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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 dy/dx=-(2x + y)/(2y + x).

(a)   y^2+ x y +x^2=2

Differentiating implicitly gives 2y y'+y+x y' +2x =0.
Solving for y ' gives
y ' (2y + x) = -2x - y  y ' = -(2x + y)/(2y + x)
THIS IS A SOLUTION


(b)   y^2+ 2x y +x^2=1

Differentiating implicitly gives 2y y'+2y+2x y' +2x =0.
Solving for y ' gives
y ' (2y + 2x) = -2x - 2y  y ' = -(x + y)/(y + x)
THIS IS NOT A SOLUTION



(2) Consider the differential equation (x^2-y)y'=y+x.   Describe the region of the xy-plane for which this differential equation would have a unique solution through any point (x_0, y_0) in the region.  Please explain your conclusion.


Here we have y'=f(x,y)=(y + x)/(x^2 - y) and ∂f/∂y=((x^2 - y) + (y + x))/( (x^2 - y)^2) .  Both of these functions are continuous except where their denominators are zero. Their denominaors are zero at (x_0, y_0) exactly when y_0= x_0^2. Thus Theoreom 1.1 asserts there is a unique solution through any point  (x_0, y_0) that is not on the parabola y=x^2. The region is thus entire plane minus the points on the prabola.


(3)  Solve:    (1+e^x)dy/dx+e^x=0

The variables can be separated.
d y = -e^x/(1 + e^x) d x  ∫d y = -∫e^x/(1 + e^x) d x  y = -ln | 1 + e^x | -ln(c)     (c>0)  y = -ln | c(1 + e^x) |


(4)   Solve:    cos(x)dy/dx+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).
dy/dx+tan(x)y =sec(x)

The integrating factor is e^(∫tan(x) dx)=e^(ln | sec x |)=|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.

sec(x) dy/dx + sec(x) tan(x) y = sec^2 (x)  d/dx[sec (x) y] = sec^2 (x)  sec (x) y = ∫sec^2(x) dx  sec (x) y = tan (x) + c  y = sin(x) + c cos(x)


(5)  Solve:    (1-2x^2-2y)dy/dx=4x^3+4x y

(4x^3+4x y)dx+(2x^2+2y-1)dy=0

Note that ∂M/∂y=4x=∂N/∂x so the D.E. is exact.
This means there is some function f(x,y) for which ∂f/∂x=4x^3+4x y
and ∂f/∂y=2x^2+2y-1, and  f(x,y)=c will be a general solution to the D.E.

Now we find f.
f (x, y) = ∫ (4x^3 + 4x y ) dx = x^4 + 2x^2y + g(y)  ∂f/∂y = 2x^2 ...  g (y) = ∫ (2y - 1) dy = y^2 - y + c  f(x, y) = x^4 + 2x^2y + y^2 - y + c

SOLUTION TO D.E. is    x^4+2x^2y+y^2-y=c


(6)   This problem concerns the differential equation dy/dx=y/x+x/y

(a) Find a general solution.

dy=(y/x+x/y)dx
Note this is homogeneous of degree 0. Set y=ux, dy = u dx + x du.
u dx + x du = (u + u^(-1)) dx  u du = 1/x  dx  u^2/2 = ln | x | + ... x |      (c>0)  u^2 = ln (cx )^2  (y/x)^2 = ln ( c^2x ^2)


(b) Find the specific solution subject to the initial condition y(1)=2.
(2/1)^2 = ln c^2  4 = ln  c^2  e^4 = c^2  c = e^4^(1/2) = e^2

(y/x)^2 = ln((e^2)^2x ^2)  (y/x)^2 = ln (e^4x )^2  (y/x)^2 = 4 + 2ln x