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  • Assignment 1MATH 2270SOLUTION

    Please write out complete solutions for each of the following 6 problems (one morewill still be added). You may, of course, consult with your classmates, the text-book or other resources, but please write up your own solutions. Copying anotherstudents solution and handing it in as your own is considered cheating. Dont doit!

    (1) Section 2.1. Problem 10 Use MAPLE (or some other computer software)to obtain a direction field for the given differential equation. Print it outand by hand sketch on the vector field approximate solution curves passingthrough each of the given points:


    dx= xey

    (a) y(0) = 2(b) y(1) = 2.5

    SOLUTION: See next page


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    (2) Section 2.1 problem 30. Consider the autonomous differential equationdy/dx = f(y) where the graph of f is given in the book (not reproducedhere). Use the graph to locate the critical points of the differential equation.Sketch a phase portrait. And by hand, sketch typical solution curves in thesubregions in the xy-plane determined by the graphs of the equilibriumsolutions.

    SOLUTION:The critical values of the differential equation are at approximately y =

    2.2, 0.5 and y = 1.5. For the phase portrait and typical solution curves,see the next page.

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    (3) Section 2.2. Problem 24: Find an explicit solution to the given initial valueproblem:



    y2 1x2 1

    , y(2) = 2.

    Solution: The solution is y = x on the interval (1,).How to find this: This is a separable differential equation, so we separate

    the variables and integrate both sides:



    y2 1=


    x2 1Using the partial fractions decomposition


    u2 1=


    (u 1)(u + 1)=




    u 1 1

    u + 1

    )We integrate both sides of equation (1) to obtain



    (y 1y + 1




    (x 1x + 1

    )+ C

    It is probably easier to find C using our constraint first, and then solve fory in terms of x: Using y = 2 when x = 2 (our initial value constraint) weobtain

    1/2(ln(1/3)) = 1/2(ln(1/3)) + C

    So, C = 0. So, solving (2) where C = 0 we get (y 1)/(y + 1) =(x 1)/(x + 1). Either you can observe that y = x is a solution, or dosome algebra to come to this conclusion. Now, returning to our original

    expression dydx =y21x21 we see that the right-hand side is continuous and has

    partial derivatives continuous for all values of (x, y) except when x = 1 orx = 1. So the largest interval containing our initial value x = 2 on whichthe differential equation is defined is (1,).

    (4) Section 2.3. Problem 10: Find the general solution to the differential equa-tion. Give the largest interval I over which the general solution is defined.


    dx+ 2y = 3.

    Solution: y = 2x+ Cx2 is a one parameter family of solutions. Either interval(, 0) or (0,) is an appropriate interval for the solution.

    How do we find this solution? Well, this is a linear first-order equation,so we can solve it by rewriting it in the standard form




    xy =



    and multiplying by the appropriate integrating factor. In this case that ise

    2xdx = x2. Doing this we obtain the differential equation x2 dydx +2xy =

    6x2. Integrating both sides with respect to x we obtain x2y = 2x3 + C.Solving for y this is y = 2x + Cx2 . This is defined on the intervals (, 0)and (0,). Either interval is a largest interval on which the solution isvalid.

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    (5) Section 2.3, Problem 28. Solve the given initial value problem. Give thelargest interval I over which the solution is defined.


    dy x = 2y2, y(1) = 5

    Solution: y =49/5+


    4 on the interval (12.005,).How to get this? Well, this equation is not linear in y but it is set up

    as a linear differential equation with independent variable y and dependentvariable x. So we can still solve by first rewriting it in the standard form


    dy x/y = 2y

    and multiplying by the integrating factor = e 1y dy = y1 we obtain




    dy x

    y2= 2

    Integrating both sides (with respect to y, treating x as a function of y)gives xy = 2y + C. Substituting in the initial value to solve for C gives

    1/5 = 10 + C. So C = 49/5. So we havexy = 2y 49/5,x = 2y2 (49/5)y2y2 (49/5)y x = 0So y =



    4 (since y(1) = 5 we need to take the positive

    square root, otherwise if y =49/5


    4 we would have y(1) = 8/5).And the largest interval on which this is defined is when (49/5)2 + 8x > 0.Solving this for x gives x > 12.005

    (6) Section 2.4. Problem 24: Solve the given initial-value problem.(3y2 t2





    2y4= 0, subject to y(1) = 1.

    Solution: Once we write this in differential form we can check to see if itis exact. (

    3y2 t2


    )dy +


    2y4dt = 0

    which is of course the same as


    2y4dt +

    3y2 t2

    y5dy = 0

    and so M(t, y) = t2y4 , and N(t, y) =3y2t2

    y5 , and it is easily checked that

    Nt = My = 2ty5 . So this is an exact equation.To solve it we integrate M(t, y) with respect to t to get f(t, y) =


    2y4 dt =t2

    4y4 + g(y).

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    To find g(y) we know that fy = N(t, y). But fy =t2y5 + g

    (y) =t2+y5g(y)

    y5 . So

    t2 + y5g(y)y5

    = N(t, y) =3y2 t2


    from which it is clear that g(y) = 3y3. So g(y) = 32y2 .Therefore our solution is of the form f(t, y) = C, so we have


    4y4 3

    2y2= C

    Our initial value constraint y(1) = 1 gives us 1/43/2 = C. So C = 5/4So an implicit solution to the IVP is


    4y4 3

    2y2= 5/4


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