A Treatise on Differential Equations, Τόμος 1Macmillan, 1872 - 85 σελίδες |
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Σελίδα 25
... Substituting these values of we have as the result and in the equation ( 1 ) dx dy dp ( x , v ) dv dx dy = 0 ..... . ( 2 ) . But , v being by hypothesis an explicit function of both dv variables x and y , dy is not identically 0. Hence ...
... Substituting these values of we have as the result and in the equation ( 1 ) dx dy dp ( x , v ) dv dx dy = 0 ..... . ( 2 ) . But , v being by hypothesis an explicit function of both dv variables x and y , dy is not identically 0. Hence ...
Σελίδα 33
... Substituting these expressions for y and dy in the given equation , we have dx + dz = 0 . x Therefore integrating and replacing z by its value y2 log z + = c . X C. 3 Ex . 2. ( y − x ) ( 1 + x2 ) 3 dy − n ( 1+ y2 ) 3 dx = 0 . Assume ...
... Substituting these expressions for y and dy in the given equation , we have dx + dz = 0 . x Therefore integrating and replacing z by its value y2 log z + = c . X C. 3 Ex . 2. ( y − x ) ( 1 + x2 ) 3 dy − n ( 1+ y2 ) 3 dx = 0 . Assume ...
Σελίδα 39
... Substituting then the above expression for y in ( 21 ) , and observing that , since C is now variable , we have = dx there results dy d dx ( Cε Spax ) : = dC -SPdx - CPE - SPdx dC fPdx 6 = dx dx 更 Q. Pg dċ Hence Spax Q. dx Therefore C ...
... Substituting then the above expression for y in ( 21 ) , and observing that , since C is now variable , we have = dx there results dy d dx ( Cε Spax ) : = dC -SPdx - CPE - SPdx dC fPdx 6 = dx dx 更 Q. Pg dċ Hence Spax Q. dx Therefore C ...
Σελίδα 51
... Substituting this value in the second member of ( 1 ) , and equating the result to an arbitrary constant , we have 3 3 - 2x3y — 2y3x + 32- y3 - C the solution required . Ex . 2. Given 3 = = 0 . + 1 - dx √ ( x2 + y2 ) 1 Here M ...
... Substituting this value in the second member of ( 1 ) , and equating the result to an arbitrary constant , we have 3 3 - 2x3y — 2y3x + 32- y3 - C the solution required . Ex . 2. Given 3 = = 0 . + 1 - dx √ ( x2 + y2 ) 1 Here M ...
Σελίδα 52
George Boole. Substituting log C for c , and then freeing the equation from logarithmic signs and from radicals , we have = C2 - 2Cx . 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 = C2 - 2Cx . 3. We may in many cases either dispense with the appli- cation of the criterion ( 1 ) , or greatly ...
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2ndly algebraic arbitrary constants arbitrary function assume C₁ C₂ Chap Chapter complete primitive condition Crown 8vo curve deduce derived determined differential coefficients dp dp dp dq dp dy dt dt dv du dv dv dv dx dx dx dy dy dx dz dx² dy dx dy dz dz dx dz dy dz dz Edition eliminating equa exact differential expressed fcap finite given equation Hence homogeneous functions independent variable integrating factor involving method Mx+Ny obtained ordinary differential equations P₁ partial differential equation particular integral pdx+qdy primitive equation reduced relation represent respect result satisfied second member second order Shew shewn singular solution substituting suppose theorem tion transformation whence X₁ y₁