By Rudolph E. Langer

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**Extra info for A First Course in Ordinary Differential Equations**

**Sample text**

A ball weighing 1 lb. is dropped from a great height. The air resistance amounts to p/32. Find the formula for s in terms of /, if f = 0, when j = 03. A bullet weighing H o lb. /scc. T h e air resistance amounts to f^/1,000,000. Find the formula for v in terms of j . 4. sec, The air resistance amounts to kv^. Find the formula for s (the height), in terms of c, while the bullet is rising. 5. In the case of the bullet of problem 4, find the formula for / in terms of v, 6. A certain particle weighs 8 lb.

It is called the linear differential equation of the first crder because it is linear, that is, of the first degree, i n y and y'. For an equation of this type an integrating factor is always easily found. Let the equation be multiplied by ${x) dx. Its resulting form is e{x) dy + {p{x)e{x)y - q{x)e{x)] dx = 0. 4, the last equation is exact if 6' = pix)6y that is, if 0'/6 = p{x). This integrates to give log $ = jp{x) dx, which is ^ = ^Jpix) di T h u s tf/p^** dx is an integrating factor, by virtue of which the 18 Some Types of Solvable Equations differential equation becomes T h e general integral is thus ^ J p ( x ) dx ^ J^(x)tf/P<^> ^dx + C.

W i t h A = ^, this is a n equation with homogeneous coefficients, whose general integral is [u -\- 2s\'^[u — 7>s\^ = c. I n terms of the original variables, the general integral is thus \Sx — y + 3j^j2jv — I p = c. Example 18. T o integrate the differential equation {2x - 6> + 3) rfx - {x - ly ~ \ \ dy 0. T h e change of variable, j = x + A, u = x — 3^ — 1, transforms the equation into {5w + 15 j ds -\- udu = 0. Here the variables are separable. W e may therefore take k ~ 0. T h e general solution, i n terms of the original variables, is 2A: — ^ — log fx — Sjy + 2 j = c.

### A First Course in Ordinary Differential Equations by Rudolph E. Langer

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