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Math 331: Differential Equations Midterm 1 (Version B) Department of Mathematics and Statistics University of Massachusetts
Instructor: Alain Bourget February 18, 2005
Question 1: Solve the following initial value problems. Indicate if the equation is linear, separable, exact or homogeneous.
(a)
y′^ = x
2 y^2 y(0) = − 2
Solution: The ODE is separable. We have
dy dx
x^2 y^2
⇒ y^2 dy = x^2 dx ⇒
y^3 3
x^3 3
- C ⇒ y^3 = x^3 + C ⇒ y(x) = 3
x^3 + C.
Since y(0) = −2, we deduce that −2 = 3
C ⇒ C = −8. Finally, the solution is y(x) = 3
x^3 − 8
(b)
ty′^ + 4y = t y(1) = 0
Solution: The ODE is linear. The integrating factor if μ(t) = e
R (^4) dt t (^) = t^4. We get
y′^ +
t y = 1 ⇒ t^4 y′^ + 4t^3 y = t^4 ⇒
d dt
t^4 y
= t^4 ⇒ t^4 y =
t^5 5
t 5
C
t^4
Since y(1) = 0, we deduce that 0 = 1/ 5 − C, so C = − 1 /5. The final solution is y(t) =
t 5
5 t^4
(c)
(ex^ sin y − 2 y sin x) + (ex^ cos y + 2 cos x + y^2 )y′^ = 0 y(2) = 0
Solution: The ODE is exact since M (x, y) = ex^ sin y − 2 y sin x and N (x, y) = ex^ cos y + 2 cos x + y^2 and
∂M ∂y
∂N
∂x
= ex^ cos y − 2 sin x.
Therefore, there exists a function ψ(x, y) such that ψx = M , ψy = N and the implicit solutions are given by ψ = C.
From ψx = M , we get
ψ =
M dx =
(ex^ sin y − 2 y sin x)dx = ex^ sin y + 2y cos x + h(y).
From ψy = N , we get
ex^ cos y + 2 cos x + h′(y) = ex^ cos y + 2 cos x + y^2 ⇒ h′(y) = y^2 ⇒ h(y) =
y^2 2
Thus, ψ(x, y) = ex^ sin y + 2y cos x + y
3 3 and the general solutions are^ e
x (^) sin y + 2y cos x + y^3 3 =^ c.
Since y(2) = 0, we deduce that C = 0. The final solution is ex^ sin y + 2y cos x +
y^3 3
(d)
y′^ = x
(^2) +3y 2 2 xy y(2) = − 2
Solution: The ODE is homogeneous. We make the substitution v = y/x to obtain the new equation
xv′^ + v = f (1, v) =
1 + 3v^2 2 v
⇒ x
dv dx
1 + 3v^2 2 v
− v =
1 + v^2 2 v
Last ODE is separable. We get
2 v 1 + v^2
dv =
dx x
⇒ ln(1+v^2 ) = ln x+ln C = ln Cx ⇒ 1+v^2 = Cx ⇒ v = ±
Cx − 1 ⇒ y(x) = ±x
Cx − 1.
Since y(2) = −2, we deduce that −2 = − 2
2 C − 1 ⇒ C = 1. So the final solution is y(x) = −x
x − 1.
Question 2: Assume that the population of some species of insects follow the differential equation given by
dy dt = 4y(1 − y/10)
where y(t) denotes the number of insects (in thousands) at time t (in years). The initial population is assumed to be 200.
(a) What are the equilibrium solutions?
Solution: The equilibrium solutions are y = 0 and y = 10.
(b) Without solving the equation, sketch the integral curve of the solution y(t).
Solution: Since y(0) = 0.2, we need to sketch the integral when 0 < y < 10. On that interval, y′(t) > 0, so y is increasing. Moreover, since y′′(t) = 4(1 − y/5)y′(t), y′′(t) > 0 on 0 < y < 5 and y′′(t) < 0 on 5 < y < 10. Thus, y is concave upward on 0 < y < 5 and it is concave downward on 5 < y < 10. The integral curve is then:
2
4
6
8
10
y
(^0 1 2) x 3 4
