Read e-book online A First Course in Partial Differential Equations: with PDF

By H. F. Weinberger

Textual content offers the final homes of partial differential equations reminiscent of features, domain names of independence, and greatest ideas. suggestions.

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Extra resources for A First Course in Partial Differential Equations: with Complex Variables and Transform Methods (Dover Books on Mathematics)

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A function F of two variables is homogeneous (of degree zero) if and only if F(tx, ty) = F(x, y) for all numbers x, y, t. The following are examples of homogeneous functions: x + y x 7 xz + xy2 2/3 + x 2y 9 xy + x 2 sin (;y/x) y2 ( * ' If F is a homogeneous function, then F(x, y) depends only on the ratio y/x, and hence F can be considered as a function of a single variable u = y/x. T o see this, put t = 1/x in the definition to obtain i-(*, s,) = f ( i | s,) _ P ( l , | ) _ i-(l, „). The homogeneous functions (3), for example, can be written x + y x 3 I 2 x + xy y 3 + x 2y 1 + (y/x)2 (y/x)3 + (y/x) ’ xy + x2 sin (y/x) y2 (y/x) + sin (y/x) (y/x)2 25 26 SPECIAL METHODS FOR FIRST ORDER EQUATIONS [CHAP.

7. Show that if every curve in the family of rectangular hyperbolas xy = c is rotated 45°, the resulting family is orthogonal to the original. 8. Show that x2/c + y2/(c — a2) = 1 is a family of ellipses and hyperbolas with foci at (± a , 0). Show that the differential equation of the family is (x + yyf)(x — y/yr) = a2, and hence that the family is self-orthogonal. A nswers 1. y = cx 2. x2 + 4y2 = c2 4. y2 = In |x|+ \x2 + c 3. x2 + (y — c) 2 = c2 5. y2 = —2x + c 2-6 Review of power series. An indicated infinite sum of the form 00 £ anxn = o 0 + axx + azx 2 H-------, ( 1) 71= 0 or of the form 00 ^ On(x — c)n = a0 + ax(x — c) + a2(x — c) 2 H------71= 0 (2) is called a power series.

If we differentiate (8), we obtain a second order equation involving only x, yf, and y " . Such an equation can be regarded as a first order equation in y'. If this derived equation can be solved for yf, then the solutions of (8) are obtained by substituting these values of y' in (8). E x a m p le 4. Consider again (cf. Example 1) the equation y = x y ' ~ W ) 2- (9) Differentiation gives the equation y' = xy" + y' — \y'y", y"[x - W ) = 0. Therefore we must have y' = c (if y " = 0), or y' = 2x. in (9) gives all solutions, y = cx — \c2, Substitution or y = (2x)x — l(2 x )2 = X2.

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A First Course in Partial Differential Equations: with Complex Variables and Transform Methods (Dover Books on Mathematics) by H. F. Weinberger

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