MATH 3350 Midterm: Math 3350 TTU Exam Exam 1Solutions

• School: Texas Tech University
• Department: Mathematics
• Course: MATH 3350
• Professor: All

Document Summary

In each part, nd the general solution of the di erential equation. Z ey dy = z x2 dx ey = x3/3 + c ln(ey) = ln(x3/3 + c) y = ln(x3/3 + c) dy dx y2 + xy + x2 x2 u = y x. Since u = y/x, we have y = xu. Using the product rule, we get y = u + xu . Plugging into (1) we get u + x du dx. We solve this by the following sequence of steps. u + x x du dx du dx. = u2 + 1 x du = (u2 + 1) dx du u2 + 1 dx x du u2 + 1. Z x tan 1(u) = ln|x| + c u = tan(ln|x| + c) y. = tan(ln|x| + c) x y = tan(ln|x| + c) .