Math 334
Assignment 1 — Solutions
1. Direction Fields. Sketch the direction field for each of the following ordinary differential equations.
(a) dy
dx =x2−y2,(b) dy
dx =x2−y2
x2+y2,(c) dy
dx =x2
y2,(d) dy
dx =x−y.
Solution
(a) (b)
(c) (d)
1
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Math 334 Assignment 1 Solutions: Direction Fields, Separable Equations, Exact Equations, Exercises of Mathematics
The solutions to assignment 1 of math 334, which covers topics such as direction fields, separable equations, exact equations, and integrating factors. The solutions include sketches of direction fields, analytical solutions to initial value problems, and the determination of domains of solutions.
Typology: Exercises
2012/2013
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Math 334
Assignment 1 — Solutions
- Direction Fields. Sketch the direction field for each of the following ordinary differential equations.
(a)
dy dx
= x^2 − y^2 , (b)
dy dx
x^2 − y^2 x^2 + y^2
, (c)
dy dx
x^2 y^2
, (d)
dy dx
= x − y.
Solution
(a) (b)
(c) (d)
- Separable equations. Solve the following initial value problems.
(a) y′^ = x^3 (1 − y), y(0) = 3; (b) dy/dθ = y sin θ, y(0) =
(c) dy/dx = ex
2 /y^2 , y(0) = 1; (d) dy/dx = (1 + y^2 )
1 + sin x, y(0) = 1.
Solution
(a) y′^ = x^3 (1 − y) =⇒ dy 1 − y
= x^3 dx =⇒ − ln | 1 − y| = x^4 4
- C 1 =⇒ y = 1 + Ce−x (^4) / 4 .
y(0) = 3 =⇒ C = 2. Therefore the solution is: y(x) = 1 + 2e−x (^4) / 4 .
(b)
dy dθ
= y sin θ =⇒
dy y
= sin θ dθ =⇒ ln |y| = − cos θ + C 1 =⇒ y = Ce−^ cos^ θ.
y(0) =
3 =⇒ C =
3 e. Therefore the solution is: y(θ) =
3 e^1 −cos^ θ^.
(c) dy dx
ex
2
y^2
=⇒ y^2 dy = ex 2 dx =⇒ y^3 3
∫ (^) x
0
et
2 dt + C.
y(0) = 1 =⇒ C =
. Therefore the solution is: y(x) =
∫ (^) x
0
et
2 dt + 1
(d) dy dx
= (1 + y^2 )
1 + sin x =⇒ dy 1 + y^2
1 + sin x dx =⇒ tan−^1 y = − 2
1 − sin x + C.
y(0) = 1 =⇒ C =
π 4
- Therefore: y(x) = tan(
π 4
1 − sin x)).
- Interval of definition. By looking at an initial value problem dy/dx = f (x, y) with y(x 0 ) = y 0 , it is not always possible to determine the domain of the solution y(x) or the interval over which the function y(x) satisfies the differential equation.
(a) Solve the equation dy/dx = xy^3. (b) Give explicitly the solutions to the initial value problem with y(0) = 1; y(0) = 1/2; y(0) = 2. (c) Determine the domains of the solutions in part (b). (d) As found in part (c), the domains of the solutions depend on the initial conditions. For the initial value problem dy/dx = xy^3 with y(0) = a, where a > 0, show that as a → 0 +, the domain approaches the whole real line (−∞, ∞), and as a → +∞, the domain shrinks to a single point.
Solution
(a) dy dx
= xy^3 =⇒ dy y^3
= x dx =⇒
2 y^2
x^2 2
- C 1 =⇒ y(x) =
C − x^2
(b) i. y(0) = 1 =⇒ C = 1 =⇒ y(x) =
1 − x^2
(a) From the equation (3x^2 + y)dx + (x^2 y − x)dy = 0
we have M (x, y) = 3x^2 + y and N(x, y) = x^2 y − x, which leads to
∂M
∂y
= 1 and
∂N
∂x
= 2xy − 1.
