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Differential Equations                              Test #1                                        March 4, 2005

Name____________________           R.  Hammack                                    Score ______
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(1) Which of the following (if any) is a solution to the differential equation =-.

(a)   + x y +=2

Differentiating implicitly gives 2y y'+y+x y' +2x =0.
Solving for y ' gives

THIS IS A SOLUTION

(b)   + 2x y +=1

Differentiating implicitly gives 2y y'+2y+2x y' +2x =0.
Solving for y ' gives

THIS IS NOT A SOLUTION

(2) Consider the differential equation (-y)y'=y+x.   Describe the region of the xy-plane for which this differential equation would have a unique solution through any point (, ) in the region.  Please explain your conclusion.

Here we have y'=f(x,y)= and = .  Both of these functions are continuous except where their denominators are zero. Their denominaors are zero at (, ) exactly when = . Thus Theoreom 1.1 asserts there is a unique solution through any point  (, ) that is not on the parabola y=. The region is thus entire plane minus the points on the prabola.

(3)
Solve:    (1+)+=0

The variables can be separated.

(4)
Solve:    cos(x)+y sin(x)=1

Notice that this is a linear D.E. Let's begin by putting it in standard form. Divide both sides by cos(x).
+tan(x)y =sec(x)

The integrating factor is ==|sec x|
It doesn't matter whether we multiply both sides by positive or negative sec x (we're still multiplying both sides by the same expression), so let's multiply both sides by sec x.

(5)
Solve:    (1-2-2y)=4+4x y

(4+4x y)dx+(2+2y-1)dy=0

Note that =4x= so the D.E. is exact.
This means there is some function f(x,y) for which =4+4x y
and =2+2y-1, and  f(x,y)=c will be a general solution to the D.E.

Now we find f.

SOLUTION TO D.E. is    +2y+-y=c

(6)
This problem concerns the differential equation =+

(a) Find a general solution.

dy=(+)dx
Note this is homogeneous of degree 0. Set y=ux, dy = u dx + x du.

(b) Find the specific solution subject to the initial condition y(1)=2.