Solve the D.E. (1-x^2)y\'\'-xy\'=0 Solution § dV § dxdydz dz z 0 x=u/y -> x=u/x/

Question

Solve the D.E. (1-x^2)y''-xy'=0

Explanation / Answer

§ dV § dxdydz dz z 0 x=u/x/v=uv/x -> x= (uv)^(1/2) v=x/y -> yv=x -> y=x/v -> y=(uv)^(1/2)/v=y=(u)^(1/2)v^(-1/2) determinante per sost. integr.doppi |dx/du dx/dv | d(x,y)/d(u,v) = | | |dy/du dy/dv | dx/du derivata di x= (uv)^(1/2) per du quindi 1/2*(u)^(-1/2)*(v)^(1/2) dx/dv derivata di x= (uv)^(1/2) per dv quindi 1/2*(u)^(1/2)*(v)^(-1/2) dy/du derivata di y=(u)^(1/2)v^(-1/2) per du quindi 1/2*(u)^(-1/2)v^(-1/2) dy/dv derivata di y=(u)^(1/2)v^(-1/2) per dv quindi (-1/2)(u)^(1/2)v^(-3/2) |1/2*(u)^(-1/2)*(v)^(1/2) 1/2*(u)^(1/2)*(v)^(-1/2) | d(x,y)/d(u,v) = | | |1/2*(u)^(-1/2)v^(-1/2) (-1/2)(u)^(1/2)v^(-3/2) | 1/2*(u)^(-1/2)*(v)^(1/2) *(-1/2)(u)^(1/2)v^(-3/2) - 1/2*(u)^(-1/2)v^(-1/2)*1/2*(u)^(1/2)*(v)… end 1/2*(-1/2)v^(-1) - 1/2*1/2*(v)^(-1) end -1/4*v^(-1) - 1/4*(v)^(-1) end -1/2*(v)^(-1) det=-1/2*1/v || 1/2*1/v I have to say, using a different path I get this determinant, 2*v as you can see here xy=u -> x=u/y -> x=u/(vx) -> x=(u/v)^(1/2) y/x=v -> y=vx -> y=v(u/v)^(1/2) -> y= v^(1/2)(u)^(1/2) |du/dx du/dy | d(x,y)/d(u,v) = | | |dv/dx dv/dy | du/dx derivata di xy=u per dx quindi y du/dy derivata di xy=u per dy quindi x dv/dx derivata di y/x=v per dx quindi (-y*1/x^2) dv/dy derivata di y/x=v per dy quindi 1/x mi viene | y x | d(x,y)/d(u,v) = | | |(-y*1/x^2) 1/x| y*1/x - (-y*1/x^2)*x y/x - (-y/x) 2*y/x 2*v executing the operations, the result is not correct, according to the solution it should be 2^(1/2)/2[2*2^(1/2) -1]. (or) ) We see that x = (u/v)^(1/2) and y = (uv)^(1/2). So, x + y = u^(1/2) * (v^(-1/2) + v^(1/2)). (ii) The given region in the xy-plane transforms to u in [1, 2] and v in [1,2]. (iii) d(u,v)/d(x,y) = | y............x| |-y/x^2....1/x| = 2y/x = 2v. Thus, d(x,y)/d(u,v) = 1/(2v). ------------------- So, V = (double) integral (x + y) dA = integral(u in [1, 2], v in [1, 2]) [u^(1/2) (v^(-1/2) + v^(1/2))] (1/(2v) * dv du) = (integral(u in [1, 2]) u^(1/2)/2 du) * (integral(v in [1, 2]) (v^(-3/2) + v^(-1/2)) dv) = [(1/3) * u^(3/2) {for u in [1, 2]}] * [(-2v^(-1/2) + 2v^(1/2)) {for v in [1, 2]}] = [(1/3) * (2^(3/2) - 1)] * [(-2 * 2^(-1/2) + 2* 2^(1/2)) - 0] = (2^(1/2) / 3) * (2^(3/2) - 1) = (2 - sqrt(2)) / 3.

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