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Resolve-y-xy-a-1-y-2-




Question Number 168742 by LEKOUMA last updated on 16/Apr/22
Resolve  y=xy′+a(√(1+(y′)^2 ))
Resolvey=xy+a1+(y)2
Answered by mindispower last updated on 19/Apr/22
⇒  y′=xy′′+y′+a((2y′y′′)/(2(√(1+y′^2 ))))  y′=z  xz′+((zz′)/( (√(1+z^2 ))))=0  z′(x+a(z/( (√(1+z^2 )))))=0  by continuity⇔ { ((z′=0)),(((z/( (√(1+z^2 ))))=−(1/a)x)) :}  z=c  or  z^2 =x^2 (1+z^2 )⇒z^2 =((((x/a))^2 )/(1−((x/a))^2 )),x∈]−1,1[  z=((−x)/( (√(a^2 −x^2 )))),⇒y=(√(a^2 −x^2 ))+c,a>0  y=((−x^2 )/( (√(a^2 −x^2 ))))+a(√(1+(x^2 /(a^2 −x^2 ))))⇒c=0  cx+d=x(c)+a(√(1+c^2 ))  d=a(√(1+c^2 ))  S={cx+a(√(1+c^2 )),(√(a^2 −x^2 ))}
y=xy+y+a2yy21+y2y=zxz+zz1+z2=0z(x+az1+z2)=0bycontinuity{z=0z1+z2=1axz=corz2=x2(1+z2)z2=(xa)21(xa)2,x]1,1[z=xa2x2,y=a2x2+c,a>0y=x2a2x2+a1+x2a2x2c=0cx+d=x(c)+a1+c2d=a1+c2S={cx+a1+c2,a2x2}

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