Homework 3 solutions

Name: Homework 3 Solutions Math 285 Problem 1 [2pts] Are the following functions linearly dependent or independent? y 1 ( t ) = 1 + 2 t - t 2 y 2 ( t ) = 2 + t - t 2 2 y 3 ( t ) = 1 - 2 t + t 2 If dependent give a non-trivial linear combination that equals zero. Answer: Assuming that A (1 + 2 t - t 2 ) + B (2 + t - t 2 / 2) + C (1 - 2 t + t 2 ) = 0 we find the three equations A + 2 B + C = 0 2 A + B - 2 C = 0 - A - B/ 2 + C = 0 The second and third equations are linearly dependent - dividing the second by - 2 gives the third. Since there are two (consistent) equations in three unknowns there are a whole family of solutions. One can solve for A, B in terms of C to find that A = 5 3 C B = - 4 3 C so (for instance) 5 y 1 - 4 y 2 + 3 y 3 = 0 Problem 2 [2pts] Solve the following equation ( 6 x 2 + 18 xy + 6 y 2 ) dy dx + ( 3 x 2 + 12 xy + 9 y 2 ) = 0 y (0) = 0 You can leave your answer in implicit form: you do not need to solve for y explicitly as a function of x . Answer: This is exact: We have ψ y = 6 x 2 + 18 xy + 6 y 2 ψ = 6 x 2 y + 9 xy 2 + 2 y 3 + C ( x ) taking the partial with respect to x gives ψ x = 12 xy + 9 y 2 + C 0 ( x ) = 3 x 2 + 12 xy + 9 y 2 so C 0 ( x ) = 3 x 2 , C ( x ) = x 3 and the solution is ψ ( x, y ) = x 3 + 6 x 2 y + 9 xy 2 + 2 y 3 = 0 Problem 3 [2pts] Suppose that y 1 ( t ) = 1; y 2 ( t ) = e - t and y 3 ( t ) = e t are three solutions to d 3 y dt 3 - dy dt = 0 Find the solution to the initial value problem d 3 y dt 3 - dy dt = 0 y (0) = 2; y 0 (0) = 3 y 00 (0) = 4 Answer: 1

Name: Homework 3 Solutions Math 285 Write the general solution as y = A + Be - t + Ce t and apply the initial conditions. We get the system y (0) = A + B + C = 2 y 0 (0) = - B + C = 3 y 00 (0) = B + C = 4 and we find that C = 7 / 2 , B = 1 / 2 , A = - 2 so y = - 2 + 1 / 2 e - t + 7 / 2 e t Problem 4 [2pts] Compute the Wronskian of the functions y 1 ( t ) = 1 y 2 ( t ) = e - t y 3 ( t ) = e t Answer: The Wronskian is given by W ( y 1 , y 2 , y 3 ) = - 2 Problem 5 [2pts] Given what you know from problem 3 what does Abel's theorem tell you about the Wronskian of y 1 ( t ) = 1 y 2 ( t ) = e - t y 3 ( t ) = e t You should not need to calculate any determinants. Answer: Since y 1 , y 2 , y 3 are all solutions to d 3 y dt 3 - dy dt = 0 it follows that the Wronskian W satisfies the equation dW dt = 0 so we know that the Wronskian is constant. The constant itself can only be found by taking the determinant. 2