Query 01

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course Mth 279

12:15PM October 5th

Section 2.2*********************************************

Question: 1. Solve the following equations with the given initial conditions:

1. y ' - 2 y = 0, y(1) - 3

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Your solution:

P(t)=-2 Integral -2dt=-2t

@& OK, but better to think of e^(-int(p(t) dt) ) from the start, so you would be integrationg -(-2) with respect to t.

Better to reconcile the - signs early.*@

General Solution: y=Ce^2t

3=Ce^2(1)

3=Ce^2

3=C(7.389056099)

C=0.4060058497

Therefore: y=0.41060058497e^2t

@& 3 = C e^2 implies C = 3 / e^2.

e^2 t means e^2 multiplied by t; what you intend is obviously e^(2 t). Sometimes it's not obvious what grouping is intended, so you need to be sure your expressions are consistent with the order of operations.

The final solution is therefore

3 / e^2 e^(2 t).

This expression is correct, if the order of operations is strictly followed.

However it's clearer if expressed as

(3 / e^2) * e^(2 t)

and can then be put in a more standard final form as

3 e^ ( 2 t - 2 ).

y=0.41060058497e^(2t) would be OK, but exact expressions are usually preferable to decimal approximations.

It never hurts to then give the decimal approximation, and in this case

y = -.41 e^(2 t) (approx.)

would be a pretty good approximation and easy to understant.

*@

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Given Solution:

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Question: 2. t^2 y ' - 9 y = 0, y(1) = 2.

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Your solution:

y’-9y/t^2=0 or y’-9yt^-2=0

p(t)=-9t^-2 Integral-9t^-2dt=-9Integralt^2dt=9t^-1/-1=9t^-1

General Solution: y=Ce^9t^-1

2=Ce^9(1)^-1

2=Ce^9(1/1)

2=C(8103.083928)

C=2.468196082*10^-4

Therefore: y=2.468196082*10^-4e^9t^-1

@& Good, but see my preceding note about the standard form of the expressions.*@

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Given Solution:

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Question: 3. (t^2 + t) y' + (2t + 1) y = 0, y(0) = 1.

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Your solution:

y’+[(2t+1)/(t^2+t)]*y=0

p(t)=(2t+1)/(t^2+t) Integral(2t+1)/(t^2+t) dt=log(t)+log(t+1)

General Solution: y=Ce^-(log(t)+log(t+1))

Log(0) does not exist

@& It isn't clear how you got that form for your integral. If you used partial fractions to get the integral, then that form could be the result.

However the correct result, in that case, would be ln(t) + ln(t + 1).

Using substitution u = t^2 + t you get du = 2 t + 1, so the integral would be integral( du / u) = ln | u | = ln | t^2 + t |.

ln(t) + ln(t+1) does simplify to ln | t ( t + 1) | = ln | t^2 + t |

You are correct that the given initial condition is impossible.*@

confidence rating #$&*:ut of 3

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Given Solution:

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Question: 4. y ' + sin(3 t) y = 0, y(0) = 2.

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Your solution:

P(t)=sin(3t) Integral sin(3t)dt=-1/3cos(3t)

General Solution: y=Ce^(1/3cos(3t))

y=Ce^(1/3cos(3(0))

2=Ce^(1/3)

2=C(1.395612425)

C=1.433062621

Therefore: y=1.433062621e^((1/3)cos(3t))

@& C = 2 e^(-1/3) would be better.

*@

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Given Solution:

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Question: 5. Match each equation with one of the direction fields shown below, and explain why you chose as you did.

y ' - t^2 y = 0

D.) slope increases more rapidly when t increases (in quadrant I)

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y ' - y = 0

A.) because the t position has no effect on the slope

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y' - y / t = 0

C.) slope decreases as you increase magnitude of t (in quadrant I & IV)

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y ' - t y = 0

B.) slope increases as you increase magnitude of either t or y (in I quadrant)

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y ' + t y = 0

F.) slopes are negative in quadrant I

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A

B

C

D

E

F

6. The graph of y ' + b y = 0 passes through the points (1, 2) and (3, 8). What is the value of b?

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Your solution:

b=-1.5 or -0.375

confidence rating #$&*:ut of 3

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Given Solution:

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Not sure how to solve this problem

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Question:

7. The equation y ' - y = 2 is first-order linear, but is not homogeneous.

If we let w(t) = y(t) + 2, then:

What is w ' ?

=y’

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What is y(t) in terms of w(t)?

y(t)=w(t)-2

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What therefore is the equation y ' - y = 2, written in terms of the function w and its derivative w ' ?

w ' - w(t) - 2 = 2

@& This should be

w’- (w(t)-2 ) = 2

which after a couple of steps becomes

w ' - w = 0.

This is easily solved for w then converted back to y.*@

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Now solve the equation and check your solution:

Solve this new equation in terms of w.

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Substitute y + 2 for w and get the solution in terms of y.

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Check to be sure this function is indeed a solution to the equation.

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Your solution:

confidence rating #$&*:out of 3

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Given Solution:

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Self-critique (if necessary):

Really confused how to solve this problem. Can’t find any similar examples in the notes or in the chapter.

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Question: 8. The graph below is a solution of the equation y ' - b y = 0 with initial condition y(0) = y_0. What are the values of y_0 and b?

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Your solution:

y_0=1

b=unsure how to solve for b value

@& y ' = b y.

You know y(0).

y ' is the slope, and you can estimate the slope at t = 0. So you can estimate b.*@

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Given Solution:

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@& You're doing pretty well here.

I've given you a couple of notes on the standard form of solutions, and exact vs. approximate forms of constants.

You did come up short on a couple of problems, so be very sure to check all of my notes.*@