Engineering II Mac Eng Class of 2015

ID # :

SAMPLE TEST 1 (a) SOLUTIONS

Instructions: You must use permanent ink. Tests submitted in pencil will not be considered 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. GOOD LUCK!

# 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

y′ _{=y(y−1) cos(y),}

y(0) = a.

Suppose that the solution y(x) satisfies

lim x→+∞

y(x) = π 2.

What can be said for sure about the number a? (Hint: Do not try to solve the ODE)

(a) 1< a < π₂

(b) 0< a <1 (c) −π₂ < a <0

→ (d) 1< a < 3₂π

(e) 0< a < π₂

Solution. The critical numbers are y = 0, y = 1 and y = π₂ +n π, n

any integer.Since the function g(y) = y(y−1) cos(y) satisfies g(y) > 0 if 1 < y < π₂ and g(y)< 0 if π₂ < y < 3₂π, it follows that π₂ is a stable critical point. Thus limx→+∞y(x) = π2 if and only if 1< a <

3π

2. (4 pts.) Find the general solution of the differential equation

y(6)−2y(3)+y= 0.

The answer (where C₁, . . . , C₆ denote arbitrary constants) is:

(a) y(x) = C1ex+C2e−x+C3 cos(x) +C4 sin(x) +C5x cos(x)

+C6x sin(x).

→ (b) y(x) =C₁ex_+C

2x ex+C3e−x/2 cos(

√

3

2 x) +C4 e−x/2sin(

√

3 2 x)

+C₅x e−x/2 _cos(√3

2 x) +C6x e−x/2sin(

√

3 2 x)

(c) y(x) =C1ex+C2x ex+C3e−x/2 cos(

√

3

2 x) +C4 e−

x/2_sin(√3 2 x)

+C5ex/2 cos(

√

3

2 x) +C6e

x/2_sin(√3 2 x)

(d) y(x) =C₁ex_+C

2x ex+C3e−x/2 cos(2x) +C4 e−x/2sin(2x)

+C₅x e−x/2 _{cos(2x) +C}

6x e−x/2sin(2x)

(e) y(x) =C₁ex+C₂x ex+C₃e−x_+C

4x e−x

+C5e−x/2 cos(

√

3

2 x) +C6e−

x/2_sin(√3 2 x)

Solution. The auxiliary equation is m6 −2m3+ 1 = 0 or (m3 −1)2 _{= 0 or}

(m−1)2_(m2_{+m+ 1)}2 _{= 0. The roots ofm}2_{+m+ 1 = 0 are m=−}1 2±

√

3 2 i.

Thus the roots arem= 1 andm =−1₂±√₂3i, all with multiplicity 2. The root

m = 1 with multiplicity 2 yields the solutions y=ex and y=x ex while the roots −1₂ ± √₂3, i with multiplicity 2 yield the solutions y = e−x/2 _cos(√3

2 x),

y = e−x/2 _sin(√3

2 x), y = x e−x/2 cos(

√

3

2 x) and y = x e−x/2 sin(

√

3

2 x). The

3. (4 pts.) Let y(x) be the unique solution of the initial value problem:

_dy

dx =y(y−1)x,

y(0) = 3₂.

Then, the largest interval containingx= 0 where the solution is defined and continuous is:

(a) (−∞,∞) (b) (−1,1)

(c) (−3₂ ln(2),3₂ ln(2)) (d) (−ln(5),∞)

→ (e) (−p

2 ln(3),p2 ln(3))

Solution. The DE is separable: 1

y(y−1)

dy dx =x.

Integrating both sides with respect to x yields:

Z ₁

y(y−1)dy =

Z ₁

y−1 − 1

ydy = x2

2 +C, C. arbitrary,

Thus

ln|y−1| −ln|y|= ln

y−1

y

= x

2

2 +C,

or

y−1

y

=ex

2

2 +C =A e

x2

4. (4 pts.) Suppose that y(x) is a solution of the differential equation

dy dx = (x

2_{−5x+ 6)e}y3

and that y(x) is defined in some interval centered at x0. Then, for which

of the following number x0 does the function y(x) necessarily reach a local

minimum at x0?

(Hint: Do not try to solve the D.E.)

