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Question Number 46694 by MJS last updated on 30/Oct/18

$$(y^{″}y-{\left(y^{\prime }\right)}^2){\mathrm{e}}^{\frac{y^{\prime }}{y}}=y^2$$$$\mathrm{not}\mathrm{sure}\mathrm{if}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{possible}\mathrm{to}\mathrm{solve}\mathrm{this}\mathrm{at}\mathrm{all}...$$

Answered by ajfour last updated on 30/Oct/18

$$\ln [y^{″}y-{\left(y^{\prime }\right)}^2]+\frac{y^{\prime }}{y}=2\ln y$$$$\Rightarrow \ln [\frac{y^{″}}{y}-{\left(\frac{y^{\prime }}{y}\right)}^2]=-\frac{y^{\prime }}{y}$$$$let\frac{y^{\prime }}{y}=t\Rightarrow \frac{{(y^{″}y-y^{\prime })}^2}{y^2}=\frac{dt}{dx}$$$$\Rightarrow \ln \frac{dt}{dx}=-t$$$$ordt=e^{-t}dx$$$$\Rightarrow \int e^{t}dt=\int dx$$$$\Rightarrow e^t=x+c_1$$$$e^{\frac{y^{\prime }}{y}}=x+c_1$$$$\Rightarrow \int \frac{dy}{y}=\int \ln (x+c_1)dx$$$$\ln y=x\ln (x+c_1)-\int \frac{xdx}{x+c_1}$$$$\ln y=x\ln (x+c_1)-x+c_1\ln (x+c_1)+c_2$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}.$$

Commented by MJS last updated on 30/Oct/18

$$\mathrm{thank}\mathrm{you}!$$

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