# Answer in Differential Equations

“displaystyletextsf{This is a Pfaffian differential equation in three}n\textsf{variables and we must verify its integrabilty}n\textsf{and determine its primitive.}\nn(1 + yz)mathrm{d}x + (xz u2212 x^2)mathrm{d}y u2212 (1 + xy)mathrm{d}z = 0 \nntextsf{The necessary and sufficient condition}\textsf{for iintegrability is} \nntextbf{X}cdot curltextbf{X} = 0 \nntextbf{X}=(1 + yz,xz u2212 x^2,-1 – xy), textsf{so that}\nn nabla times X = begin{vmatrix}ntextbf{i} &textbf{j} &textbf{k} \nfrac{partial}{partial x} &frac{partial}{partial y} &frac{partial}{partial z}\n1 + yz &xz u2212 x^2 &-1 – xynend{vmatrix} = \nnbegin{aligned}n&textbf{i} left(frac{partial}{partial y}(-1 – xy) – frac{partial}{partial z}(xz – x^2)right) – n\&textbf{j} left(frac{partial}{partial x}(-1 – xy) – frac{partial}{partial z}(1 + yz)right) + n\&textbf{k} left(frac{partial}{partial x}(xz – x^2) – frac{partial}{partial y}(1 + yz)right)nend{aligned} \nnbegin{aligned}n&=textbf{i}(-x – x) – textbf{j}(-y – y) + textbf{k}(z – 2x – z) \&= -2xtextbf{i} + 2ytextbf{j} – 2xtextbf{k}nend{aligned} \nnbegin{aligned}ntherefore &(1 + yz,xz u2212 x^2,-1 -xy) cdot (-2x, 2y, -2x) \&= -2x(1 + yz) + 2y(xz – x^2) + 2x(1 + xy) \&= -2x – 2xyz + 2xyz – 2yx^2 + 2x + 2x^2y = 0nend{aligned} \nntextsf{Thus, the given equation is integrable.}\nntextsf{Solving by Inspection} \nn(1 + yz)mathrm{d}x + (xz u2212 x^2)mathrm{d}y u2212 (1 + xy)mathrm{d}z = 0\nn(1 + yz)mathrm{d}x + (xz u2212 x^2)mathrm{d}y u2212 (1 + xy)mathrm{d}z = 0\nn-(1 + yz)mathrm{d}x – (xz u2212 x^2)mathrm{d}y u2212 (1 + xy)mathrm{d}z = 0\nn(1 + yx)mathrm{d}z – (1 + yx)mathrm{d}x – x(z – x) mathrm{d}y – y(z – x)mathrm{d}x = 0\nn(1 + yx)mathrm{d}(z – x) – (xmathrm{d}y + ymathrm{d}x)(z – x) = 0\nn(1 + yx)mathrm{d}(z – x) – mathrm{d}(1 + xy)(z – x) = 0 \nnimplies frac{mathrm{d}(z – x)}{z – x} – frac{mathrm{d}(1 + xy)}{1 + xy} = 0\nntextsf{Integrating both sides, we have}\nnintfrac{mathrm{d}(z – x)}{z – x} – intfrac{mathrm{d}(1 + xy)}{1 + xy} = int0, mathrm{d}x\nnln(z – x) – ln(1 + xy) = C_1 \nnln(z – x) = ln(A(1 + xy)), hspace{1cm} {C_1 = ln(A)}\nnimplies z = x + A(1 + xy) \nnnntherefore z = x + A (1 + yx)hspace{0.1cm}textsf{is a solution to the Pfaffian differential equation}”

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