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Question Number 53426 by Necxx last updated on 21/Jan/19

$$Thegeneralsolutionoftheequation$$$$\frac{dy}{dx}+ylnx=x^{-x}$$$$a)x^x(1-{ce}^x)b)-x^{-x}(1+{ce}^{2x})$$$$c)-x^{-x}(1-{ce}^x)$$

Commented by maxmathsup by imad last updated on 26/May/19

$$theequationis{y}^{{}^{\prime }}+yln\left(x\right)=x^{-x}$$$$\left(he\right)\to y^{{}^{\prime }}+yln\left(x\right)=0\Rightarrow y^{{}^{\prime }}=-yln\left(x\right)\Rightarrow \frac{y^{{}^{\prime }}}{y}=-ln\left(x\right)\Rightarrow ln\mid y\mid =-\int ln\left(x\right)dx+c$$$$=-(xlnx-x)+c=-xln\left(x\right)+x+c\Rightarrow y\left(x\right)=k{e}^x{x}^{-x}=k{e}^{x-xlnx}$$$$mvcmethod\to y^{{}^{\prime }}=k^{{}^{\prime }}{e}^{x-xlnx}+k{(x-xlnx)}^{{}^{\prime }}{e}^{x-xlnx}$$$$=(k^{{}^{\prime }}+k(1-(lnx+1))e^{x-xlnx}=(k^{{}^{\prime }}-kln\left(x\right))e^{x-xln\left(x\right)}$$$$\left(e\right)\Rightarrow (k^{{}^{\prime }}-kln\left(x\right))e^{x-xln\left(x\right)}+k{e}^x{x}^{-x}ln\left(x\right)=e^{-xln\left(x\right)}\Rightarrow$$$$k^{{}^{\prime }}{e}^{x-xln\left(x\right)}-kln\left(x\right)e^{x-xln\left(x\right)}+kln\left(x\right)e^{x-xln\left(x\right)}=e^{-xln\left(x\right)}\Rightarrow$$$$k^{{}^{\prime }}{e}^{x-xln\left(x\right)}=e^{-xln\left(x\right)}\Rightarrow k^{{}^{\prime }}=e^{-x}\Rightarrow k\left(x\right)=-e^{-x}+\lambda \Rightarrow$$$$y\left(x\right)=(-e^{-x}+\lambda )e^x{x}^{-x}=(\lambda e^x-1)x^{-x}sothetrueansweris\left(c\right).$$

Answered by tanmay.chaudhury50@gmail.com last updated on 21/Jan/19

$$\frac{dy}{dx}+ylnx=x^{-x}$$$$intregatingfactor=e^{\int lnx}=e^{xlnx-x}.$$$$e^{xlnx-x}\frac{dy}{dx}+ylnx\times e^{xlnx-x}=x^{-x}\times e^{xlnx-x}.$$$$x^x{e}^{-x}\frac{dy}{dx}+ylnx\times x^x{e}^{-x}=e^{-x}$$$$x^x{e}^{-x}\frac{dy}{dx}+y\times \left(x^x{e}^{-x}\times lnx\right)=e^{-x}$$$$\frac{d}{dx}\left(y\times x^x{e}^{-x}\right)=e^{-x}$$$$d\left({yx}^x{e}^{-x}\right)=e^{-x}dx$$$${yx}^x{e}^{-x}=\frac{e^{-x}}{-1}+c$$$$y=\frac{-e^{-x}}{x^x{e}^{-x}}+\frac{c}{x^x{e}^{-x}}$$$$y=\frac{1}{x^x}(-1+{ce}^x)=x^{-x}(-1+{ce}^x)$$$$plscheck...$$$$$$$$$$$$$$

Commented by Necxx last updated on 22/Jan/19

$$oh.....thankssomuch$$

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