Download AMS 361- Applied Calculus IV Homework 3

AMS 361- Applied Calculus IV
Homework 3 - Solution
Due Date: February 26th
1. Chapter 1.6 Problem 11, Page 71
Find general solution of the differential equation.
(x2 − y 2 ) y ′ = 2xy
Solution:
y′ =
2xy
2 xy
2xy
x2
=
=
x2 − y 2
x2 − y 2 x2
1 − ( xy )2
Homogenous DE: Use the substitution u = y/x to solve the Homogenous DE.
dy
du
=u+x
dx
dx
After substituting the transformation into the DE, we get
y = xu
u+x
du
2u
=
dx
1 − u2
⇒
x
⇒
du
u3 + u
2u − u(1 − u2 )
=
=
dx
1 − u2
1 − u2
It is transformed into the Seperable DE.
Z
Z
dx
1 − u2
du
=
u3 + u
x
Using the partial fraction:
1 − u2
A
Bu
1 − u2
=
= + 2
3
u +u
u(u2 + +1)
u
u +1
To find the values of A and B, set the coefficients of the polynomial terms individually.
1 − u2 = A(u2 + 1) + (Bu)u ⇒ A = 1
&
Z
Z dx
1
2u
du =
−
u u2 + 1
x
Integrating the both sides of the equation gives:
lnu − ln(u2 + 1) = lnx + lnC
u = Cx(u2 + 1)
⇒
A + B = −1
⇒
y = C(y 2 + x2 )
2. Chapter 1.6 Problem 18, Page 71
Find general solution of the differential equation.
(x + y)y ′ = 1
Solution: By using the substitution u = x + y, ⇒ u′ = 1 + y ′
1
1
⇒ u′ − 1 =
x+y
u
Z
Z
Z 1
du
u
1+u
1−
du = dx
=
⇒
du =
dx
u
1+u
1+u
Integration of the both sides of equations gives us
y′ =
u − ln(1 + u) = x + c
⇒
x + y − ln(1 + x + y) = x + c
y = ln(1 + x + y) + c
1
3. Chapter 1.6 Problem 24, Page 71
Find general solution of the differential equation.
2xy ′ + y 3 e−2x = 2xy
Solution:
e−2x 3
y
2x
Bernoulli DE is solved by using the substitution u = y 1−3 = y −2 .
y′ − y = −
u′ = −2y −3 y ′
The substitution transform the equation into the Linear DE.
u′ + 2u =
e−2x
x
Step1: Find the integrating factor ρ(x)where P (x) = 2 and Q(x) = e−2x /x.
ρ(x) = e
R
2dx
= e2x
Step2: Multiply the both sides of the DE by ρ(x).
e2x u′ + 2e2x u =
Step3:
Dx (e2x u) =
1
x
1
x
Step4: Integrate the both sides:
Z
e2x
1
2x
dx = lnx + c ⇒ u = e−2x (lnx + c) ⇒ y =
e u=
x
(lnx + c)
4. Chapter 1.6 Problem 51, Page 72
Find a general solution of second-order differential equation. Assume x, y, and/or y ′
positive where helpful(as in Example 11).
y ′′ = 2y(y ′ )3
Hint: Use the substitution p = y ′ defined in equation (36).
Solution:
y ′ = p,
y ′′ = pp′ = p(dp/dy)
Substitute into the DE gives:
p
dp
= 2yp3
dy
⇒
dp
= 2ydy
p2
Seperable equation is solved by integrating the both sides of equation.
Z
Z
1
dy
1
dp
= 2ydy ⇒ − = y 2 + c ⇒ p =
=− 2
p2
p
dx
y +c
(y 2 + c)dy = −dx ⇒
y3
+ cy = −x + d ⇒ y 3 + 3x + Ay + B = 0
3
2
5. Chapter 1.6 Problem 58, Page 72
Solve the differential equation
x
dy
− 4x2 y + 2y lny = 0
dx
Hint: The substitution v = lny transform the differential equation into the Linear First
Order Differential Equation.
Solution:
dy
dv
= ev
dx
dx
v = lny ⇒ y = ev ⇒
Substitute these into the DE gives:
xev
dv
− 4x2 ev + 2ev v = 0
dx
dv
2
+ v = 4x
dx x
The resulting Linear 1st order differential equation is solved by using the integrating factor
ρ(x). Step1: Find the integrating factor.
ρ(x) = e
R
2
x dx
2
= e2lnx = elnx = (x2 )lne = x2
Step2: Multiply the both sides of DE by integrating factor ρ(x).
x2
dv
2
+ x2 v = (4x)x2
dx
x
Step3: Left-hand side of DE is equal to Dx (x2 v).
Dx (x2 v) = 4x3
Step4: Integrating the both sides of equation gives
Z
x2 v = 4x3 dx = x4 + c ⇒ v = x2 + cx−2
y = ex
2
+cx−2
6. Chapter 1.6 Problem 59, Page 72
Solve the differential equation
dy
x−y−1
=
dx
x+y+3
by finding h and k so that the substitutions x = u + h, y = v + k transform it into the
homogeneous equation
dv
u−v
=
.
du
u+v
Solution:
x = u + h ⇒ dx = du,
y = v + k ⇒ dy = dv
u−v+h−k−1
u−v
dv
=
=
du
u+v+h+k+3
u+v
3
h and k should satisfy the equations h − k = 1 and h + k = −3 to transform the given
DE into the homogeneous equation. So h = −1 and k = −2. The substitutions should be
x = u − 1 and y = v − 2.
u−v
dv
u−v
u
=
= u+v
du
u+v
u
1 − uv
dv
=
du
1 + uv
To solve Homogenous DE, use the substitution w = v/u.
dv
dw
=w+u
du
du
1−w
dw
1 − w − w(1 + w)
−(w2 + 2w − 1)
dw
=
⇒ u
=
=
w+u
du
1+w
du
1+w
1+w
Z
Z
1+w
−1
dw =
du
w2 + 2w − 1
u
v = uw ⇒
1
ln(w2 + 2w − 1) = −lnu + lnC ⇒ ln(w2 + 2w − 1) = −2lnu + 2lnC = −lnu2 + lnC 2
2
(w2 + 2w − 1)u2 = c ⇒ v 2 + 2uv − u2 = c
(y + 2)2 + 2(x + 1)(y + 2) − (x + 1)2 = c ⇒ y 2 + 2xy − x2 + 2x + 6y = c
7. Chapter 1.6 Problem 63, Page 72
The equation dy/dx = A(x)y 2 + B(x)y + C(x) is called a Riccati equation. Suppose
that one particular solution y1 (x) of this equation is known. Show that the substitution
1
v
transforms the Riccati equation into the linear equation
y = y1 +
dv
+ (B + 2Ay1 )v = −A
dx
Solution: First remind ourselves that the particular solution of the DE satisfies the DE.
That means
dy1
= A(x)y12 + B(x)y1 + C(x)
dx
From the substitution
y = y1 + v −1 ⇒ y ′ = y1′ − v −2 v ′
After substituting
dv
dy1
− v −2
dx
dx
=
A(x)(y1 + v −1 )2 + B(x)(y1 + v −1 + C(x)
=
A(x)(y12 + v −2 + 2y1 v −1 ) + B(x)(y1 + v −1 ) + C(x)
If we use the DE for the particular solution, we will get
−v −2
dv
= A(x)(v −2 + 2y1 v −1 ) + B(x)v −1
dx
dv
= A(x)(−1 − 2y1 v) − B(x)v = −(B + 2Ay1 )v − A
dx
So;
dv
+ (B + 2Ay1 )v = −A
dx
4