Differential Equations
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In our eReader you can find the full English version of the book. Read Differential Equations Online - link to read the book on full screen. Our eReader also allows you to upload and read Pdf, Txt, ePub and fb2 books. In the Mini eReder on the page below you can quickly view all pages of the book - Read Book Differential Equations
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The equation also gives the locus of points of inflexion on integral curves, because the equation -? + « l$ = o is satisfied d 2 y dx r dy when ^ = 0.
The first example of a singular solution was given by Leibnitz in 1694. Brook Taylor showed in 1715 that a singular solution can be deduced directly from the differential equation in the case of the equation 9(3. V> P) = 0- + % 2 )p z - ^xyp + y 2 - 1 = This equation is of Clairaut's type, and so its general solution is /(*, V, c) = (1 + x 2 )c 2... - 2xyc +y 2 -l=0 The ^-discriminant is (1 + x 2 )(y 2 - 1) - x 2 y 2 = or y 2 — x 2 = 1 Also =| = 2xp 2 - 2yp ^ = 2y - 2xp and so -^ + p ^ = ox * dy Hence y 2 - x 2 = 1 is a solution of the differential equation, and it cannot be derived from f(x, y, c) = by giving a special value to c ; consequently it is a singular solution.
EXAMPLES.
1. Find the singular solutions of the following differential equations : a 2 [x + y + p(y - a;)] 2 = (xp - y) 2 (x* + y*) p 2 (2x* + 1) + p(x* + 2xy + y 2 + 2) + 2if + 1 = p 3 - 4xyp + 8y 2 = Show that in the last example y = is included in the general integral and is also a true ' singular solution ' in the sense that it is included in the envelope (Cauchy).
What to read after Differential Equations?
You can find similar books in the "Read Also" column, or choose other free books by Harry Bateman to read onlineMoreLess
To quickly assess the difficulty of the text, read a short excerpt:
The equation also gives the locus of points of inflexion on integral curves, because the equation -? + « l$ = o is satisfied d 2 y dx r dy when ^ = 0.
The first example of a singular solution was given by Leibnitz in 1694. Brook Taylor showed in 1715 that a singular solution can be deduced directly from the differential equation in the case of the equation 9(3. V> P) = 0- + % 2 )p z - ^xyp + y 2 - 1 = This equation is of Clairaut's type, and so its general solution is /(*, V, c) = (1 + x 2 )c 2... - 2xyc +y 2 -l=0 The ^-discriminant is (1 + x 2 )(y 2 - 1) - x 2 y 2 = or y 2 — x 2 = 1 Also =| = 2xp 2 - 2yp ^ = 2y - 2xp and so -^ + p ^ = ox * dy Hence y 2 - x 2 = 1 is a solution of the differential equation, and it cannot be derived from f(x, y, c) = by giving a special value to c ; consequently it is a singular solution.
EXAMPLES.
1. Find the singular solutions of the following differential equations : a 2 [x + y + p(y - a;)] 2 = (xp - y) 2 (x* + y*) p 2 (2x* + 1) + p(x* + 2xy + y 2 + 2) + 2if + 1 = p 3 - 4xyp + 8y 2 = Show that in the last example y = is included in the general integral and is also a true ' singular solution ' in the sense that it is included in the envelope (Cauchy).
What to read after Differential Equations?
You can find similar books in the "Read Also" column, or choose other free books by Harry Bateman to read onlineMoreLess
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