A Treatise on Differential Equations, Volumen1Macmillan and Company, 1877 - 496 páginas |
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Página 25
... Substituting these values of dp ( x , v ) dv dv dy dV d V and dx dy in the equation ( 1 ) we have as the result dp ( x , v ) dv dx = = 0 ( 2 ) . dy But , being by hypothesis an explicit function of both dv variables x and y , is not ...
... Substituting these values of dp ( x , v ) dv dv dy dV d V and dx dy in the equation ( 1 ) we have as the result dp ( x , v ) dv dx = = 0 ( 2 ) . dy But , being by hypothesis an explicit function of both dv variables x and y , is not ...
Página 33
... Substituting these expressions for y and dy in the given d * + * dzo equation , we have dx + dz = 0 . Therefore integrating and replacing z by its value loga + c . = C. Ex . 2. ( y − x ) ( 1 + x2 ) 3 dy — n ( 1 + y2 dx = 0 . - Assume x ...
... Substituting these expressions for y and dy in the given d * + * dzo equation , we have dx + dz = 0 . Therefore integrating and replacing z by its value loga + c . = C. Ex . 2. ( y − x ) ( 1 + x2 ) 3 dy — n ( 1 + y2 dx = 0 . - Assume x ...
Página 39
... Substituting then the above expression for y in ( 21 ) , and observing that , since C is now variable , we have = dy dx dx d dC ( Ce - Spax ) : = -fPda — CPc - JPdz dx there results d C dx € ̃Spax = Q. dC Hence dx = € SPaz Q. Therefore ...
... Substituting then the above expression for y in ( 21 ) , and observing that , since C is now variable , we have = dy dx dx d dC ( Ce - Spax ) : = -fPda — CPc - JPdz dx there results d C dx € ̃Spax = Q. dC Hence dx = € SPaz Q. Therefore ...
Página 51
... Substituting this value in the second member of ( 1 ) , and equating the result to an arbitrary constant , we have - 3 the solution required . 2x3y — 213x + y2 = C , 3 Ex . 2. Given dx √ ( x2 + y2 ) + { 1 - ন х dy = 0 . √ ( x2 + y2 ) ...
... Substituting this value in the second member of ( 1 ) , and equating the result to an arbitrary constant , we have - 3 the solution required . 2x3y — 213x + y2 = C , 3 Ex . 2. Given dx √ ( x2 + y2 ) + { 1 - ন х dy = 0 . √ ( x2 + y2 ) ...
Página 52
George Boole. Substituting log C for c , and then freeing the equation from logarithmic signs and from radicals , we have y2 = C2 - 20x . 3. We may in many cases either dispense with the appli- cation of the criterion ( 1 ) , or greatly ...
George Boole. Substituting log C for c , and then freeing the equation from logarithmic signs and from radicals , we have y2 = C2 - 20x . 3. We may in many cases either dispense with the appli- cation of the criterion ( 1 ) , or greatly ...
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Términos y frases comunes
2ndly algebraic arbitrary constant arbitrary function assume C₁ C₂ Cambridge Chap Chapter College complete primitive condition Crown 8vo curve d²u deduce derived determined differential coefficients dp dq dp dy dv dv dx dx dx dy dx dz dx² dy dy dx dy dz dz dx dz dz Edition eliminating equa exact differential expressed Extra fcap fcap finite given equation Hence homogeneous functions Illustrations independent variable infinite integrating factor involving Mdx+Ndy method Mx+Ny obtained ordinary differential equations partial differential equation particular integral primitive equation Professor Prop reduced relation represent respect result satisfied second member second order Shew shewn singular solution substituting suppose theorem tion transformation V₁ V₂ whence X₁ Y₁ аф
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