +} 1
2

·Z

cos(x2_{)}_{x dx}

¸

sin(x2_{)}

Using the substitution *u*=*x*2_{,} _{du}_{= 2}_{x dx,}

Z

sin(x2_{)}_{x dx}_{=} 1
2

Z

sin(u)*du*=*−*1

2 cos(u) +*C* =*−*
1
2 cos(x

2_{) +}_{C}

and Z

cos(x2)*x dx*= 1
2

Z

cos(u)*du*= 1

2 sin(u) +*C* =
1
2 sin(x

2_{) +}* _{C.}*
Hence,

*yp* =

1 4 cos(x

2_{) cos(x}2_{) +} 1
4 sin(x

2_{) sin(x}2_{) =} 1
4

¡

SCRATCH

*yp*(x) = *−*

·Z

*y*2(x)*f*(x)
*W*(y1*, y*2)

*dx*

¸

*y*1(x) +

·Z

*y*1(x)*f*(x)
*W*(y1*, y*2)

*dx*

¸