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1. Divide by t so that the leading term has coefficient 1. Then the equation is of the form y'''' + (1/t) y''' + ..., so the Wronskian is of the form C e^(integral (-1/t) dt) = C/t, by Abel's theorem.
2. These functions are linearly dependent (as the first is the second plus the third), so the Wronskian is 0.
3. No, they can't; if they could, the Wronskian would be either always 0, or never 0 (on I); see Theorem 3.3.3.
4. e^t is another solution. Here's why.
Rewrite the equation as y'' + (t/(1-t)) y' - (1/(1-t)) y = 0. By Abel's theorem,
W(y_1, y_2) = C e^{- integral (t/1-t)} = C e^{t + ln(1-t)} = C(1-t) e^t.
We will find y_2 so that W(y_1, y_2) = t e^t, i.e.
y_1 y_2' - y_1' y_2 = (1-t) e^t
t y_2' - y_2 = (1-t) e^t
y_2' - (1/t ) y_2 = ((1-t)/t) e^t.
Now use an integrating factor of 1/t, to get
(y_2/t)' = ( (1-t)/t^2 ) e^t
Integrating, we get y_2/t = -e^t/t + constant.
Taking the constant = 0, we get y_2 = -e^t is a solution.
5. The answer can be found using the "Variation of Parameters" formula, see Thereom 3.7.1.
Here is the argument, explicitly. phi_1(x) = cos(x), phi_2(x)=sin(x) be a basis for the space of solutions. We seek a single solution to the differential equation, as we already know the solutions to the homogeneous version.
We look for a solution phi = u_1 phi_1 + u_2 phi_2, such that
u_1' phi_1 + u_2' phi_2 = 0 (*)
Then
(*) phi' = u_1 phi_1' + u_2 phi_2',
and
phi'' = (u_1 phi_1'' + u_2 phi_2'') + (u_1' phi_1' + u_2' phi_2').
As phi_1'' = -phi_1 and phi_2" = - phi_2,
(**) phi'' = - (u_1 phi_1 + u_2 phi_2) + (u_1' phi_1' + u_2' phi_2').
Now phi'' + phi = f, so from (**),
u_1' phi_1' + u_2' phi_2' = f.
Solving the system of equations
u_1' phi_1 + u_2' phi_2 = 0
u_1' phi_1' + u_2' phi_2' = f(x)
i.e.
u_1' cos x + u_2' sin x = 0
u_1' cos x + u_2' sin x = f(x)
we find u_2' = f(x) cos x, u_1' = -f(x) sin x, so we take u_2 = integral f(s) cos s ds, u_1 = - integral f(s) sin s ds. The answer is:
u_1 phi_1 + u_2 phi_2 = integral ( - f(s) sin s cos x + f(s) cos s sin x ) ds = integral f(s) sin (x-s) ds as desired.
as desired.
You can check that this satisfies the differential equation.
u_1' phi_1' + u_2' phi_2' = x (**).
Rewriting (*) and (**):
u_1' x^2 + u_2' /x = 0
u_1' (2x) + u_2' (-1/x^2) = x
and solving this systems gives u_1' = 1/3, u_2' = -1/3 x^2.
Take u_1 = 1/3 x, u_2 = -1/12 x^4. Then
phi = 1/3 x (x^2) -1/12 x^4 (1/x) = 1/4 x^3.
To be safe, we check that phi(x) = x^3/4 really does satisfy the differential equation.
6. Suppose you have a set S, a distance function on it, and a map f: S --> S.
(i) Suppose |x-y| > |f(x)-f(y)| for x <> y. Then there is at most one fixed point f(x) = x.
This applies in the case where S is an interval of R, f is a differentiable function, and f'(x)<1 on the interval.
(ii) Suppose there is a number r<1 such that r|x-y| > |f(x)-f(y)|, and S is complete, then there is exactly one fixed point, and it can be obtained by starting with any x in S, and applying f repeatedly.
S is complete if: given any sequence x_1, x_2, ... in S where for every e>0 there is an N so that |x_i-x_j| is less than epsilon for i, j = N, there is some x such that x_1, x_2, ... has limit x. i.e.
This applies in the case where S is a _closed_ interval of R, and f is differentiable with f'(x) less than r on the interval.
7. Find a function like f(x) = (x+2)/(x+1). For this function, note that
(i) f'(x) = -1/(x+1)^2,
(ii) For the interval I = (.01,infinity), |f'(x)| is less than r = 1/ 1.01^2,
(iii) root(3) is in the interval, and f(root(3)) = root(3).
So by the contraction mapping theorem, iterating f(x) (starting with any number in the interval) will give a sequence with limit root(3).
8. (See BD 2.11.1.) Consider the initial value problem y' = f(t,y), y(0)=0. If f and f_y are continuous in a rectangle |t| <= a, |y| <= b, then there is some interval |t| <= h <= a in which there exists a unique solution y=phi(t).
(You can also have a different initial condition, y(t_0) = y_0, in which case you should center the rectangle around (t_0, y_0).)
9. Answer: dw/ds = 1-(w+3)^3, w(0)=0. This is problem 2.11.2.
10. (From Apr. 3 recitation.) Check (as stated in the problem) that w(t) is a solution to the initial value problem y' = y with initial condition w(0)=1. By the existence and uniqueness theorem, there is only one such function, so w(t)=z(t), i.e. z(t+a)/z(a)=z(t). Now plug in t=b.
11. The eigenvalues are lambda_1 = 3 (with eigenvector v_1=(1 -2)), and lambda_2 = 7 (with eigenvector v_2 = (1 2)). The answer is
3^{100} (1 -2) /2 + 7^{100} (1 2)/2.