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24) e(x)=-( 1XI )n/2-1Kn/2_1(KIxI) x > 0 , with the modified Hankel function Kv(z). 24) more directly, observing that e(x)=y(Ixl) solves (A+X)e=0, hence y(r) solvn-1, we obtain es the ODE y"+n-1y'-x2y=0. 24). A partial integration shows that fe(A+X)tpdx= g(x) for all TEED , fixing the remaining multiplicative constant. =k2=-x2=it. ) , and we then must obtain the inverse Laplace transform. It is more practical, however, to first obtain Ft Note has inverse Laplace transform 04(t)- 8(t)=(it+a)-1 ta0, =0, t<0 , 2ne-at, f e-ate-ittdt.

Also, we get = fR f , at right, by the pro- perties of the (analytic) integrand. 16) follows from Fourier inversion for functions in D. 17) involving the Laplace transform f- of f . Problems. 1) Obtain the Laplace transforms of the following functions (Each is extended zero for x<0). , b) eax; c) cos bx ; d) eaxsin bx ; e) sin In each case, discuss the . , the linear functional on Z. 2) Obtain the inverse Laplace transform of a) ; b) log(1+Z ). ) 3) For uE D'(ien) with supp uC {xix0}= 0. ,xn). 19), may be defined for general distributions u,vE D'(&n) under a support restriction -for example (i) if supp U= 1n, supp v general, or (ii) if supp vC {xlzo}, supp v C {jxjscx,} One then defines (w,(p>=ffdxdyu(x)v(y)T(x+y), .

5) Also we know that F(x(s,t),w(s,t),p(s,t))= 0. 15) Thus indeed we solved the Cauchy problem. 2) . 18) x = Flp(x,u(x),ulx(x)) , n'n = d/dt of n first order ODE's in n unknowns x(t). 18) and the initial cdn's y(s,O)=x(s). Let K(s,t)= u(y(s,t)), q(s,t)= ulx(y(s,t)). 5) , so that we have uniqueness of the solution of this Cauchy problem. 18) get y =Flp(y,x,q), assuming u E C2. 1). 19) Flx(x,u,ulx) + Flu(x,u,ulx) + Flp(x,u,ulx)ulxx = 0 . 18) let x = y(s,t). 6) for y,K,q follows We have proven the result below.

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Group analysis of ODEs and the invariance principle in mathematical physics (Russ.Math.Surv. 47, n.4, 89-156) by Ibragimov N.Kh.

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