Examples of the Processes of the Differential and Integral CalculusJ. and J.J. Deighton, 1846 - 529 стор. |
З цієї книги
Результати 1-5 із 100
Сторінка 30
... equation in which y is the independent variable . The result is d'x + x - € = 0 . dy ( 4 ) Change the variable in du u + = a dy ( 1+ y2 ) from y to x , when x = log { y + ( 1 + y2 ) 3 } . du a The result is dx + u = · o 2 ( e * + € ̄ ...
... equation in which y is the independent variable . The result is d'x + x - € = 0 . dy ( 4 ) Change the variable in du u + = a dy ( 1+ y2 ) from y to x , when x = log { y + ( 1 + y2 ) 3 } . du a The result is dx + u = · o 2 ( e * + € ̄ ...
Сторінка 43
... given by ( 2 ) in ( 1 ) ; dy then y = 2xy + b . dx To eliminate both a and b , differentiate ( 2 ) again ; then d2 y y dx dy 2 + = 0 . dx ( 2 ) Eliminate a from the equation dy dx - mx y = x2 + a € TM x ; · my = ( n − mx ) x - 1 ...
... given by ( 2 ) in ( 1 ) ; dy then y = 2xy + b . dx To eliminate both a and b , differentiate ( 2 ) again ; then d2 y y dx dy 2 + = 0 . dx ( 2 ) Eliminate a from the equation dy dx - mx y = x2 + a € TM x ; · my = ( n − mx ) x - 1 ...
Сторінка 50
... functions 2 ' dz a2 - - b2 dx2 Jd2 ≈ dy - 1 \ dy ( d ) " } = 0 . ( 23 ) Eliminate the arbitrary functions from ( 1 ) x ƒ ( a ) + y $ ( a ) + ≈ † ( a ) = 1 , where a is a function of x , y , and ≈ given by the equation ( 2 ) x ƒ ' ( a ) ...
... functions 2 ' dz a2 - - b2 dx2 Jd2 ≈ dy - 1 \ dy ( d ) " } = 0 . ( 23 ) Eliminate the arbitrary functions from ( 1 ) x ƒ ( a ) + y $ ( a ) + ≈ † ( a ) = 1 , where a is a function of x , y , and ≈ given by the equation ( 2 ) x ƒ ' ( a ) ...
Сторінка 63
... functions , is to be found in the Theorems of Lagrange and Laplace , to which we now proceed . SECT . 3. Theorems of Lagrange and Laplace . If y be given in an equation of the form y = x + xp ( y ) , and if u = ƒ ( y ) , ƒ and being any ...
... functions , is to be found in the Theorems of Lagrange and Laplace , to which we now proceed . SECT . 3. Theorems of Lagrange and Laplace . If y be given in an equation of the form y = x + xp ( y ) , and if u = ƒ ( y ) , ƒ and being any ...
Сторінка 68
... from the given equation we have u + k = 0 , or k = - u , and therefore dx d2 x u2 If we put h = dx du - = du u + du2 1.2 d3 x du3 1 U3 2 • 3 + & c . • d2 x dv v , then = v du dx > d a du3 = v d dv d 2 v = - บ dx dx x d u ? v , and so on ...
... from the given equation we have u + k = 0 , or k = - u , and therefore dx d2 x u2 If we put h = dx du - = du u + du2 1.2 d3 x du3 1 U3 2 • 3 + & c . • d2 x dv v , then = v du dx > d a du3 = v d dv d 2 v = - บ dx dx x d u ? v , and so on ...
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a² b2 a²x² angle arbitrary constant asymptote becomes C₁ c²x² Cambridge circle co-ordinates condition Crelle's Journal curvature curve cycloid determine differential coefficients differential equation dx dx dx dy dx dy dx dx² dy dx dy dy dy dy dz dz dz eliminate ellipse equal Euler factor formula fraction function Geometry gives Hence hypocycloid infinite intersection John Bernoulli Let the equation lines of curvature locus logarithmic logarithmic spiral Multiply negative origin parabola perpendicular plane of reference radius SECT singular solution spiral Substituting subtangent surface tangent plane theorem triangle vanish whence x²)³