Solution to Homework 1

Solution to Homework 1

Differential Equations

I-Hsiang Wang

Solution to Homework 1

1. (Practice of Different Methods)

[15]

Solve the following initial-value problems (y: dependent variable)

(a)

dy dx

=

x4

1 -

, 1

y(0)

=

1.

[5]

(b) dy = x3 , y(2) = 1.

[5]

dx (2y + 1)

(c) (x2 - 1) dy = xy + 1, y(0) = 1.

[5]

dx

Solution.

(a)

{

}

dy

1

1

11

1

=

=

=

dx x4 - 1 (x2 - 1)(x2 + 1) 2

x2 - 1 - x2 + 1

1

1

1

=

x

4

-

1

-

x

4

+1

-

x2

2

+1

Hence,

y = 1 ln |x - 1| - 1 ln |x + 1| - 1 tan-1 x + c.

4

4

2

Plug in the initial condition x = 0, y = 1, we get c = 1.

= y = 1 ln(1 - x) - 1 ln(x + 1) - 1 tan-1 x + 1 .

4

4

2

Interval of definition: x (-1, 1) .

(b)

dy

x3

=

= (2y + 1)dy = x3dx = y2 + y = 1 x4 + c

dx (2y + 1)

4

Plug in the initial condition x = 2, y = 1, we get c = -2.

=

y2 + y = 1 x4 - 2

=

( 1 )2 y+ =

1

(x4

-

) 7

=

1(

)

y = -1 ? x4 - 7

4

2

4

2

1

Solution to Homework 1

Differential Equations

I-Hsiang Wang

Plug in the initial condition x = 2, y = 1, we know that we have to choose

1(

)

y = -1 + x4 - 7 .

2

()

Interval of definition:

x

1

74 ,

.

(c)

(x2

dy - 1)

=

xy

+1

dx

=

dy xy + 1 x

1

=

=

y+

,

dx x2 - 1 x2 - 1 x2 - 1

x = ?1.

We shall introduce an integrating factor ?(x) to solve this linear equation, which has to satisfy the following auxiliary DE:

d?

x

dx

=

- x2

-

? 1

=

d?

x

1 d(x2)

?

=

- x2

-

dx 1

=

-2

x2

-

1

=

ln

|?|

=

1 -

ln

|x2

-

1|.

2

Based on the initial condition x = 0, we choose the domain of x to be x (-1, 1) and hence we get an integrating factor

1

?=

.

1 - x2

Finally, plug in the integrating factor and we get

(

)(

)

d(?y)

x

1

x

?

1

dx

=?

y+ x2 - 1 x2 - 1

+y

- x2 - 1 ?

=

x2

-

1

=

- (1

-

. x2)3

To solve ?y, we need to compute the following integral:

- 1

dx x==sin

( 1 - x2)3

- cos

cos3 d =

- sec2 d = - tan + c

x

= -

+ c.

1 - x2

Hence,

1

x

?y =

y = -

+ c.

1 - x2

1 - x2

Plug in the initial condition x = 0, y = 1 we get c = 1.

= y = -x + 1 - x2 .

Singular points x = ?1 cannot be added back to the interval of definition, because

dy

x

= -1 -

dx

1 - x2

is not defined at the singular points. Interval of definition: x (-1, 1) .

2

Solution to Homework 1

Differential Equations

I-Hsiang Wang

2. (Discontinuous Coefficients)

[10]

Solve

dy + P (x)y = x

dx

{

subject to y(0) = 0, where P (x) =

1, -1

x0 x ................
................

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