Then it may be seen that xy' = dy/dt and (x^2)y'' = (d^2/dt^2)y (dy/dt) The equation becomes (D^2 2D 1)y = (1e^t)^(2), where the operator D denotes differentiation with respect to t The homogeneous equation corresponding to the above equation is (D^2 2D 1)y = 0,
X^2 y^2 z^2-xy-yz-zx=0 215105-X^2+y^2+z^2-xy-yz-zx=0
X^2y^2z^2xyyzzx=0 multiplying the RHS and LHS by 2 we get , 2 x^2y^2z^2xyyzzx =0 or, (xy)^2(yz)^2(zx)^2=0 since in LHS therePranabguchhait71 is waiting for your help Add your answer and earn points Factonizex^2y^2z^2_xy_ yz_zx Get the answers you need, now!
Find The Minimum Value Of X 2 Y 2 Z 2 Subject To The Conditions Xy Yz Zx 3a 2 Sarthaks Econnect Largest Online Education Community
X^2+y^2+z^2-xy-yz-zx=0
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