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

                  (dy/dx)−y = xy^5

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

Answered by liberty last updated on 13/Jan/21

 let v = y^(−4)  ; (dv/dx) =−4y^(−5 ) (dy/dx)   (dy/dx) = −(1/4)y^5  (dv/dx) (substitute into diff eq )  give −(1/4)y^5  (dv/dx)−y = xy^5    (dv/dx) + 4v = −4x ; put integrating factor   μ =e^(∫ 4 dx) = e^(4x)  , then we get    v = ((∫ −4x.e^(4x)  dx + C)/e^(4x) )   (1/y^4 ) = ((−xe^(4x) +(1/4)e^(4x)  +C)/e^(4x) )   y^4  = (e^(4x) /(e^(4x) (−x+(1/4))+C))

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

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