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Question Number 129126 by benjo_mathlover last updated on 13/Jan/21

$$\frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{y}={\mathrm{xy}}^5$$

Answered by liberty last updated on 13/Jan/21

$$\mathrm{let}\mathrm{v}={\mathrm{y}}^{-4};\frac{\mathrm{dv}}{\mathrm{dx}}=-{4\mathrm{y}}^{-5}\frac{\mathrm{dy}}{\mathrm{dx}}$$$$\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{1}{4}{\mathrm{y}}^5\frac{\mathrm{dv}}{\mathrm{dx}}\left(\mathrm{substitute}\mathrm{into}\mathrm{diff}\mathrm{eq}\right)$$$$\mathrm{give}-\frac{1}{4}{\mathrm{y}}^5\frac{\mathrm{dv}}{\mathrm{dx}}-\mathrm{y}={\mathrm{xy}}^5$$$$\frac{\mathrm{dv}}{\mathrm{dx}}+4\mathrm{v}=-4\mathrm{x};\mathrm{put}\mathrm{integrating}\mathrm{factor}$$$$\mu ={\mathrm{e}}^{\int 4\mathrm{dx}}={\mathrm{e}}^{4\mathrm{x}},\mathrm{then}\mathrm{we}\mathrm{get}$$$$\mathrm{v}=\frac{\int -4\mathrm{x}.{\mathrm{e}}^{4\mathrm{x}}\mathrm{dx}+\mathrm{C}}{{\mathrm{e}}^{4\mathrm{x}}}$$$$\frac{1}{{\mathrm{y}}^4}=\frac{-{\mathrm{xe}}^{4\mathrm{x}}+\frac{1}{4}{\mathrm{e}}^{4\mathrm{x}}+\mathrm{C}}{{\mathrm{e}}^{4\mathrm{x}}}$$$$\underset{―}{}{\mathbf{y}}^4=\frac{{\mathbf{e}}^{4\mathbf{x}}}{{\mathbf{e}}^{4\mathbf{x}}(-\mathbf{x}+\frac{1}{4})+\mathbf{C}}\underset{―}{}$$

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