(a) x₀ = 0 (b) x0 = 1

(c) x₀ = 2

→ (d) x₀ = 3 (e) x0 = 4

Solution. To find the local extrema, we must set the first derivative of y(x) equal to 0. Since dy_dx = (x − 2) (x −3)ey3, the only points x where the derivative of the solution vanish are x= 2 andx= 3. Since

d2_y

dx2 = (2x−5)e

y3

+ (x−2) (x−3)) (3y2)ey3 dy dx,

we have

d2y dx2

x=2

=−ey03 <0, d 2_y

dx2

x=3

=ey03 >0,

5. (4 pts.) An electromotrice force E(t) = 12−10e−t _{volt is applied to} an LR-circuit in which the inductance is L= 1

2 henry and the resistance is

R = 10 ohms. Determine the limiting current as t → ∞ if i(0) = 0. The differential equation for an LR-circuit is

Ldi

dt +R i=E(t).

The value of limt_→∞ i(t) is:

→ (a) 6 5

(b) 2₃ (c) 4

3

(d) 0 (e) 2

Solution. The differential equation fori(t) is 1

2

di

dt + 10i= 12−10e

−t _or di

dt + 20i= 24−20e

−t_.

The integrating factor is

u(t) =eR 20dt =e20t.

We obtain thus

(di

dt + 20i)e

20t₌ d

dt

e20ti = (24−20e−t_)e20t_{= 24e}20t_−20e19t

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

6. (10 pts.) Compute in explicit form the solution of the initial value problem

(1 +x2_)y′_{−2x y=x(1 +x}2₎2_ex_,

y(0) = 1.

Solution. (Integrating factors) In standard form the DE is written as

y′₋ 2x

1 +x2 y=x(1 +x 2_)ex_.

The integrating factor is

u(x) = e− R 2x

1+x2dx =e−ln(1+x

2

)₌ 1

1 +x2.

Muliplying both sides of the DE in standard form by the integrating factor, we obtain

y′ 1 +x2 −

2x

(1 +x2₎2 y=

d dx

y

1 +x2

=x ex.

Therefore,

y

1 +x2 =

Z

x exdx=

Z

x(ex)′_{dx=x e}x₋

Z

exdx= (x−1)ex+C.

The condition y(0) = 1 yields C = 2. Therefore, the solution is

7. (10 pts.) Use the method of undetermined coefficientsto compute in explicit form the solution of the initial value problem

y′′_{+ 4y = 8 sin(2x),}

y(0) =y′_{(0) = 0.}

Solution. The auxiliary equation is m2 + 4 = 0 with roots m = ±2i. The complementary solution is thus yc(x) =C1 cos(2x) +C2 sin(2x). Because 2i

(or −2i) is a root of multiplicity one of the auxiliary equation, we know that a particular solution exists of the form

yp(x) =A x cos(2x) +B x sin(2x).

Therefore,

y′

p(x) = A[cos(2x)−2x sin(2x)] +B[sin(2x) + 2x cos(2x)]

and

y′′

p(x) = A[−4 sin(2x)−4x cos(2x)] +B[4 cos(2x)−4x sin(2x)].

It follows that

y′′

p + 4yp =A[−4 sin(2x)−4x cos(2x)] +B[4 cos(2x)−4x sin(2x)] + 4 [A x cos(2x) +B x sin(2x)]

=−4A sin(2x) + 4B cos(2x) = 8 sin(2x),

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

y′′_{+ 9y=} 1

cos(3x), for −

π

6 < x <

π

6.

Solution.Since this is a linear equation with constant coefficients we can find the general solution of the associated homogeneous equation by solving the auxiliary equationm2_{+9 = 0 which has rootsm=±3i. The complementary}

solution is thusyc(x) =C₁ cos(3x)+C₂ sin(3x) andy₁(x) = cos(3x),y₂(x) = sin(3x) are a fundamental system of solutions for the associated homogeneous equation. The associated Wronskian is

W(y₁, y₂)(x) =

y₁ y₂ y′

1 y2′

=

cos(3x) sin(3x)

−3 sin(3x) 3 cos(3x)

= 3 [cos2(3x) + sin2(3x)] = 3.

Since the DE is already in standard form, letting f(x) = _cos(31 _x), we know that there exists a particular solution yp(x) of the form

yp(x) =−

Z _f

(x)y₂(x)

W(y1, y2)(x)

dx y₁(x) +

Z _f

(x)y₁(x)

W(y1, y2)(x)

dx y₂(x)

=− Z

(cos(3x))−1 _sin(3x)

3 dx cos(3x) +

Z

(cos(3x))−1 _cos(3x)

3 dx sin(3x)

=−1

3

Z

sin(3x)

cos(3x)dx cos(3x) + 1 3

Z

1dx sin(3x)

= 1

9 ln|cos(3x)| cos(3x) + 1

3x sin(3x).

The general solution is thus

y(x) =yp(x) +yc(x)

= 1

9 ln|cos(3x)| cos(3x) + 1

3x sin(3x) +C1 cos(3x) +C2 sin(3x),