Since
∂M
∂y
∂N
∂x
the equation is not exact. However,
N
∂M
∂y
∂N
∂x
x
which implies that
there exists an integrating factor, depending only on x, which satisfies
μ
dμ dx
x
. An integrating
factor is therefore given by μ(x) =
x^2
. Multiplying through by μ leads to the equation
y x^2
)dx + (y −
x
)dy = 0,
which is exact. There exists a function F which satisfies ∂F ∂x
y x^2
∂F
∂y
= y −
x
The first of these implies F (x, y) = 3x −
y x
- g(y). Differentiating with respect to y and comparing
the result with the second equation in (1) leads to g′(y) = y or g(y) =
y^2 2
The solution to the original equation is
3 x − y x
y^2 2
= C.
(b) From the equation (y^2 + 2xy)dx − x^2 dy = 0
we have M (x, y) = y^2 + 2xy and N(x, y) = −x^2 , which leads to
∂M
∂y
= 2x + 2y and
∂N
∂x
= − 2 x.
Since
∂M
∂y
∂N
∂x
the equation is not exact. However,
M
∂N
∂x
∂M
∂y
y
which implies that
there exists an integrating factor, depending only on y, which satisfies
μ
dμ dy
y
. An integrating
factor is therefore given by μ(y) =
y^2
. Multiplying through by μ leads to the equation
2 x y^2 )dx −
x^2 y^2 dy = 0,
which is exact. There exists a function F which satisfies ∂F ∂x
2 x y
∂F
∂y
x^2 y^2
The first of these implies F (x, y) = x + x^2 y
- g(y). Differentiating with respect to y and comparing the result with the second equation in (2) leads to g′(y) = 0 or g(y) = constant. The solution to the original equation is
x +
x^2 y = C or y(x) =
x^2 C − x
- Integrating factors II. Find integrating factors of the form xnym^ to solve the following equations:
(a) (2y^2 − 6 xy)dx + (3xy − 4 x^2 )dy = 0; (b) (12 + 5xy)dx + (6xy−^1 + 3x^2 )dy = 0.
Solution
(a) Multiply the equation by μ(x, y) to get
(2xnym+2^ − 6 xn+1ym+1) dx + (3xn+1ym+1^ − 4 xn+2ym) dy = 0.
For this equation to be exact requires
∂y
(2xnym+2^ − 6 xn+1ym+1) =
∂x
(3xn+1ym+1^ − 4 xn+2ym), which implies that
2(m + 2)xnym+1^ − 6(m + 1)xn+1ym^ = 3(n + 1)xnym+1^ − 4(n + 2)xn+1ym. (3)
Upon rearrangement we get [(2m − 3 n + 1)y − 2(3m − 2 n − 1)x]xnym^ = 0. This is true if 2 m − 3 n + 1 = 0 and 3m − 2 n − 1 = 0, which yields m = n = 1. Therefore, an integrating factor which renders the equation exact is μ(x, y) = xy and equation (3) becomes
(2xy^3 − 6 x^2 y^2 ) dx + (3x^2 y^2 − 4 x^3 y) dy = 0,
which is easily solved to give x^2 y^3 − 2 x^3 y^2 = C.
(b) Multiply the equation by μ(x, y) to get
(12xnym^ + 5xn+1ym+1) dx + (6xn+1ym−^1 + 3xn+2ym) dy = 0.
For this equation to be exact requires
∂y
(12xnym^ + 5xn+1ym+1) =
∂x
(6xn+1ym−^1 + 3xn+2ym), which implies that
12 mxnym−^1 + 5(m + 1)xn+1ym^ = 6(n + 1)xnym−^1 + 3(n + 2)xn+1ym. (4)
Upon rearrangement we get [6(2m − n − 1)y−^1 + (5m − 3 n − 1)x]xnym^ = 0. This is true if 2 m − n − 1 = 0 and 5m − 3 n − 1 = 0, which yields m = 2, n = 3. Therefore, an integrating factor which renders the equation exact is μ(x, y) = x^3 y^2 and equation (4) becomes
(12x^3 y^2 + 5x^4 y^3 ) dx + (6x^4 y + 3x^5 y^2 ) dy = 0,
which is easily solved to give 3 x^4 y^2 + x^5 y^3 = C.
- Integrating factors III. Show that if (∂N/∂x − ∂M/∂y)/(xM − yN) depends only on the product xy, that is ∂N/∂x − ∂M/∂y xM − yN = H(xy),
then the equation M (x, y)dx + N(x, y)dy = 0 has an integrating factor of the form μ = μ(xy). Give the general formula for μ(xy).