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Ordinary Differential Equations. Math 22B-002, Spring 2017. Final Exam: Solutions. 1. [15 pts.] Solve the initial value problem. 6y// − y/ − y = 0,.
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Ordinary Differential Equations Math 22B-002, Spring 2017 Final Exam: Solutions
6 y′′^ − y′^ − y = 0, y(0) = 10, y′(0) = 0.
Solution.
y(x) = c 1 ex/^2 + c 2 e−x/^3.
c 1 + c 2 = 10,
c 1 −
c 2 = 0
whose solution is c 1 = 4, c 2 = 6, so
y(x) = 4ex/^2 + 6e−x/^3.
y′^ − 2 xy = x, y(0) = y 0.
(b) For what initial value y 0 does the solution remain bounded as x → +∞? What is the solution in that case?
Solution.
μ(x) = exp
2 x dx
= e−x
2 .
2 y
= xe−x
2 ,
so e−x
2 y(x) =
xe−x
2 dx + c = −
ex
2
and the general solution is
y(x) = −
2 .
y(x) = −
y 0 +
ex
2 .
Remark. Alternatively, we could use variation of parameters: the solution of the homogeneous equation is y(x) = cex 2 , so write y(x) = u(x)ex 2 and solve for u(x); or we could observe that yp(x) = − 1 /2 is a particular solution, and write y(x) = yp(x) + cex 2 .
y′′^ + y = ex^ + x^2.
(b) Find a particular solution of the ODE
y′′^ − y = ex^ + x^2.
Solution.
yp(x) = Aex^ + Bx^2 + C.
ex^ + x^2 − 2.
(b) The function ex^ is a solution of the homogeneous equation, so we look for a particular solution of the form
yp(x) = Axex^ + Bx^2 + C.
xex^ − x^2 − 2.
λ 1 = − 2 , ~r 1 =
; λ 2 = 1, ~r 2 =
(a) Write down the general solution for (x(t), y(t))T^ of the system
( x y
x y
(b) Sketch the phase plane of this system. Include in your sketch the trajec- tory that passes through the point (0, 2).
(c) Sketch the graphs of x(t) and y(t) versus t for the solution of this system that satisfies the initial condition x(0) = 0, y(0) = 2.
Solution.
= c 1 e−^2 t
where c 1 , c 2 are arbitrary constants.
Figure 3: Phase plane for 6(b).
( x y
x y
Write your answer in complex form.
(b) Sketch the phase plane of this system.
Solution.
1 − λ 2 − 5 − 1 − λ
∣ =^ λ
so the eigenvalues satisfy λ^2 + 9 = 0, and λ = ± 3 i.
p q
One solution is (^) ( p q
−1 + 3i
Any non-zero (complex) scalar multiple of this vector would do just as well.
= c 1 e^3 it
−1 + 3i
− 1 − 3 i
where c 1 , c 2 are arbitrary (complex) constants. To get real-valued solutions, we take c 1 and c 2 to be complex conjugates.
= −kT 4 , T (0) = T 0 ,
where k > 0 is a constant.
(a) Solve for T (t).
(b) Let t 1 / 2 be the time it takes for the bar to cool to half its initial temper- ature. Find an expression for t 1 / 2 in terms of k and T 0.
(c) Does t 1 / 2 increase, decrease, or stay the same as T 0 increases? How does this compare with the behavior of t 1 / 2 for an object that cools according to Newton’s law of cooling (dT /dt = −kT )?
Solution.
k, dt + c
Integration gives −
3 T 3 (t)
= −kt + c.
The initial condition gives c = − 1 / 3 T 03 , and solving for T , we find that
T (t) =
(1 + 3T 03 kt)^1 /^3
Solving for t 1 / 2 , we get that
t 1 / 2 =
3 T 03 k
y′^ + sin^2 (x + y) = 0. (1)
(a) What is the order of this ODE? Is it linear or nonlinear? Separable or non-separable?
(b) What can you say from general theorems about the existence and unique- ness of solutions of the initial value problem for (1) with the initial condition y(x 0 ) = y 0?
(c) Show that the substitution u(x) = x + y(x) gives the following ODE:
u′^ = cos^2 u. (2)
(d) Solve (2) for u(x) and write down the general solution of (1) for y(x). For what x-values do your solutions exist?
(e) What are the solutions of (1) with the initial conditions : (i) y(0) = 0; (ii) y(0) = π/2?
Solution.
u′^ − 1 + sin^2 u = 0.
Since 1 − sin^2 u = cos^2 u, it follows that u′^ = cos^2 u.
dx + c.
Since 1/ cos^2 u = sec^2 u and ∫ sec^2 u du = tan u + c,
