qa 1

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course 279

2/17 12

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:

dy = 2y

dy/y = 2

Int(dy/y ) = Int(2 dt)

Int(dy/y ) = ln|y|

Int(2 dt) = 2t

ln|y| = 2t + c

y = e^(2t + c) = e^(2t)*e^(c)

= Ae^(2t), for y(1) = -3

-3 = Ae^(2*1) = Ae^(2)

A = -3/(e^2) = approx. -.406, so solution is…

y = -.406e^(2t)

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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:

t^2 dy = 9 y

dy/y = 9/t^2

Int(dy/y ) = Int(9/t^2 dt)

Int(dy/y ) = ln|y|

Int(9/t^2 dt)=9*Int(1/t^2 dt)=9*Int(t^-2 dt) = 9*(-1/t) = -9/t + c

ln|y| = -9/t + c

y = e^(-9/t + c) = e^(-9/t)*e^(c)

= Ae^(-9/t), for y(1) = 2

2 = Ae^(-9/1) = Ae^(-9)

A = 2/(e^-9) = approx. 16,206.1679, so solution is…

y = 16,206.1679e^(-9/t)

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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:

(t^2 + t) dy = -(2t + 1) y

dy/y = -(2t + 1)/ (t^2 + t)

Int(dy/y ) = Int([-(2t + 1)/ (t^2 + t)] dt)

Int(dy/y ) = ln|y|

Int([-(2t + 1)/ (t^2 + t)] dt)= -Int([ (2t + 1)/ t*(t+ 1)] dt)

using u sub, u = t*(t+ 1), du = (t*(t+ 1)) dt

(t*(t+ 1))’ = t’*(t+1) + t*(t+1)’

=1*(t+1) + t*(1+0) = (t+1)+t

=2t+1, so (t*(t+ 1)) dt = numerator

-Int(1/u du) = -ln|u| + c, so we get

Int([-(2t + 1)/ (t^2 + t)] dt) = -ln| (t^2+ t)| +c

ln|y| = -ln| (t^2+ t)| +c

y = t^2+t+e^c = t^2+t+A, for y(0) = 1

1 = 0^2+0+A

A=1, so final soln.

y = t^2+t+1

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@& @& Dividing both sides of

(t^2 + t) y' + (2t + 1) y = 0

by t^2 + t we get

y ' + (2 t + 1) / (t^2 + t) * y = 0,

This is linear first-order homogeneous with p(t) = (2 t + 1) / (t^2 + t). The integral of p(t) is ln ( t^2 + t), so the solution is

y = A e^( - ln(t^2 + t) ) = A / (t^2 + t)

The initial condition y(0) = 1 is not possible since y(0) is undefined.*@

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@& Note that

Int([-(2t + 1)/ (t^2 + t)] dt) = ln (t^2 + 2)

(use substitution u = t^2 + t).*@

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Given Solution:

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Self-critique (if necessary):

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

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Your solution:

dy = sin(3 t) y

dy/y = sin(3t)

Int( dy/y ) = ln|y|

Int( sin(3t) dt ) = -cos(3t)/3 + c

ln|y| = -cos(3t)/3 + c

y = e^(-cos(3t)/3 + c)

=Ae^(-cos(3t)), so general soln..

y = Ae^(-cos(3t))

Init conditions y(0) = 2

2 = Ae^(-cos(3*0)) = Ae^(-1)

A = 2/(e^(-1) = approx. 5.4366 so…

y(t) = 5.4366e^(-cos(3t))

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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

y’ = t^2*y, seems to be E

Slope = t^2*y, would be 0 along t and y axis because of either t or y values being equal to zero. Because t is squared either moving in Quad 3 and 4 slope is neg because of neg y value, but as fn approaches y axis slope is decreasing. Then if that is so then in Quads 1 and 2 would be increasing under same logic( there is no neg values for t because it’s squared.).

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y ' - y = 0

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y’ = y, seems to be A

Slope = y, so slope is only dependent on y value. Giving us horizontal lines of slope that are constant. Only place where slope is 0 would be along x axis.

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y' - y / t = 0

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y’ = y/t, seems C

Slope = y/t, would be 0 along t axis because y values being equal to zero. Because t is in the denominator slope approaches limit of 1 as t approaches zero. When fn moves away from y axis but keeps the same y value (straight line moving to left or right on graph) slope would decrease because t value gets greater. Same goes for keeping the same t value and either moving in pos or neg y direction (straight line up or down graph)

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y ' - t y = 0

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y’ = t*y, seems to be B

Slope = t*y, would be 0 along t or y axis because y or t values would be equal to zero. When fn moves away from y axis but keeps the same y value (straight line moving to left or right on graph) slope would increase in Quad 1 and 3( y and t are both neg in Quad 3 that’s why slope is pos) . Slope is neg for Quad 2 and 4 because one of axis values is neg while the other is pos.

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y ' + t y = 0

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y’ = -(t*y), seems to be F

Slope = t*y, would be 0 along t or y axis because y or t values would be equal to zero. This is the same as previous question but with neg sign in front of expression (t*y) then slope is pos in Quad 2 and 4 and neg in 1 and 3. Approaches a value of as fn approaches y and t axis

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@& Given Solution:

@& y ' - t^2 y = 0 gives us

y ' = t^2 y.

t^2 is always positive.

y ' is therefore positive when y is positive, negative when y is negative.

