Solutions of the Examples in a Treatise On Differential Equations

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As the equation is p. 429 z=px+qy+f(p,q), the subsidiary equations are dp _dq _ !T - ~0~ _ " - " Two integrals are p = a, q = b, each consistent with the original equation, and also consistent with one another (§ 208). Elimi- nating p and q, we have z — ax + by +f(a, b).
Ex. 4. The subsidiary equations are dp _dq _ P ~ a ~"\ so we can take p = aq as an integral. Using this, we have q(ax+y) = q n f(a, 1), i . ( ax + y so that q = i ~f-, — t~, I/O. 1) _ z — b n — 1 ( ax + y )' 1 ' 1 Then f(a, 1)
...n \f {a, 1)\ For the second equation, the subsidiary equations are dp _dq _ p~ 2 ~~f ' so we can take = - as an integral. Hence p q a p=a--{l + (I-xy)K q=-a + l[l-(l-xy)i}, x y and so z = a(x-y) -olog^-2o(l-ay)*-olog ~ i~ +*>• y (1 — xyy + 1 142 chapter ix. §§ 219, 228 p. 437 p. 449 §219.
Ex. Proceeding as in § 217, we form the equations dF r djh d_Fr d_Fr dx l + Pl dz dp 1 dx x ' ' dp n d%!
d Jj +p dF l+ d Zl d Pl + + dJrZPn = () .
dx 1 ^ dz dpx 3«i '" dp n dx x eliminating J-l , we have + P.


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