By Lothar Collatz

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**Additional resources for Differential Equations: An Introduction with Applications **

**Example text**

In H. V. Mangold-K. Knopp: Einfuhrung in die hoheren Mathematik, vol. ; Stuttgart 1958, p. ; and E. , Leipzig 1956, p. 115f. 47 Fig. 28. 62) yields ayZy'-x=0. Elimination of y' from F= 0 and Fy, = 0 gives, in fact, y = x2, that is, the parabola is the singular solution, and here it is the envelope of the tangents, see Fig. 28. Example 3. 62) does not always give the envelope. Let us examine the family of semi-cubical parabolas, Fig. 29, y=(x-c)3"3 We have y' =3(x-c)-%3=3 y-'12 or F(x,y,y')= y,2- 4 =0.

Xo We can estimate the integrals, because the absolute value of a definite integral of a function is at most equal to the definite integral over the same interval of the absolute value of the function. 33), Yn+I(X)Yn(X)I I Zn+1(X) - zn(x) I s dE, k the same bound. Instead of proving the convergence of the two integrals separately we can prove, more simply, the convergence of their sum, which we denote by Sn(x). 36) Sn(X) x <2k xo xa Sn-1( d and show that the sum of all the Sn(a) converges as n -+ co.

Comparing the two solutions for z shows that we must have wl (y) = 2 y2 + const and w2(x) = const = Ci. Hence the general solution of our original equation reads y2+x2y4=C. In general the determination of a particular solution of the partial differential equation for a may be very difficult. Equations which are solvable by means of an integrating factor can often be solved more easily in a different way. Our example above could be regarded as a Bernoulli differential equation for the function x(y) inverse to the required function y(x) (this artifice is often worthwhile): dx + 2 y x= - y3 x 1 _,dy 22 Here the roles of x and y have been interchanged.