Eng Math. Mid Term (4/23/2008)

Eng Math. Mid Term (4/23/2008)

(Closed book and note: 120 min.)

1. Solve the differential equation [ points]:

a)

dy y

⁵

, 0

dx − = x x x >

Sol) ⁵

1 5

y x C

− + x =

b)

xdy − ydx − y dx

²

= 0

Sol) By dividing with y², the equation above simply becomes

x 0

d dx

y

− ⎛ ⎞ ⎜ ⎟ − =

⎝ ⎠

By integration,

y x

c x

= −

2. Solve the nonhomogeneous equation [ points]:

a) k ^kx

1

"

n

k

y y a e

=

+ = ∑

Sol)

kx

kx k

k 2

kx k

2

kx k

2

1 1

" ,

1 1

1

k

k

n n

k

k k

y Ae

y y a e A a

k y a e

k y y a e

= =

k

=

+ = =

+

= +

∴ = =

∑ ∑ +

b)

x y

³

"' 3 − x y

²

" 6 + xy ' 6 − y = x

⁴

ln x

Sol) p.119

3. Prove that

e

^λx

, xe

^λx

, x e

^{2 λx}

..., x

^m−1

e

^λx will be the linearly independent solutions of a homogeneous linear ODE with constant coefficients (order n) if the characteristic equation of the high order ODE has a multiple real root of order m (note: n>m) [20 points]

Sol)

L[y]=[ Dⁿ + a_n-1D^n-1 +. . . + a₀]y

For y=e^λx, L[e^λx]=[ λⁿ + an-1λ^n-1 +. . . + a0]e^λx Let λ₁ be an mth order of the polynomial, then L[e^λx]=(λ－λ₁)^m h(λ) e^λx

Differentiate with respect to λ

x m-1 x m

1 1

[e ]= m( - ) h( ) e + ( - ) [ ( ) ^x]

L ^λ λ λ λ ^λ λ λ h λ e^λ

λ λ

∂ ∂

∂ ∂ --- 1)

] e [ ^{λ x} λ^L

∂

∂

=L

] [ e^λx

λ

∂

∂

= L[xe^λx]

When λ=λ_{1 ,} eqn 1) is equal to zero, and xe^λ1x is a solution of [Dⁿ + a_n-1D^n-1+. . . + a0]y= 0

we can repeat this step and produce x²e^λ1x, . . . , x^m-1e^λ1x by another m–2 such diffentiations!

4. Solve the systems of ODE by finding the eigenvalues and eigenvectors of the

coefficient matrix:

3 1 3

' 2 4

^t

X X t

e

⁻

⎡ − ⎤ ⎡ ⎤

= ⎢ ⎣ − ⎥ ⎦ + ⎢ ⎥ ⎣ ⎦

[ points]

Sol)

3 1

det( ) 0, 2 or 5

2 4

A I λ

λ λ

λ

− = − − = = − −

− −

And the corresponding eigenvectors are

1 1

⎡ ⎤ ⎢ ⎥

⎣ ⎦

^and

1

2

⎡ ⎤ ⎢ ⎥ −

⎣ ⎦

, respectively.

So, the solution vectors of the system are then

2

1 2

t t

X e

e

−

−

⎡ ⎤

= ⎢ ⎥

⎣ ⎦

^and

5

2 5

2

t t

X e

e

−

−

⎡ ⎤

= ⎢ ⎣ − ⎥ ⎦

^.

Let

X

^{( )h}

= Φ ⋅ ( ) t c

, then

2 5

( ) 1

2 5

2

( ) 2

t t

h

t t

e e c

X t c

e e c

− −

− −

⎡ ⎤ ⎡ ⎤

= Φ ⋅ = ⎢ ⎣ − ⎥ ⎢ ⎥ ⎦ ⎣ ⎦

^.

When

X

^{( )p}

= Φ ⋅ ( ) t u

, then

1

3

' ( ) t

_t

u t

e

−

−

= Φ ⋅ ⎢ ⎥ ⎡ ⎤

⎣ ⎦

by substituting into the problem.

So,

( )

6 27

5 50 4

3 21

5 50 2

t

p

t

t e

X

t e

−

−

⎡ ⎤

− +

⎢ ⎥

⎢ ⎥

= ⎢ ⎢ ⎣ − + ⎥ ⎥ ⎦

.

As a result,

2 5

1

2 5

2

6 27

5 50 4

2 3 21

5 50 2

t

t t

t t t

t e e e c

X e e c e

t

−

− −

− − −

⎡ ⎤

− +

⎢ ⎥

⎡ ⎤ ⎡ ⎤ ⎢ ⎥

= ⎢ ⎣ − ⎥ ⎢ ⎥ ⎦ ⎣ ⎦ + ⎢ ⎢ ⎣ − + ⎥ ⎥ ⎦