{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

sol_set1

# sol_set1 - Solutions to Problem Set 1 ARE 261 Question 1...

This preview shows pages 1–3. Sign up to view the full content.

Solutions to Problem Set 1 ARE 261 Question 1 Begin with the assumption that the solution to the second order differential equation is of the form y = k exp xt , where k and x are constants. Then ˙ y = kx exp xt and ¨ y = kx 2 exp xt . Substitute these expressions for y, ˙ y and ¨ y into the differential equation and divide through by k exp xt to get x 2 + ax + b = 0 (1) This gives a quadratic in x which we assume has two distinct root, x 1 and x 2 given by x 1 = - a 2 + a 2 - 4 b 2 (2) and x 2 = - a 2 - a 2 - 4 b 2 (3) The solution to the second order differential equation is then given by y = k 1 exp x 1 t + k 2 exp x 2 t (4) where k 1 and k 2 are constants to be determined by boundary or initial conditions. In the next subpart we are given two such initial conditions, y (0) = 0 and ˙ y (0) = 1. Substitute these conditions 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
into equation (4) and its first derivative with respect to time to get the following two equations k 1 + k 2 = 0 (5) k 1 x 1 + k 2 x 2 = 0 (6) These can be solved for k 1 and k 2 to get the complete solution for the differential equation. The
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}