Homework #6 Solutions
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1 Problems Section.1: 6, 4, 40, 46 Section.:, 8, 10, 14, 18, 4, 0 Homework #6 Solutions.1.6. Determine whether the functions f (x) = cos x + sin x and g(x) = cos x sin x are linearly dependent or linearly independent on the real line. Solution: We compute the Wronskian of these functions: W( f, g) = f g f g = cos x + sin x cos x sin x sin x + cos x sin x cos x = ( cos x + sin x)( sin x cos x) ( cos x sin x)( sin x + cos x) = ( 1 cos x sin x 9 sin x 4 cos x) + (1 cos x sin x 9 cos x 4 sin x) = 1(sin x + cos x) = 1 Since the Wronskian is the constant function 1, which is not the 0 function, these functions are linearly independent on the real line (and in fact on any subinterval of the real line) Find the general solution of the DE y + y 15y = 0. Solution: Guessing the solution y = e rx, we obtain the characteristic equation r r + 15 = 0, which factors as (r 5)(r + ) = 0. Therefore, r = 5 and r = are roots, so y 1 = e 5x and y = e x are solutions. Furthermore, by Theorem 5 in.1, the general solution to the DE is y = c 1 e 5x + c e x Find the general solution of the DE 9y 1y + 4y = 0. Solution: As above, the characteristic equation for the DE is 9r 1r + 4 = 0, which factors as (r ) = 0. Therefore, this equation has a double root at r = /. By Theorem 6 in.1, the general solution is then y = c 1 xe x/ + c e x/. 1
2 Find a homogeneous second-order DE ay + by + cy = 0 with general solution y = c 1 e 10x + c e 100x. Solution: This constant coefficient DE must have y 1 = e 10x and y = e 100x as solutions, so we expect r 10 and r 100 to be factors of its characteristic polynomial. Then we may take ar + br + c = a(r 10)(r 100) = a(r 110r ), so taking a = 1, we have the corresponding DE y 110y y = Show directly that the functions f (x) = 5, g(x) = x, and h(x) = x are linearly dependent on the real line. Solution: We find a nontrivial linear combination c 1 f + c g + c h of these functions identically equal to 0. Since all functions are polynomials in x, the function is 0 exactly when the coefficients on all the powers of x are 0. Since c 1 f + c g + c h = c 1 (5) + c ( x ) + c ( x ) = (5c 1 + c + 10c ) + ( c + 15c )x, we require that 5c 1 + c + 10c = 0 and c + 15c = 0. From the second equation, c = 5c. Substituting this into the first, 5c 1 + c + 10c = 5c 1 + (5c ) + 10c = 5c 1 + 0c = 0. Then c 1 = 4c, and there are no more constraints on the c i. Choosing to set c = 1, c 1 = 4 and c = 5. We check that this nontrivial linear combination of functions is 0: ( 4)(5) + (5)( x ) + (1)( x ) = x x = 0. Rearranging this equation, we can express any single one of these functions as a linear combination of the other two: for example, x = 4(5) 5(10 x )...8. Use the Wronskian to prove that the functions f (x) = e x, g(x) = e x, and h(x) = e x are linearly independent on the real line. Solution: We compute W( f, g, h): f g h f g h f g h = e x e x e x e x e x e x e x 4e x 9e x ( = e x e x e x 4e x 9e x e x e x 4e x 9e x + e x e x e x e x ) = e x e x e x (( 9 4 ) ( ) + (1 1 )) = e 6x ( ) = e 6x. Since W(x) = e 6x, which is not zero on the real line (and in fact nowhere 0), these three functions are linearly independent.
