MATH 240 Midterm: MATH 240 KSU Test 1SoluTest ions14

• School: Kansas State University
• Department: Mathematics
• Course: MATH 240
• Professor: All

Document Summary

Problem 2. (16 pts) solve the initial value problem dy dx. (1/2) ln |3 2y| = x + c; ln |3 2y| = 2x + c; Problem 3. (16 pts) solve the initial value problem dy dx. Multiply by the integrating factor = ex: ex dy dx d dx. + exy = 4xex; (exy) = 4xex; exy = z 4xexdx = 4xex 4z exdx. = (4x 4)ex + c, where we used integration by parts. Since y(1) = 0, c = 0, and. Problem 4. (16 pts) find all solutions of the equation dy dx y 4x x y. This is a homogeneous equation, which can be re-written as dy dx (y/x) 4. Change of variables v = y/x, y = xv, dy = xdv + vdx leads to x dv dx. 1 v which is a separable equation. The general solution: x dv dx v 4. Z 1 v dv +z 3/4 v + 2 x v 2.