MATHEMATICS 215 MIDTERM ANSWERS February 10, 2015 1. [5

MATHEMATICS 215 MIDTERM ANSWERS February 10, 2015 1. [5

MATHEMATICS 215
MIDTERM ANSWERS
February 10, 2015
1. [5 points each] Solve the following differential equations. In each case express y as a function of x.
dy
2x + 1
=
y(0) = −2 Answer: This is a separable equation with implicit solution y 2 = x2 + x + C.
dx
2y
The condition y(0) √
= −2 implies that C = 4 and that we need to take the negative square root when
solving for y: y = − x2 + x + 4.
dy
(b)
− y = 2e2x .
Answer: This is a linear equation with P (x) = −1 and Q(x) = 2e2x . The integration
dx
R
R
R
factor is H(x) = e P dx = e−x . Then H(x)Q(x) dx = 2ex dx = 2ex + C giving the solution y =
2e2x + Cex . This can also be solved using the method of undetermined coefficients.
dy
y
y2
(c)
= + 2 x > 0.
Hint: Substitute y = xv where v is a function of x.
Answer: Differentiating
dx
x x
y = xv, gives y 0 = v+xv 0 . Plugging this into the differential equation yields the separable equation xv 0 = v 2 ,
with solution v = −1/(ln x + C). Finally y = xv = −x/(ln x + C).
(a)
2. [5 points each] Solve the following differential equations. If possible, express y as a function of x.
(a) (cos x)y 0 = (sin x)y
Answer: Written as (cos x) dy −(sin x)y dx this equation is exact. Solution is y cos x =
1
C or y = C sec x. This equation is also separable, dy = tan x dx, with the same solution.
y
2
dy
+ y = 4x. Integrating factor is H = x2 so
(b) xy 0 + 2y = 4x2 . Answer: Linear when written as
dx x
Z
1
1
C
y = 2 (x2 )(4x) dx = 2 (x4 + C) = x2 + 2 .
x
x
x
(c) y 3 dx + (xy 2 − 1) dy = 0. Hint: Multiply first by the integrating factor 1/y 2 .
Answer: This equation
∂M
∂N
is exact when multiplied by 1/y 2 with M = y, N = x − 1/y 2 , and
=
= 1. Integration gives
∂y
∂x
F = xy + 1/y, so the implicit solution is xy + 1/y = C.
3. [5 points each] Give the general solutions of the following differential equations.
(a) (D3 −D2 +2)y = 0 Answer: Auxiliary equation: m3 −m2 +2 = (m+1)(m2 −2m+2). Roots: m = −1, 1±i.
Solution: y = c1 e−x + ex (c2 cos x + c3 sin x).
(b) y 000 = x
Answer: y = x4 /24 + c1 + c2 x + c2 x2
(c) y (4) − 2y 00 + y = 0 Answer: y = (c1 + c2 x)ex + (c3 + c4 x)e−x
4. [5 points each] Find one solution of each of the following equations.
(a) y 00 + y 0 = 4x Answer: Case 2: We guess yp = Ax2 + Bx. Since (D2 + D)yp = 2Ax + 2A + B, we set
A = 2, B = −4 and C = −2 to get yp = 2x2 − 4x.
1
1
1
(b) y 0 − 4y = e4x cos x Answer: yp =
e4x cos x = e4x
cos x = e4x cos x = e4x sin x
D−4
(D + 4) − 4
D
D
+
4
1
1
1
(c) y 0 − 4y = cos x Answer: yp =
cos x = 2
cos x = − (D + 4) cos x = − (− sin x + 4 cos x)
D−4
D − 16
17
17
dy
y
+ P (x)y = Q(x) ln x
x > 0. Show that v =
is
dx
ln x
dy
dv v
a solution of a first order linear equation. Answer: Differentiating y = v ln x gives
= ln x + . Plugging
dx
dx x
dv v
these equations into the differential equation we get ln x
+ + P (x)v ln x = Q(x) ln x. Then dividing by ln x
dx x
dv
1
gives the linear differential equation
+ P (x) +
v = Q(x).
dx
x ln x
5. [5 points] Suppose that y is a solution of the equation