Comparing this equation with Dudu + D. d ̧u+ D1y.d1u, The first two give tx = § and y = tn, and these being used in the third to eliminate x and y, we have t, (§, ʼn are quasi-constants); n2 t3 2 + F (E, n) = 2 + F (tx, Y). Ex. 8. To integrate du - 2.du+2 du In this example, ᎠᏆ t У y3 = tx Dix=-2, Dy=2, and Du=? t' ; tx bu = 0. ht+kx Comparing this with the usual formula we have Dx = a, and (ht+kx) Du=-bu. Hence x=at+; and D1 (ht + kx) = h+ka. It still remains to integrate (ht + kx) Du+bu = 0. Multiply by (ht+7x)-1, m being at present an unknown constant; m-1 :. (ht+kx)TM Du+b (ht+kx)-1 u = 0. We see at a glance that this will be an exact differential, if m be such that b = mD; (ht + kx), that is, if du+ad ̧u+bd ̧u=c, and du+a'd„u+b'd„u= c' & multaneously. Proceeding with the former of these as before, we have which being integrated give x= at + §, y=bt+n, and u = ct + F′ (§, n) = ct+F(x — at, y—bt). Now substitute the value of u, given in the last equation, in the second of the simultaneous equations, using F for F(x-at, y-bt) for brevity, :: (a− a') d¿F+ (b − b') d, F = c-c'. This is an ordinary differential equation of the first order with constant coefficients, and being integrated as in Examples preceding gives ... u= 1 c'n F= ƒ {(b − b') § + (a' — a) n} + a-a-b-b'' bc' b α + ) t+ƒ{(b−b')x+(a'—a) y+(ab′ — ba') t}. 10. In equations of the first order, if the coefficients be of a symmetrical form, it will sometimes be found to simplify the process of integration if we use the formula of Art. 4, instead of that of Art. 7. Ex. 12. To integrate (x+t) d¿u + (x − t) d ̧u = D1u = d1u. D‚t+d ̧u. D‚x, Comparing this with we have the following equations, Du=au, D,t=x+t, D,x=x-t. From the last two we obtain = au. and tD,t + xD,x= t2 + x2, xD‚t-tD‚x = ť2 + x2 ; -1 .. log √t2 + x2 = 7+ 4,5, and tan¬12=7+ $25. Now the right-hand members of these equations being arbitrary functions of §, the left-hand members must also be arbitrary functions of each other; -1 2 ... u = (t+x) F(tan Ex. 13. To integrate Comparing this with the formula of Art. 4, we have Dt-bu-cx, D,x= ct-au, and D,uax-bt; :: aD,t+bDx + cD‚u = 0, and tD‚t + xD,x+uD‚u = 0. :. at+ bx + cu = §, and t2+x2 + u2 = F (§) ; ... t2 + x2 + u2. = F (at + bx + cu), which is the integral required. Ex. 14. To integrate (x+y+u) du + (y + t + u) d ̧u + (t + x+u) d ̧u=t+x+y. and D, (t+x+y+u) = 3 (t+x+y+u). Integrating these equations, and writing Q for t+x+y+u for brevity, we have t− x = €1F1 (§, n), x−y=e ̃ ̄F,(§, n), y—u=='F。 (§, n), 1 2 Eliminating from these integrals, we find that (t - c) Vẽ ( −y) (y - u) Q are (each of them) arbitrary functions of , n; consequently each of them is an arbitrary function of the other two, .. 0 = F{(t − x) IQ, (x−y) IT, (y—u) JQ}. Ex. 15. To integrate t'du+x3d ̧u = atx. Comparing this with the formula of Art. 4, we have From the last two of these we find xD,t=tDx, .. D‚u= ax2. = |