Differential Equations Exam One NAME:
1. Solve (explicitly) the separable Differential Equation
dy
dx =y2+1
y(x+1) with y(0) = 2.
Separate: ydy
y2+1 =dx
x+1 .
Integrate: 1
2ln(y2+ 1) = ln(x+ 1) + c
Simplify: y2+ 1 = k(x+ 1)2.
Initial Value: 5 = k(1)
Final Answer: y=p5(x+ 1)2−1.
2. Solve (implicitly) the Exact Differential Equation
2x+y3sec2x)dx + (1 + 3y2tan x)dy = 0.
Integrate: [x2+y3tan x]S[y+y3tan x] = c.
Final Answer: x2+y3tan x +y=c.
3. Use Euler’s method with h=.2, for 3 steps, to find an approximate value
y(1.6) for the solution to the initial value problem dy/dx =y−xwith y(1) = 3.
x0= 1 and y0= 3 and yn+1 =yn+hf (xn, yn) = yn+ (.2)(yn−xn).
x1= 1.2 and y1= 3 + (.2)(3 −1) = 3.4.
x2= 1.4 and y2= 3.4+(.2)(3.4−1.2) = 3.84.
x3= 1.6 and y3= 3.84 + (.2)(3.84 −1.4) = 4.33.
Final Answer: y(1.6) ∼4.33.
4. Solve the equation y0=y2(y−2)(y−5) graphically.
(a) Find the critical values csuch that y=cis a solution.
(b) Find the sign of y0and the limiting value L on each interval.
(c) Sketch solutions with y(0) = −1, y(0) = 1 and y(0) = 6.
(a) y= 0, y= 2, and y= 5.
(b) y0>0 on (−∞,0), (0,2) and (5,∞); y0<0 on (2,5).
The limiting value is +∞on (5,∞).
The limiting value is 2 on (0,5).
The limiting value is 0 on (−∞,0].
