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(MATH152)midterm2.pdf
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Solutions of Math 152 Midterm Exam

1 (30 points)
(1)
Find the general solution of equation xdx + ydy =0.

(2)
Solve the initial value problem: y . . t2 y = t2 ,y(0) = 0.


Solution:
(1) Multiply the equation by 2, we get d(x 2 + y 2)=0. Therefore, we have x 2 + y 2 = C (.)
where C is arbitrary and positive constant. Here I assume x and y are real variables, then C must be positive, otherwise (*) would not de.ne an implicit function.
(2) The integration factor is . 1
exp[(.t2)dt] = exp[. t3].
3 Multiply the equation by this integrating factor, we get
(exp[. 1 3 t3]y). = t2 exp[. 1 3 t3],
so exp[. 1 3 t3]y = . t2 exp[. 1 3 t3]dt = . exp[. 1 3 t3] + C.
So
y = .1 + C exp[ 1 3 t3].
Now y(0) = 0 implies that C = 1, so the solution is
y = .1 + exp[ 1 3 t3].

1

2 (25 points) Consider the di.erential equation
(.) t2 y .. + ty. . y =1, t> 0
a) Find a solution to equation (*) by inspection.
b) Find a nontrivial (i.e., not identically zero) solution to equation

t2 y .. + ty. . y =0 by inspection. c) Find the general solution of equation (*). Solution: a) By inspection yp = .1 is a solution to equation (*). b) By inspection y1 = t is a solution to equation
t2 y .. + ty. . y =0.
b) Let y = tv, then y. = v + tv. and y.. =2v. + tv... Plug y = tv into the equation
t2 y .. + ty. . y =0

we get t2(tv.. +2v .)+ t(tv. + v) . tv =0, or t3 v .. +3t2 v . =0, or (t3 v .). =0. .. C
So t3v= C, or v= t3 . So v = C1 + C2t.2 , or yh = C1t + C2t.1 . Here C1, C2 are arbitrary constants. So the general solution to equation (*) is y = .1+ C1t + C2t.1 .
3 (25 points) Find the general solution of the equation

.. . 2y .
y + y = (t . 1).
Solution: The characteristic equation of the corresponding homogeneous equation is
2 . 2 +1=0,
which has only one solution = 1. So the general solution of the corresponding homoge-neous equation is
yh = C1e t + C2tet ,
where C1 and C2 are arbitrary constants.
We use Laplace transform to .nd a particular solution yp. For that purpose we assume that yp(0) = 0and yp. (0) = 0. (You can make any assumption, but this one makes computations easy.)
Let F (s)= L{yp(t)}. Applying Laplace transform to the equation, we get
s 2F (s) . 2sF (s)+ F (s)= e .s .
So
1
F (s)= e .s .
(s . 1)2 Then yp(t)= L.1{F (s)} = u1(t)(t . 1)e t.1 . So the general solution is
tt t.1
y = C1e + C2te+ u1(t)(t . 1)e.
(Remark: You can assume y(0) = c1 and y.(0) = c2 and let c1 and c2 be arbitrary constants, then use Laplace transform to solve the equation to get the general solution.)
4 (20 points) Let

(.) y .. +2y . + cy = P (t)e at
be a linear second order nonhomogeneous di.erential equation. Here c, a are constants and P(t) is a polynomial of degree n. Use the Method of Variation of Parameters to show that there is a solution to equation
(*) which is of the form y = Q(t)eat where Q is a polynomial and the degree of Q(t) is 1) n if a is not a root of 2 +2 + c