math 104-02 (de) - assignment 02 (section 2-1) - first order des (solutions)

7/25/2019 Math 104-02 (de) - Assignment 02 (Section 2-1) - First Order DEs (Solutions)

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Math 104-02 Differential EquationsAssignment 2: Section 2.1 (Solutions)

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Section 2.1: First Order Differential Equations

For the following problems:a) Find the general solution of the given DE; and

b) Use it to determine how the solution behaves as t .

1. 2ty ' 3y t e + = +

Solution Answer: a) General Solution: ( ) ( ) 2t 3t 1 13 9 1y t e c e

= + +

b) Solution Behavior: As t , ywill become asymptotic to the line ( ) ( )1 13 9t

Explanation: a) The General Solution

Since the DE is already in standard form i.e., ( ) ( )y ' p t y g t + = , where ( )p t 3=

and ( ) 2tg t t e = + -- we will move directly to determining the integrating factor,

( )t .

( ) ( )( )

( ) ( ) ( ) ( )

( )

3 dt

3t

Given : t exp p t dt

And : p t 3

Then : t exp 3 dt t e

t e

==

= =

=

Next, we apply the integrating factor ( ) 2tt e = and solve the equation:

( )

( )

( )( ) ( )

( ) ( )

3t

2t 3t 3t 3t 2 t

3t 3t 3t t

3t 3t t

3t 3t 3t t 1 13 9 1

2t 3t 1 13 9 1

Given : t e

Then : y' 3y t e e y ' e 3y e t e

e y ' 3e y te e

e y te e dt e y te e e c

y t e c e

=

+ = + + = +

+ = +

= +

= + +

= + +

So the General Solution: ( ) ( ) 2t 3t 1 13 9 1y t e c e

= + + .

b) Solution Behavior as t

As t , ywill become asymptotic to the line ( ) ( )1 13 9t .

2. ( )ty ' 2y sin t + =

Solution Answer: a) General Solution: ( ) ( ) ( )1 2 21y t cos t t sin t c t = + +

or( ) ( ) 1

2

t cos t sin t c y

t

+ +

=

b) Solution Behavior: As t , y 0 .



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First we re-write the DE in standard form i.e. ( ) ( )y ' p t y g t + = :

( ) ( ) ( ) ( ) ( )

( ) ( ) ( )

1 1 1ty ' 2y sin t ty ' 2y sin t t t t

2 1y ' y sin t t t

+ = + =

+ =

Now that it is in standard form, where ( ) 2p t t= and ( ) ( ) ( )1g t sin t t= , we

determine the integrating factor, ( )t .

( ) ( )

( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( )

2

2 dtt

2Ln t Ln t

2


2And : p t t

2Then : t exp dt t e t

t e t e

t t

=

=

= =

= =

=

Next, we apply the integrating factor ( ) 2t t = and solve the equation:

( )

( ) ( ) ( )

( ) ( ) ( )( )

( )

( ) ( )

( ) ( ) ( )

2

2 2 2

2

2

2

1

1 2 2

1

Given : t t

2 1Then : y ' y sin t t t

2 1t y ' t y t sin t t t

t y ' 2ty t sin t

t y t sin t dt

t y t cos t sin t c

y t cos t t sin t c t

=

+ =

+ =

+ =

=

= + +

= + +

So the General Solution: ( ) ( ) ( )1 2 2

1y t cos t t sin t c t

= + +

, or( ) ( ) 1

2

t cos t sin t c y

t

+ +

= .


As t , y 0 .

3. 2y ' y 3t + =

Solution Answer: a) General Solution:t2

1y 3t 6 c e

= +

b) Solution Behavior: As t , ywill become asymptotic to the line 3t 6

.



( ) ( ) ( )

( ) ( )

1 1 12y ' y 3t 2y ' y 3t 2 2 2

31y ' y t 2 2

+ = + =

+ =


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Now that is in standard form, where ( ) 1p t 2= and ( ) ( )3g t t2= , we determinethe integrating factor, ( )t .

( ) ( )

( )

( ) ( ) ( ) ( )

( )

1 dt2

t2


1And : p t 2

1Then : t exp dt t e 2

t e

=

=

= =

=

Next, we apply the integrating factor ( )t2t e = and solve the equation:

( )

( ) ( ) ( ) ( )

( ) ( )

( )

t2

t t t2 2 2

t t t2 2 2

t t2 2

t t t2 2 2

1

t2

1

Given : t e

3 31 1Then : y ' y t e y ' e y e t 2 2 2 2

31e y ' e y te 2 2

3e y te dt 2

e y 3te 6e c

y 3t 6 c e

=

+ = + =

+ =

=

= +

= +

So the General Solution:t2

1y 3t 6 c e

= + .


As t , ywill become asymptotic to the line 3t 6 .

For the following problems, find the solution to the given IVP.

4. 2ty ' y 2te = ; ( )y 0 1=

Solution Answer: 2t 2 t t y 2te 2e 3e = + or ( )t t ty e 2te 2e 3 = +

Explanation: Step #1: Determine The General Solution.

The DE is already in standard form i.e., ( ) ( )y ' p t y g t + = , where

( )p t 1= and ( ) 2tg t 2te = -- so we move directly to determining the

integrating factor, ( )t .

