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 anotherstudent’s solution and handing it in as your own is considered cheating. Don’t 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:

dy

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:

dy

dx=

y2 − 1

x2 − 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:

(1)

∫dy

y2 − 1=

∫dx

x2 − 1

Using the partial fractions decomposition

1

u2 − 1=

1

(u− 1)(u + 1)=

1

2

(1

u− 1− 1

u + 1

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

(2)1

2ln

(y − 1

y + 1

)=

1

2ln

(x− 1

x + 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 = y2−1

x2−1 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.

xdy

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

dy

dx+

2

xy =

3

x

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

2xdx = x2. Doing this we obtain the differential equation x2 dy

dx +2xy =

6x2. Integrating both sides with respect to x we obtain x2y = 2x3 + C.Solving for y this is y = 2x + C

x2 . 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.

ydx

dy− x = 2y2, y(1) = 5

Solution: y =49/5+

√(49/5)2+8x

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

dx

dy− x/y = 2y

and multiplying by the integrating factor = e∫− 1

y dy = y−1 we obtain

1

y

dx

dy− x

y2= 2

Integrating both sides (with respect to y, treating x as a function of y)gives x

y = 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 = 0

So y =49/5+

√(49/5)2+8x

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

square root, otherwise if y =49/5−

√(49/5)2+8x

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

y5

)dy

dt+

t

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

y5

)dy +

t

2y4dt = 0

which is of course the same as

t

2y4dt +

3y2 − t2

y5dy = 0

and so M(t, y) = t2y4 , and N(t, y) = 3y2−t2

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) =∫

t2y4 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

y5

from which it is clear that g′(y) = 3y−3. So g(y) = − 32y2 .

Therefore our solution is of the form f(t, y) = C, so we have

t2

4y4− 3

2y2= C

Our initial value constraint y(1) = 1 gives us 1/4−3/2 = C. So C = −5/4So an implicit solution to the IVP is

t2

4y4− 3

2y2= −5/4