A Treatise on Differential EquationsMacmillan, 1865 - 496 σελίδες |
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Σελίδα 3
... Hence the solution becomes du dk = C. In applying this theory to the reduction of the general solution ( 2 ) in the case in which m1 = m2 , it must be observed that the numerator of the first member is the same function of m ,, x , y ...
... Hence the solution becomes du dk = C. In applying this theory to the reduction of the general solution ( 2 ) in the case in which m1 = m2 , it must be observed that the numerator of the first member is the same function of m ,, x , y ...
Σελίδα 5
... ƒn ( X , Y ̧ ) ( — x ̧ ) * → 1.2 ... ( nr ) + & c . , ( 32 ) ........ the summation extending from n = r to no . Forming and hence the differential coefficients of A , with respect ART . 5. ] 5 ADDITIONS TO CHAPTER II .
... ƒn ( X , Y ̧ ) ( — x ̧ ) * → 1.2 ... ( nr ) + & c . , ( 32 ) ........ the summation extending from n = r to no . Forming and hence the differential coefficients of A , with respect ART . 5. ] 5 ADDITIONS TO CHAPTER II .
Σελίδα 6
George Boole. and hence the differential coefficients of A , with respect to x yo , and reducing by ( 28 ) , we shall find dA . dA , dx 。 + f1 ( x 。, Yo ) = 0 , dy 。 whence in particular dA . dA . + f1 ( xo , Yo ) dyo = = 0 ...
George Boole. and hence the differential coefficients of A , with respect to x yo , and reducing by ( 28 ) , we shall find dA . dA , dx 。 + f1 ( x 。, Yo ) = 0 , dy 。 whence in particular dA . dA . + f1 ( xo , Yo ) dyo = = 0 ...
Σελίδα 11
... Hence x = 0 , provided n is between 0 and 1 , or y = 0 . Consider these separately : - First . Let n be between 0 and 1 , and x = 0. This is by the test a singular solution . Substituting it in the com- plete primitive we get y = c ...
... Hence x = 0 , provided n is between 0 and 1 , or y = 0 . Consider these separately : - First . Let n be between 0 and 1 , and x = 0. This is by the test a singular solution . Substituting it in the com- plete primitive we get y = c ...
Σελίδα 13
... Hence - y m2 dc х Both and dy dx vanish then if dc de x2 + y2 — m2 = 0 . This therefore is the singular solution and it satisfies both the tests , as both x and y are contained in its expression . Of the partial tests do аф do = 0 ...
... Hence - y m2 dc х Both and dy dx vanish then if dc de x2 + y2 — m2 = 0 . This therefore is the singular solution and it satisfies both the tests , as both x and y are contained in its expression . Of the partial tests do аф do = 0 ...
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arbitrary constants Cambridge Chap Chapter College complete primitive condition Crelle's Journal Crown 8vo deduce derived determine dF dF dF dF dF dF dp dF dx dFdF differential coefficients dp dF dp dp dx dp dq dp dy dp₁ dq dp dv dv dx dp dp dx dx dx dy dy dx dz dx₁ dx² dy dp dy dx dy dz dz dy dz dz Edition eliminate equa Eton College expression Extra fcap factor function given equation Hence J. P. MAHAFFY Jacobi Last Multiplier linear partial differential m₁ memoir ordinary differential equations Owens College P₁ partial differential equations particular integral Professor Boole reduced represent result revised School shewn singular solution system of ordinary theorem theory tion transformation u₁ u₂ values vanish whence X₁ Y₁ аф
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