DIFFERENTIAL EQUATIONS Note: Differential equations are equations containing a derivative. They can be solved by integration to obtain a general solution with +C. To obtain a specific solution requires additional information. Ex1 dy / dx + 7 = 2x dffntl eqtn dy / dx = 2x - 7 y =  (2x – 7) dx y = x 2 – 7x + C General Soln Integrate with respect to x

Ex2 Given that dy / dx = x x 2 and y = f(x) passes through (-2,2) then find the exact solution to this differential equation. ********* dy / dx = x x 2 = x x 2 x 2 = 1 + 4x -2 So y =  (1 + 4x -2 ) dx y = x + 4x -1 + C -1 y = x – 4 + C x At (-2,2) this becomes 2 = -2 – (-2) + C ie C = 2 Solution is y = x – x

Ex3 (from Physics) Newton’s 1 st law of motion states that v = u + at v – final velocity, u - initial velocity, a – acceleration, t- time In Physics the symbol for distance is s Velocity is basically speed Given thatspeed = difference in distance difference in time We now have v = ds / dt ctd

So v = u + at becomes ds / dt = u + at Integrate with respect to t givings =  (u + at) dt ors = ut + 1 / 2 at 2 + C Assuming that s, u, t, a are all zero simultaneously gives C = 0 and we now get s = ut + 1 / 2 at 2 which is Newton’s 2 nd law of motion.