( ) ( )

( )

( ) ( ) ( ) ( )

( )

1 dt

t


And : p t 1

Then : t exp 1 dt t e

t e

=

= = =

=

Next, we apply the integrating factor ( ) tt e = and solve the equation:


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( ) t

2t t t t 2 t

t t t

t t

t t t 1

2t 2 t t

1

Given : t e

Then : y ' y 2te e y ' e y e 2te

e y ' e y 2te

e y 2te

e y 2te 2e c

y 2te 2e c e

=

= =

=

=

= +

= +

So the General Solution: 2 t 2t t 1y 2te 2e c e = + .

Step #2: Determine The Solution to The IVP

( )

( ) ( ) ( )

( ) ( )

2 0 2 0 2 t 2t t 0

1 1

1

1 1

Given : y 0 1

Then : y 2te 2e c e 1 2 0 e 2e c e

1 0 2 1 c 1

1 2 c c 3

=

= + = +

= +

= + =

Therefore, IVP Solution: 2 t 2t t y 2te 2e 3e = + or ( )t t ty e 2te 2e 3 = + .

5. 2ty ' 2y t t 1+ = + ; ( ) 12y 1 = , t 0>

SolutionAnswer: ( ) ( ) ( ) ( )2 21 1 1 14 3 2 12 y t t t

= + + , or4 3 2

2

3t 4t 6t 1y

12t

+ +=



( ) ( ) ( )( )

( )

2 21 1 1ty ' 2y t t 1 ty ' 2y t t 1t t t

2 1y ' y t 1t t

+ = + + = +

+ = +

Now that it is in standard form i.e., ( ) ( )y ' p t y g t + = , where ( ) 2p t t= and

( ) 1g t t 1 t= + , we determine the integrating factor, ( )t .

( ) ( )

( )

( ) ( ) ( ) ( )

( ) ( )

( ) ( ) ( )2

2 dtt

2 Ln t

Ln t 2


2And : p t t


t e

t e t t

=

=

= =

=

= =



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( )

( )

( ) ( )

( )

( ) ( ) ( )

( ) ( ) ( )

2

2 2 2

2 3 2

2 3 2

2 4 3 2

1

2 2

1

Given : t t

2 1Then : y' y t 1t t

2 1t y ' t y t t 1t t

t y ' 2ty t t t

t y t t t

1 1 1t y t t t c 4 3 2

1 1 1y t t c t 4 3 2

=

+ = +

+ = +

+ = +

= +

= + +

= + +

So the General Solution: ( ) ( ) ( )2 21 1 14 3 2 1y t t c t

= + + .

Step #2: Determine The solution to The IVP

( )

( ) ( ) ( )

( )( ) ( )( ) ( ) ( )

12

2 21 1 14 3 2 1

2 2

1

1

1 1

Given : y 1

Then : y t t c t

1 1 1 11 1 c 12 4 3 2

1 1 1 1 c2 4 3 2

51 1c c2 12 12

=

= + +

= + +

= + +

= + =

Therefore, IVP Solution: ( ) ( ) ( ) ( )2 21 1 1 14 3 2 12 y t t t

= + + , or4 3 2

2

3t 4t 6t 1y

12t

+ += .

6. ( )ty ' 2y sin t + = ; ( )y 12 = , t 0>

Solution Answer: ( ) ( ) ( ) ( ) ( )2

2 241 1y cos t sin t

t t 4t

= + + , or ( ) ( )

2

24t cos t 4sin t 4 y

4t

+ + =



( ) ( ) ( ) ( ) ( )

( ) ( ) ( )

1 1 1ty ' 2y sin t ty ' 2y sin t t t t

2 1y ' y sin t t t

+ = + =

+ =

Now that it is in standard form i.e., ( ) ( )y ' p t y g t + = , where ( ) 2p t t= and

( ) ( ) ( )1g t sin t t= , we determine the integrating factor, ( )t .

( ) ( )

( )

( ) ( ) ( ) ( )

( ) ( )

( ) ( ) ( )2

2 dtt

2 Ln t

Ln t 2


2And : p t t


t e

t e t t

=

=

= =

=

= =


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( )

( ) ( ) ( )

( ) ( ) ( )( )

( )

( ) ( )

( ) ( ) ( ) ( ) ( )

2

2 2 2

2

2

2

1

2 21

Given : t t

2 1Then : y ' y sin t t t

2 1t y ' t y t sin t t t

t y ' 2ty t sin t

t y t sin t

t y t cos t sin t c

1 1 1y cos t sin t c t t t

=

+ =

+ =

+ =

=

= + +

= + +

So the General Solution: ( ) ( ) ( ) ( ) ( )2 211 1 1y cos t sin t c t t t= + + .Step #2: Determine The Solution to The IVP

( )

( ) ( ) ( ) ( ) ( )

( )( )

( )( )

( ) ( ) ( ) ( ) ( )( )( ) ( )( ) ( )

( )

( ) ( ) ( )

2 21

12 2

2 21

2 21

2 21

2 2

2 21 1

Given : y 12

1 1 1Then : y cos t sin t c t t t

1 1 11 cos sin c

2 22 2 2

2 4 41 cos sin c 2 2

2 4 41 0 1 c

4 41 c

4 44c c4

=

= + +

= + +

= + +

= + +

= +

= =

Therefore, IVP Solution: ( ) ( ) ( ) ( ) ( )2

2 2

41 1y cos t sin t t t 4t

= + + , or

( ) ( ) 2

2

4t cos t 4sin t 4 y

4t

+ +

= .

math 104-02 (de) - assignment 02 (section 2-1) - first order des (solutions)

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