Homework 5-solutions - feuge(ejf557 Homework 5 staron(53940...

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feuge (ejf557) – Homework 5 – staron – (53940) 1 This print-out should have 40 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points Consider the graph 2 2 4 2 4 6 2 x y Using the above graph of y ( x ), choose the equation whose solution satisfies the initial condition y (0) = 4 . 5. 1. y = y 1 2. y = y x 2 3. y = y 2 x 2 4. y = y 3 x 3 5. y = y x correct Explanation: Let us begin by approximating the tangent line about y (0). 2 2 4 2 4 6 2 x y From the above graph, we can see the tan- gent line about y (0) has a slope of about 4 . 5. This tells us that y (0) 4 . 5. Now, only two of our answer choices match this property. Those are y 1 ( x ) = y ( x ) x y 2 ( x ) = y ( x ) x 2 . Now we’ll pick a different value of x and use the same method. Let us choose x = 2. 2 2 4 2 4 6 2 x y Looking at this tangent, we can see y ( 2) 1 . 5. We can also see that y ( 2) ≈ − 0 . 5. If we use this value in our remaining two answer choices we get y 1 ( 2) = y ( 2) + 2 ≈ − 0 . 5 + 2 = 1 . 5 y 2 ( 2) = y ( 2) 4 ≈ − 0 . 5 4 = 4 . 5 .
feuge (ejf557) – Homework 5 – staron – (53940) 2 Now we can see only y 1 ( x ) can be the deriva- tive of our function. To show that it works, let us now look at a direction field of y 1 . 2 2 4 2 4 6 2 x y From this, we can see that the given graph is indeed a solution to the differential equation y = y x . 002 10.0points Consider the graph 2 2 4 2 4 6 2 x y Using the above graph of y ( x ), choose the equation whose solution is graphed and satis- fies the initial condition y (0) = 2 . 1. y = y 2 x 2 2. y = y 2 x 3. y = y 1 correct 4. y = y + 2 . 5 5. y = y x 1 6. y = y 3 x 2 7. y = y x Explanation: Let us begin by approximating the tangent line about y (0). 2 2 4 2 4 6 2 x y From the above graph, we can see the tan- gent line about y (0) has a slope of about 1. This tells us that y (0) 1. Now, only three of our answer choices match this property. Those are y 1 ( x ) = y ( x ) 1 y 2 ( x ) = y ( x ) x 1 y 3 ( x ) = y ( x ) + 2 . 5 Now we’ll look similarly at x = 4. Notice that y ( 4) 0 and y ( 4) 1. If we use this value in our remaining 3 answer choices, we get y 1 ( 4) = y ( 4) 1 1 1 = 0 y 2 ( 4) = y ( 4) + 4 1 1 + 4 1 = 4 y 3 ( 4) = y ( 4) + 2 . 5 ≈ − 1 + 2 . 5 = 1 . 5 .
feuge (ejf557) – Homework 5 – staron – (53940) 3 Now we can see only y 1 ( x ) can be the deriva- tive of our function. To show that it works, let us now look at a direction field of y 1 . 2 2 4 2 4 6 2 x y From this, we can see that the given graph is indeed a solution to the differential equation y = y 1. 003 10.0points Use the direction field of the differential equation y = y 4 to sketch a solution curve that passes through the point (0 , 1). 1. 2 2 2 2 x y 2. 2 2 2 2 x y 3. 2 2 2 2 x y correct 4. 2 2 2 2 x y 5. 2 2 2 2 x y
feuge (ejf557) – Homework 5 – staron – (53940) 4 6. 2 2 2 2 x y Explanation: The vector field for y = y 4 with a sample curve through our initial condition y (0) = 1 looks like the following 2 2 2 2 x y 004 10.0points Consider the following graph of y ( x ) 2 4 6 x y π 2 π 2 π π Choose the equation whose solution is graphed and satisfies the initial condition y (0) = 1 . 2.

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