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Question Number 121028 by bramlexs22 last updated on 05/Nov/20

Answered by liberty last updated on 05/Nov/20

$$\mathrm{no}.\mathrm{this}\mathrm{not}\mathrm{separable}\mathrm{but}\mathrm{it}\mathrm{is}\mathrm{a}\mathrm{homogenous}$$$$\mathrm{diff}\mathrm{equation}.$$$$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{{\mathrm{x}}^4+{2\mathrm{y}}^4}{{\mathrm{xy}}^3};[\mathrm{put}\mathrm{y}=\mathrm{zx}]$$$$\Rightarrow \mathrm{z}+\mathrm{x}\frac{\mathrm{dz}}{\mathrm{dx}}=\frac{{\mathrm{x}}^4(1+{2\mathrm{z}}^4)}{{\mathrm{z}}^3{\mathrm{x}}^4}=\frac{1+{2\mathrm{z}}^4}{{\mathrm{z}}^3}$$$$\Rightarrow \mathrm{z}+\mathrm{x}\frac{\mathrm{dz}}{\mathrm{dx}}=\frac{1}{{\mathrm{z}}^3}+2\mathrm{z};\mathrm{x}\frac{\mathrm{dz}}{\mathrm{dx}}=\frac{1+{\mathrm{z}}^4}{{\mathrm{z}}^3}$$$$\Rightarrow \frac{{\mathrm{z}}^3}{1+{\mathrm{z}}^4}\mathrm{dz}=\frac{\mathrm{dx}}{\mathrm{x}};\int \frac{\mathrm{d}(1+{\mathrm{z}}^4)}{1+{\mathrm{z}}^4}=4\int \frac{\mathrm{dx}}{\mathrm{x}}$$$$\Rightarrow \ln (1+{\mathrm{z}}^4)=\ln \left({\mathrm{Cx}}^4\right)$$$$\Rightarrow 1+{\left(\frac{\mathrm{y}}{\mathrm{x}}\right)}^4={\mathrm{Cx}}^4\Rightarrow {\mathrm{x}}^4+{\mathrm{y}}^4={\mathrm{Cx}}^8$$

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