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Q. One of the relation of the D.E. \( \frac{d x}{3 z-4 y}=\frac{d y}{4 x-2 z}=\frac{d z}{2 y-3 x} \), Using a set of multipliers \( x, y, z \) is |
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Answer» \(\frac{dx}{3z-4y}\)=\(\frac{dy}{4x-2z}\)=\(\frac{dz}{2y-3x}\) = \(\frac{xdx+ydy+zdz}{(3xz-4xy)+(4xy-2yx)+(2yz-3xz)}\) ⇒ \(\frac{dx}{3z-4y}\)=\(\frac{dy}{4x-2z}\)=\(\frac{dz}{2y-3x}\) = \(\frac{xdx+ydy+zdz}{0}\) ⇒ xdx + ydy + zdx = 0 ⇒ ∫(xdx + ydy + zdz) = \(\frac{c}{2}\) Where \(\frac{c}{2}\) is an integral constant. ⇒ \(\frac{x^2}{2}\)+\(\frac{y^2}{2}\)+\(\frac{z^2}{2}\) = \(\frac{c}{2}\) ⇒ x2 + y2 +z2 = c Hence, The solution of given differential equation is x2 + y2 +z2 = c. |
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