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Question Number 117029 by bemath last updated on 09/Oct/20

$$({\mathrm{x}}^2+{2\mathrm{y}}^2)\mathrm{dx}+(4\mathrm{xy}-{\mathrm{y}}^2)\mathrm{dy}=0$$

Answered by bemath last updated on 09/Oct/20

$$\mathrm{letting}\mathrm{y}=\phi \mathrm{x}\Rightarrow \mathrm{dy}=\phi \mathrm{dx}+\mathrm{x}\mathrm{d}\phi$$$$\Rightarrow ({\mathrm{x}}^2+2{\phi }^2{\mathrm{x}}^2)\mathrm{dx}+({4\mathrm{x}}^2\phi -{\phi }^2{\mathrm{x}}^2)(\phi \mathrm{dx}+\mathrm{x}\mathrm{d}\phi )=0$$$$(1+2{\phi }^2)\mathrm{dx}+(4\phi -{\phi }^2)(\phi \mathrm{dx}+\mathrm{xd}\phi )=0$$$$\Rightarrow (1+2{\phi }^2+4{\phi }^2-{\phi }^3)\mathrm{dx}=\mathrm{x}({\phi }^2-4\phi )\mathrm{d}\phi$$$$\Rightarrow (1+6{\phi }^2-{\phi }^3)\mathrm{dx}=\mathrm{x}({\phi }^2-4\phi )\mathrm{d}\phi$$$$\frac{\mathrm{dx}}{\mathrm{x}}=\frac{({\phi }^2-4\phi )}{1+6{\phi }^2-{\phi }^3}\mathrm{d}\phi$$$$\int \frac{\mathrm{dx}}{\mathrm{x}}=-\frac{1}{3}\int \frac{\mathrm{d}(1+6{\phi }^2-{\phi }^3)}{1+6{\phi }^2-{\phi }^3}$$$$\Rightarrow -3\int \frac{\mathrm{dx}}{\mathrm{x}}=\int \frac{\mathrm{d}(1+6{\phi }^2-{\phi }^3)}{1+6{\phi }^2-{\phi }^3}$$$$\Rightarrow -3(\ln \mid \mathrm{x}\mid +\mathrm{c})=\ln (1+6\phi -{\phi }^3)$$$$\Rightarrow \ln {\left(\frac{1}{\mathrm{Cx}}\right)}^3=\ln (1+6\phi -{\phi }^3)$$$$\Rightarrow 1+6\left(\frac{\mathrm{y}}{\mathrm{x}}\right)-{\left(\frac{\mathrm{y}}{\mathrm{x}}\right)}^3={\left(\frac{1}{\mathrm{Cx}}\right)}^3$$$$\Rightarrow {\mathrm{x}}^3+{6\mathrm{x}}^2\mathrm{y}-{\mathrm{y}}^3=\mathrm{K};\mathrm{where}\mathrm{K}=\frac{1}{{\mathrm{C}}^3}$$

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