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Question Number 67719 by aliesam last updated on 30/Aug/19

Answered by mind is power last updated on 30/Aug/19

$$\iff \frac{d(y+x)}{dx}+x(x+y)=x^3{(x+y)}^5$$$$letz=y+x$$$$\Rightarrow \frac{dz}{dx}+xz=x^3{z}^5$$$$\Rightarrow z^{{}^{\prime }}+xz=x^3{z}^5$$$$\Rightarrow \frac{z^{\prime }}{z^5}+\frac{x}{z^4}=x^3$$$$s=\frac{1}{z^4}\Rightarrow s^{\prime }=-\frac{4z^{{}^{\prime }}}{z^5}\Rightarrow \frac{z^{\prime }}{z^5}=-\frac{s^{\prime }}{4}$$$$\Rightarrow -\frac{s^{\prime }}{4}+xs=x^3.....1$$$$s\left(x\right)=k\left(x\right)e^{2x^2}solutionofhomgenequqtion$$$$-\frac{k^{\prime }{e}^{2x^2}}{4}=x^3$$$$\Rightarrow k^{\prime }=-4x^3{e}^{-2x^2}$$$$k=-\int 4x^3{e}^{-2x^2}dx$$$$u=x^2$$$$du=2xdx$$$$k=-2\int {ue}^{-2u}du$$$$k=({ue}^{-2u}+\frac{1}{2}{e}^{-2u}+c)$$$$s=((x^2+\frac{1}{2})e^{-2x^2}+c)e^{2x^2}={ce}^{2x^2}+x^2+\frac{1}{2}$$$$z=\frac{1}{s^{\frac{1}{4}}}=\frac{1}{{({ce}^{2x^2}+x^2+\frac{1}{2})}^{\frac{1}{4}}}$$$$y=z-x=\frac{1}{{({ce}^{2x^2}+x^2+\frac{1}{2})}^{\frac{1}{4}}}-x$$$$$$$$$$

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