As we move away from the y axis t^2 gets larger, so on any horizontal line the directions will get steeper as we move away from this axis.

The only graph with these characteristics is E.

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y ' - y = 0 gives us

y ' = y

The slope doesn't depend on t, so on any horizontal line the slope will be constant. The only graph with this characteristic is A.

The slope will be negative for points below the x axis, where y is negative, and positive above the x axis. Graph A also has this characteristic.

We conclude that the correct match is graph A.

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y' - y / t = 0 gives us

y ' = y / t.

As we approach the y axis, t approaches 0 the slope therefore approaches +- infinity, and as we move away from t = 0 the slope approaches zero.

The only graph with these characteristics is C.

As y increases in magnitude, y / t increases in magnitude, so the graph should get steeper as we move away from the t axis on any vertical line. Graph C has this characteristic.

y / t is positive in quadrants 1 and 3, negative in quadrants 2 and 4. So the slopes will be positive in quadrants 1 and 3, negative in quadrant 4. Graph C has this characteristic.

We conclude the graph C is the correct match.

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y ' - t y = 0 gives us

y ' = t y.

As t or y approaches zero, the slope therefore approaches zero. At least three graphs have this characteristic.

t y is positive in quadrants 1 and 3, negative in quadrants 2 and 4. The slopes must share this description. The only graphs with the first characteristic and this one is B.

We conclude that B is the appropriate graph.

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y ' + t y = 0 gives us

y ' = - t y.

As t or y approaches zero, the slope therefore approaches zero. At least three graphs have this characteristic.

t y is positive in quadrants 2 and 4, negative in quadrants 1 and 3. The slopes must share this description. The only graphs with the first characteristic and this one is F.

We conclude that F is the appropriate graph.

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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:

slope = (8-2)rise/(3-1)run = 6/2 = 3

@& This is the average slope between two points, not the slope at any specific point.

*@

y’= -by = -b(8-2) = -b*6, where y’ = 3

3 = -b*6, so…

b = -.5

@&

@& The general solution of this equation is easily found to be

y = A e^(-b t).

The given (t, y) points yield the equations

2 = A e^(- b)

and

8 = A e^(-3 b).

We solve these equations for A and b.

Dividing the second by the first we obtain

4 = e^(-2 b)

which gives us

-2 b = ln(4)

so that

b = - ln(4) / 2.

Substituting this into either of the original equations we get A. We choose the first equation:

2 = A e^(- (-ln(4) )/ 2) = A * 4^(1/2) = 2 A

so that

A = 1.

Our solution is

y = e^( ln(4) / 2 * t),

which could also be expressed as

y = e( ln(4) ) ^ ( t / 2 ) = 4^(t / 2) = 4^(1/2) * t = 2^t.

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Given Solution:

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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 ' ?

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

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

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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:

What is w ' ?

This derivative, by the Chain rule, is dw / dt = y ' (t) * w ' ( y (t) )

y’(t)*(y(t)+2)’ = y(t)’*(y(t)’ + 2’) = y(t)’*(y(t)’+0)

w(t)’ = y(t)’^2

@& This would be dw/ddt = dw/dy * dy/dt.

You appear to be thinking of both derivatives as being with respect to t.

That makes things much more difficult, as in nonlinear.

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

For w(t) = y(t) + 2

y(t) = w(t) - 2

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

w = y + 2

w’ = y’^2

Now solve the equation and check your solution:

w = y+2

y = w-2

w(t)’ = 2y(t)’

Int( w(t) dt) = Int(2y(t) dt)

w = y^2 + c

@& It's way easier than you made it. You'll understand:

@& Letting w = y + 2, w ' = y ' and y = w - 2 so the equation

y ' - y = 2 (which is first-order linear nonhomogeneous)

becomes

w ' - (w - 2) = 2,

which simplifies to

w ' - w = 0,

a first-order linear homogeneous equation.

We easily solve this to obtain

w = A e^t.

It follows that

y + 2 = A e^t

so that

y = A e^t - 2.

We see that a simple change of variable made the equation (which wasn't all that difficult to solve in the first place) easier to solve.*@

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confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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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 = 2.5/1 = 2.5

@& It's not clear on what you based your estimate of b, though 2.5 is a plausible estimate.

Compare your thinking with the following:

@& y(0) is just the y-intercept, which we see to be 1.

From the graph the slope at this point appears to be about y ' (0) = 1. If this is accurate, then at t = 0 we have

y ' = b y = 1, and y = 1,

so that b = 1.

This has the advantage of not requiring us to solve the equation. However it's difficult to estimate the slope accurately, since this involves sketching a good tangent line then determining its slope.

We might improve our accuracy if we first solve the equation, then fit our solution to two points on the graph.

The solution is y(t) = A e^( b t).

y(0) = 1 implies that A = 1, so

y(t) = e^(b t).

The t = -1 value is a bit lower than .5, but probably a bit higher than .4. Using .45 as the approximate t = -1 value we get

-45 = e^(b * (-1) )

so that

-b = ln(.45) = -.8, approx..

Thus, based on our estimate of y(-1), we have b = .8 and our function would be

y = e^(.8 t).

We can check this against the y = 1 value on our graph, which appears to be around 2.3. This compares pretty well with the value y(1) = e^(.8 * 1) = 2.23.

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Given Solution:

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@& This looks pretty good, but be sure to see my notes and let me know if you have questions.*@