5525. Math April 14, 2010. Midterm Exam 2. Problems and Solutions. Problem 1. Find the general solution of the equation xy - (2x + 1)y + (x + 1)y = 0. Note that one of two linearly independent solutions is y1 (x) = ex . Solution. By Abel's formula, the Wronskian W (x) = W (y1 , y2 )(x) satisfies W =- 2x + 1 p1 W = W = p0 x 2+ 1 x W,
so that W (x) = W (x0 ) xe2x . On the other hand,
2 W (x) = y1
y2 y1
= e2x
y2 y1
.
Therefore, (y2 /y1 ) = const x, y2 /y1 = const x2 . We can take y2 = x2 y1 = x2 ex . Then the general solution has the form y = C1 y1 + C2 y2 = (C1 + C2 x2 )ex . Problem 2. Let y = y(x) be a solution of the problem y =1+y+ Show that y(x) - ln(1 - x) Proof. The function z = - ln(1 - x) satisfies z = 1 z2 = ez 1 + z + 1-x 2 for 0 x < 1, z(0) = 0. for 0 x < 1. y2 , 2 y(0) = 0.
Note that the comparison theorem (Theorem 11.1) can be applied in the case of non-strict inequalities, if the function f (x, y) is Lipschitz. Therefore, we have z(x) y(x) for 0 x < 1. Problem 3. Let y = y(x, ) be a solution of the problem y = y 2 + 2 x-1 , Find the derivative z = z(x, ) = Solution. At the point = 0, we have y = y2, The function z(x) satisfies z = 2yz + 2x-1 = -2x-1 z + x-1 , Therefore, z(x) = Cx-2 + 1 = 1 - x-2 . 1 z(1) = 0. -y -1 = x + C = x, y = -x-1 . y y(1) = -1.
at the point = 0.
Problem 4. Solve the problem u = 2 4 3 6 u+ -8 0 , u(0) 1 = 4 ; where u = u(x) = u1 (x) u2 (x) .
Solution. We know that the solution of the problem u (x) = Au(x)+b(x), u(0) = u0 , is represented in the form x eA(x-t) b(t) dt. u(x) = eAx u0 +
0
In our case, the characteristic polynomial p() = |I - A| = has roots 1 = 0 and 2 = 8. Denote p1 () = Then e We have eAx u0 = eAx eAx b = eAx
x Ax 2
-2 -4 -3 - 6
= 2 - 8
p() = - 8, - 1
p2 () = -6 4 3 -2 1 8 = -10 5 -6 3
p() = . - 2 + e8x 8 e8x 8 2 4 3 6 18 27 -2 -3 , .
=
k=1
ek x 1 pk (A) = pk (k ) -8 1 4 -8 0
t=x
=
+
,
+ e8x
e8(x-t) -8 0 x -6x eA(x-t) b(t) dt = 3x 0 e8(x-t) dt = Finally, u(x) = e u +
0 Ax 0 x
1 e8x , =- + 8 8 t=0 1 2 e8x + + 8 3 8 -6x - 1 3x + 1
-2 -3 + e8x
.
eA(x-t) b(t) dt =
2 3
.
Problem 5. Let A be a constant 2 2 matrix. Show that all solutions of the system u (x) = Au(x) satisfy the equality u(2) + c1 u(1) + c2 u(0) = 0, with constants c1 and c2 depending only on the matrix A, and not depending on the solution u(x). Hint. Apply the Cayley-Hamilton theorem to the matrix B = eA . Proof. Let c1 and c2 be the coefficients of the characteristic polynomial of the matrix B = eA : p() = |I - B| = 2 + c1 + c2 . By the Cayley-Hamilton theorem, p(B) = B 2 + c1 B + c2 I = e2A + c1 eA + c2 I = 0. Since any solution of u = Au has the form u(x) = eAx u0 , where u0 = u(0), we get u(2) + c1 u(1) + c2 u(0) = e2A u0 + c1 eA u0 + c2 u0 = p(B)u0 = 0. 2