MATH 246 Final: Final No Solutions Spring 2006

• School: University of Maryland
• Department: Mathematics
• Course: MATH 246
• Professor: All

Document Summary

[3 15 = 45 points] solve the following di erential equations. (hint: identify the types of the equations rst. ) (1. 1) y = 1. This is a bernoulli equation with n = 3. We rst check that y = 0 is a solution. If y is not the zero function, we can let u = u(x) = y1 n = y 2. u = 2y 3y = 2y 3( 2 y3) = y 2 + x = u + x. This means u = u(x) is a solution to the rst order linear equation u +u : the integrating facto is v = er 1dx = ex. Multiplying by the integrating factor, we have (uv) = xex. So y 2 = u(x) = v 1z xexdx = e x(xex ex + c) = x 1 + ce x. So y = 1/ x 1 + ce x, where c is an arbitrary constant, or y = 0.