3 ..10. Use the Wronskian to prove that the functions f (x) = e x, g(x) = x, and h(x) = x ln x are linearly independent on the interval x > 0. Solution: We compute W( f, g, h). First, we compute derivatives of h: h (x) = x ln x + x 1 = (1 ln x)x x h (x) = ( )(1 ln x)x 4 + x x = (6 ln x 5)x 4 Plugging these into the Wronskian, we have f g h f g h f g h = e x x x ln x e x x (1 ln x)x e x 6x 4 (6 ln x 5)x 4. Rather than expand this directly, we make use of some additional properties of the determinant. One of these is that the determinant is unchanged if a multiple of one column is added to or subtracted from a different column. We subtract ln x times the second column from the third to cancel the ln x terms there: e x x 0 e x x x e x 6x 4 5x 4 Using another property of the determinant, we factor the scalar e x out of the first column, so that it multiplies the determinant of the remaining matrix: e x 1 x 0 1 x x 1 6x 4 5x 4 With these simplifications, we expand along the first row, which conveniently contains a 0 entry: ( e x x x 6x 4 5x 4 1 x ) x 1 5x ( ) = e x 10x 7 6x 7 x ( 5x 4 x ) = e x x 7 (x + 5x + 4) = ex (x + 1)(x + 4) x 7. This function is defined and continuous for all x > 0. Furthermore, none of the factors in its numerator is 0 for x > 0, so it is in fact nowhere 0 on this interval. Since their Wronskian is not identically 0, these functions are linearly independent on this interval.
4 ..14. Find a particular solution to the DE y () 6y + 11y 6y = 0 matching the initial conditions y(0) = 0, y (0) = 0, y (0) = that is a linear combination of y 1 = e x, y = e x, and y = e x. Solution: We let y = c 1 e x + c e x + c e x. Then y = c 1 e x + c e x + c e x and y = c 1 e x + 4c e x + 9c e x, so evaluating these functions at x = 0 and matching them to the initial conditions, we obtain the linear system c 1 + c + c = 0 c 1 + c + c = 0 c 1 + 4c + 9c = We solve this linear system by row reduction of an augmented matrix to echelon form: R R R 1, R R R R R R R 1 R, R R R, R 1 R 1 R R 1 R 1 R Then c 1 = c = and c =, so y = ex e x + ex is the solution to the IVP Find a particular solution to the DE y () y + 4y y = 0 matching the initial conditions y(0) = 1, y (0) = 0, y (0) = 0 that is a linear combination of y 1 = e x, y = e x cos x, and y = e x sin x. Solution: We let y = c 1 e x + c e x cos x + c e x sin x. Then y = c 1 e x + c e x (cos x sin x) + c e x (sin x + cos x) y = c 1 e x c e x sin x + c e x cos x Evaluating these functions at x = 0 and matching them to the initial conditions, we obtain the linear system c 1 + c = 1 c 1 + c + c = 0 c 1 + c = 0 4
5 By the third equation, c 1 = c. Substituting this into the second, c c = 0, so c = c. Finally, in the first equation, c + c = 1, so c = 1, c = 1, and c 1 = ( 1) =. Then y = e x e x cos x e x sin x = e x ( cos x sin x) is the solution to the IVP...4. The nonhomogeneous DE y y + y = x has the particular solution y p = x + 1 and the complementary solution y c = c 1 e x cos x + c e x sin x. Find a solution satisfying the initial conditions y(0) = 4, y (0) = 8. Solution: The general solution to this DE is of the form Then y = y p + y c = x c 1 e x cos x + c e x sin x. y = 1 + c 1 e x (cos x sin x) + c e x (sin x + cos x). Evaluating at x = 0 and applying the initial conditions, y(0) = 1 + c 1 = 4 and y (0) = 1 + c 1 + c = 8. Then c 1 = and c = 4, so the solution to the IVP is y = x e x ( cos x + 4 sin x)...0. Verify that y 1 = x and y = x are linearly independent solutions on the entire real line of the equation x y xy + y = 0, but that W(x, x ) vanishes at x = 0. Why do these observations not contradict part (b) of Theorem? Solution: We first check that these are solutions: x y 1 xy 1 + y 1 = x (0) x(1) + (x) = x( + ) = 0 x y xy + y = x () x(x) + (x ) = x ( 4 + ) = 0 We then compute their Wronskian: W(x, x ) = x x 1 x = x(x) 1(x ) = x. This function is not identically 0, so the two functions x and x are linearly independent on the real line, but it is 0 at precisely x = 0. We note that Theorem applies only to normalized homogeneous linear DEs. Normalizing this DE, we obtain y x y + x y = 0, the coefficient functions of which are continuous for x = 0. Thus, any interval on which the theorem applies does not include x = 0, the only point at which W(x) = 0, so W(x, x ) is nonzero on every such interval. This is consistent with the linear independence of the solutions x and x. 5
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