Math 3301 Homework Set 1 10 Points

#1. (2 pts) The derivative will be zero at 3y = − , 1y = − , and 4y = . A sketch of the direction field

and some solutions is to the right. From this we can see the following long term behavior and dependence on ( )0y .

( ) ( )( ) ( )

( ) ( )( ) ( )

0 4

0 4 4

4 0 1 1

0 1 3

y y t

y y t

y y t

y y t

> →∞

= =

< ≤ − → −

< − → −

#4. (3 pts) Don’t forget that the y′ needs a coefficient of 1 before we do any of the work.

2 2 3ln 3 23 dt tt t ttµ

− −−∫= = =e e e

( ) ( ) ( ) ( )3 2 10 1012 2 2 10sin 4sin 2 cost t tt t tt y dt t dt t c− ′ = − = − − +⌠

⌡ ∫e e e

( ) ( ) ( )( )3 2 3 8 3 212 2 104sin 2 cost tt t ty t t t t ct− −= − − +e e e

Applying the initial condition gives,

( ) ( ) ( )( )3 2 3 8 3 2 101 110 100 4 1 2 0 4y c cπ π π ππ π π π π− −= = − − + → = −e e e e

( ) ( ) ( )( ) ( )3 2 3 8 10 3 21 12 2 10 104sin 2 cos 4t tt t ty t t t t tπ− −= − − + −e e e e

#5. (2 pts) We know from Calc I that relative extrema occur at critical points and critical points are those points where the derivative is zero or doesn’t exist. However, we’ve been told that the derivative exists and is continuous everywhere so that means that if the critical point that gives the relative maximum is

ct then we also have ( ) 90cy t = and ( ) 0cy t′ = so all we need to do is plug ct into the differential

equation and then solve for ct .

( ) ( ) ( )6 6 36716 24 7 2 360 7 2 ln 0.8687c c

c c ct ty t y t t′ − = − ⇒ − = − ⇒ = =e e

#7. (3 pts) Here’s the solution to the differential equation. I’ll leave the details to you to check.

( )

( ) ( )

8 2 3 82 15 5

2 3 2 82 1 215 5 5 2

t t t t

t t t

y t c

y t

µ

α

− −

−

= = − + +

= − + + −

e e e e

e e e

Okay, the first two terms will always go to positive infinity. The last term will vary depending the sign of the coefficient. Note however that if the third term isn’t there (i.e. the coefficient is zero) then the second term will determine long term behavior. If the third term goes to positive infinity (i.e. the coefficient is positive) then the solution will go to positive infinity. Finally, if the third term goes to

Math 3301 Homework Set 1 10 Points

negative infinity (i.e. the coefficient is negative) then the solution will go to negative infinity (the exponent on the third exponential is larger than the exponent on the second exponential).

So, all we need to do is set the coefficient of the third term equal to zero and determine where it is positive/negative. Here’s a table giving all these results.

( ) ( )( ) ( )

21 21 21 2110 10 10 10,

1.4491 1.4491 1.4491, 1.4491

y t y t

y t y t

α α α

α α α

− ≤ ≤ →∞ > < − → −∞

− ≤ ≤ →∞ > < − → −∞

Not Graded #2. The derivative will be zero at 1y = − and

2y = . A sketch of the direction field and some solutions is to the right. From this we can see the following long term behavior and dependence on ( )0y .

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

0 2

0 2 2

0 2 1

y y t

y y t

y y t

> →∞

= =

< → −

#3. Don’t forget that the y′ needs a coefficient of 1 and the y needs to be moved over to the left side before we do any of the work.

( ) ( )2

26

33 1 21 ln 1x dx xx xµ

++∫= = = +e e

( )( ) ( ) ( ) ( ) ( )3 13 32 2 2 212 2

51 1 1 1y x dx x x dx y x x c x− − −′

+ = + ⇒ = + + +⌠ ⌠ ⌡⌡

Applying the initial condition gives,

( ) ( ) ( ) ( )1 32 21 11 1 112

5 5 5 52 0 1 1y c c y x x x− −

− = = + → = − ⇒ = + − +

#6. (3 pts) Here’s the solution to the differential equation. I’ll leave the solution details to you.

( ) ( ) ( )4 2 4 23 31 102 2 2 2

2 3 3 1t t t t ty t t c y t t yµ − − − −= = − + + ⇒ = − + + +e e e e e

Now, just like in #5 we know we have a relative minimum and we can see from the solution that the

derivative will exist everywhere and so we know that ( )1 0y′ = and we can use the differential

equation to determine the value of the relative minimum, ( )1y .

Math 3301 Homework Set 1 10 Points

( ) ( ) ( )4 4122 1 6 1 6 2.99084y y− −= − ⇒ = − =e e

Find the solution to the following IVP in terms of 0y . Determine the value of 0y for which the solution

will have a relative minimum at 1t = .

( )402 6 0ty y t y y−′ + = − =e

All we need to do at this point is plug this into our solution and solve for 0y .

( ) ( )4 23 10 02 22.99084 1 3 1 9.94825y y y− −= = − + + + ⇒ =e e