Question Number 15797 by tawa tawa last updated on 14/Jun/17
$$\mathrm{If}\:\:\left(\mathrm{a}\:+\:\mathrm{bx}\right)\mathrm{e}^{\mathrm{y}/\mathrm{x}} \:=\:\mathrm{x},\:\:\:\mathrm{where}\:\:\mathrm{a}\:\mathrm{and}\:\mathrm{b}\:\mathrm{are}\:\mathrm{constant},\: \\ $$$$\mathrm{prove}\:\mathrm{that},:\:\:\:\mathrm{x}^{\mathrm{3}} \mathrm{y}''\:=\:\left(\mathrm{xy}'\:−\:\mathrm{y}\right)^{\mathrm{2}} \\ $$
Commented by tawa tawa last updated on 14/Jun/17
$$\mathrm{please}\:\mathrm{any}\:\mathrm{answer}\:\mathrm{on}\:\mathrm{this}\:?.\:\mathrm{please}\:\mathrm{help}.\:\mathrm{Thanks}\:\mathrm{in}\:\mathrm{advamce}. \\ $$
Answered by ajfour last updated on 14/Jun/17
![(a+bx)e^(y/x) =x (d/dx)[(a+bx)e^(y/x) ]=(d/dx)(x) (a+bx)e^(y/x) [((y′)/x)−(y/x^2 )]+be^(y/x) =1 ⇒ x[((y′)/x)−(y/x^2 )]+be^(y/x) =1 or y′−(y/x)+be^(y/x) =1 (1−y′+(y/x))e^(−y/x) =b differentiating again: (−y′′+((y′)/x)−(y/x^2 ))e^(−y/x) + (1−y′+(y/x))e^(−y/x) [−((y′)/x)+(y/x^2 )] =0 multiplying by e^(y/x) : −y′′+((y′)/x)−(y/x^2 ) =−(1−y′+(y/x))(−((y′)/x)+(y/x^2 )) multiplying by (−x^3 ): x^3 y′′−x^2 y′+xy=(x−xy′+y)(−xy′+y) x^3 y′′−x^2 y′+xy =−x^2 y′+xy+(xy′)^2 −xyy′−xyy′+y^2 or x^3 y′′= (xy′−y)^2 .](https://www.tinkutara.com/question/Q15833.png)
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({a}+{bx}\right){e}^{{y}/{x}} ={x} \\ $$$$\:\:\:\:\:\:\frac{{d}}{{dx}}\left[\left({a}+{bx}\right){e}^{{y}/{x}} \right]=\frac{{d}}{{dx}}\left({x}\right) \\ $$$$\:\left({a}+{bx}\right){e}^{{y}/{x}} \left[\frac{{y}'}{{x}}−\frac{{y}}{{x}^{\mathrm{2}} }\right]+{be}^{{y}/{x}} =\mathrm{1} \\ $$$$\:\Rightarrow\:\:\:\:\:\:\:{x}\left[\frac{{y}'}{{x}}−\frac{{y}}{{x}^{\mathrm{2}} }\right]+{be}^{{y}/{x}} =\mathrm{1} \\ $$$${or}\:\:\:\:\:\:\:{y}'−\frac{{y}}{{x}}+{be}^{{y}/{x}} =\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\left(\mathrm{1}−{y}'+\frac{{y}}{{x}}\right){e}^{−{y}/{x}} ={b} \\ $$$$\:\:\:{differentiating}\:{again}: \\ $$$$\left(−{y}''+\frac{{y}'}{{x}}−\frac{{y}}{{x}^{\mathrm{2}} }\right){e}^{−{y}/{x}} + \\ $$$$\:\:\:\:\:\left(\mathrm{1}−{y}'+\frac{{y}}{{x}}\right){e}^{−{y}/{x}} \left[−\frac{{y}'}{{x}}+\frac{{y}}{{x}^{\mathrm{2}} }\right]\:=\mathrm{0} \\ $$$${multiplying}\:{by}\:{e}^{{y}/{x}} : \\ $$$$−{y}''+\frac{{y}'}{{x}}−\frac{{y}}{{x}^{\mathrm{2}} }\:=−\left(\mathrm{1}−{y}'+\frac{{y}}{{x}}\right)\left(−\frac{{y}'}{{x}}+\frac{{y}}{{x}^{\mathrm{2}} }\right) \\ $$$${multiplying}\:{by}\:\:\left(−{x}^{\mathrm{3}} \right): \\ $$$$\:{x}^{\mathrm{3}} {y}''−{x}^{\mathrm{2}} {y}'+{xy}=\left({x}−{xy}'+{y}\right)\left(−{xy}'+{y}\right) \\ $$$${x}^{\mathrm{3}} {y}''−{x}^{\mathrm{2}} {y}'+{xy}\:=−{x}^{\mathrm{2}} {y}'+{xy}+\left({xy}'\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−{xyy}'−{xyy}'+{y}^{\mathrm{2}} \\ $$$${or}\:\:\:\boldsymbol{{x}}^{\mathrm{3}} \boldsymbol{{y}}''=\:\left(\boldsymbol{{xy}}'−\boldsymbol{{y}}\right)^{\mathrm{2}} \:\:. \\ $$