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Question-7216

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Question Number 7216 by peter james last updated on 16/Aug/16
Answered by Yozzia last updated on 16/Aug/16
ln(x+y)−ln(x+3)+ln∣((3−y)/(x+3))∣=B−ln∣x+3∣ Let u=((x+y)/(x+3))⇒y=u(x+3)−x ∴ y′=u′(x+3)+u−1 ⇒y′+1=u+u′(x+3). ∴ (u+u′(x+3))lnu=u u+u′(x+3)=(u/(lnu)) u′(x+3)=(u/(lnu))−u=((u(1−lnu))/(lnu)) or (du/dx)(x+3)=((u(1−lnu))/(lnu)) ⇒∫((lnu)/(u(1−lnu)))du=∫(dx/(x+3)) Let k=lnu⇒dk=(1/u)du ∴∫((k dk)/(1−k))=ln∣x+3∣+D ∫(−1+(1/(1−k)))dk=ln∣x+3∣+D −k−ln∣1−k∣=ln∣x+3∣+D ⇒k+ln∣1−k∣=B−ln∣x+3∣ lne^k +ln∣1−k∣=B−ln∣x+3∣ ∵ ln∣1−k∣=ln∣k−1∣ ⇒ln∣e^k (k−1)∣=B−ln∣x+3∣ ⇒ee^(k−1) (k−1)=e^(B−ln∣x+3∣) (k−1)e^(k−1) =e^(−1) ×A×(1/(x+3))=(F/(x+3)) ∴ W{(k−1)e^(k−1) }=W((F/(x+3))) k−1=W((F/(x+3))) k=1+W((F/(x+3))) ∴ lnu=1+W((F/(x+3))) u=exp(1+W((F/(x+3)))) ∴((y+x)/(x+3))=exp(1+W((F/(x+3)))) ⇒y=(x+3)exp(1+W((F/(x+3))))−x W=Lambert W function F=constant −−−−−−−−−−−−−−−−−−− Alternative to the result that uses the W−function, an implicit equation in x and y can be obtained. k+ln∣1−k∣=B−ln∣x+3∣ ∴lnu+ln∣1−lnu∣=B−ln∣x+3∣ ∴ ln((x+y)/(x+3))+ln∣1−ln((x+y)/(x+3))∣=B−ln∣x+3∣ ln(x+y)−ln(x+3)+ln∣1+ln(x+3)−ln(x+y)∣=B−ln∣x+3∣ For ln(x+3)∈R, x+3>0⇒∣x+3∣=x+3 ∴ ln(x+y)−ln(x+3)+ln∣1+ln(x+3)−ln(x+y)∣=B−ln(x+3) ln(x+y)+ln∣1+ln(x+3)−ln(x+y)∣+C=0 C=constant
$${ln}\left({x}+{y}\right)−{ln}\left({x}+\mathrm{3}\right)+{ln}\mid\frac{\mathrm{3}−{y}}{{x}+\mathrm{3}}\mid={B}−{ln}\mid{x}+\mathrm{3}\mid \\ $$$${Let}\:{u}=\frac{{x}+{y}}{{x}+\mathrm{3}}\Rightarrow{y}={u}\left({x}+\mathrm{3}\right)−{x} \\ $$$$\therefore\:{y}'={u}'\left({x}+\mathrm{3}\right)+{u}−\mathrm{1} \\ $$$$\Rightarrow{y}'+\mathrm{1}={u}+{u}'\left({x}+\mathrm{3}\right). \\ $$$$\therefore\:\left({u}+{u}'\left({x}+\mathrm{3}\right)\right){lnu}={u} \\ $$$${u}+{u}'\left({x}+\mathrm{3}\right)=\frac{{u}}{{lnu}} \\ $$$${u}'\left({x}+\mathrm{3}\right)=\frac{{u}}{{lnu}}−{u}=\frac{{u}\left(\mathrm{1}−{lnu}\right)}{{lnu}} \\ $$$${or}\:\frac{{du}}{{dx}}\left({x}+\mathrm{3}\right)=\frac{{u}\left(\mathrm{1}−{lnu}\right)}{{lnu}} \\ $$$$\Rightarrow\int\frac{{lnu}}{{u}\left(\mathrm{1}−{lnu}\right)}{du}=\int\frac{{dx}}{{x}+\mathrm{3}} \\ $$$${Let}\:{k}={lnu}\Rightarrow{dk}=\frac{\mathrm{1}}{{u}}{du} \\ $$$$\therefore\int\frac{{k}\:{dk}}{\mathrm{1}−{k}}={ln}\mid{x}+\mathrm{3}\mid+{D} \\ $$$$\int\left(−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}−{k}}\right){dk}={ln}\mid{x}+\mathrm{3}\mid+{D} \\ $$$$−{k}−{ln}\mid\mathrm{1}−{k}\mid={ln}\mid{x}+\mathrm{3}\mid+{D} \\ $$$$\Rightarrow{k}+{ln}\mid\mathrm{1}−{k}\mid={B}−{ln}\mid{x}+\mathrm{3}\mid \\ $$$${lne}^{{k}} +{ln}\mid\mathrm{1}−{k}\mid={B}−{ln}\mid{x}+\mathrm{3}\mid \\ $$$$\because\:{ln}\mid\mathrm{1}−{k}\mid={ln}\mid{k}−\mathrm{1}\mid \\ $$$$\Rightarrow{ln}\mid{e}^{{k}} \left({k}−\mathrm{1}\right)\mid={B}−{ln}\mid{x}+\mathrm{3}\mid \\ $$$$\Rightarrow{ee}^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)={e}^{{B}−{ln}\mid{x}+\mathrm{3}\mid} \\ $$$$\left({k}−\mathrm{1}\right){e}^{{k}−\mathrm{1}} ={e}^{−\mathrm{1}} ×{A}×\frac{\mathrm{1}}{{x}+\mathrm{3}}=\frac{{F}}{{x}+\mathrm{3}} \\ $$$$\therefore\:{W}\left\{\left({k}−\mathrm{1}\right){e}^{{k}−\mathrm{1}} \right\}={W}\left(\frac{{F}}{{x}+\mathrm{3}}\right) \\ $$$${k}−\mathrm{1}={W}\left(\frac{{F}}{{x}+\mathrm{3}}\right) \\ $$$${k}=\mathrm{1}+{W}\left(\frac{{F}}{{x}+\mathrm{3}}\right) \\ $$$$\therefore\:{lnu}=\mathrm{1}+{W}\left(\frac{{F}}{{x}+\mathrm{3}}\right) \\ $$$${u}={exp}\left(\mathrm{1}+{W}\left(\frac{{F}}{{x}+\mathrm{3}}\right)\right) \\ $$$$\therefore\frac{{y}+{x}}{{x}+\mathrm{3}}={exp}\left(\mathrm{1}+{W}\left(\frac{{F}}{{x}+\mathrm{3}}\right)\right) \\ $$$$\Rightarrow{y}=\left({x}+\mathrm{3}\right){exp}\left(\mathrm{1}+{W}\left(\frac{{F}}{{x}+\mathrm{3}}\right)\right)−{x} \\ $$$${W}={Lambert}\:{W}\:{function} \\ $$$${F}={constant} \\ $$$$−−−−−−−−−−−−−−−−−−− \\ $$$${Alternative}\:{to}\:{the}\:{result}\:{that}\:{uses} \\ $$$${the}\:{W}−{function},\:{an}\:{implicit}\:{equation} \\ $$$${in}\:{x}\:{and}\:{y}\:{can}\:{be}\:{obtained}. \\ $$$${k}+{ln}\mid\mathrm{1}−{k}\mid={B}−{ln}\mid{x}+\mathrm{3}\mid \\ $$$$\therefore{lnu}+{ln}\mid\mathrm{1}−{lnu}\mid={B}−{ln}\mid{x}+\mathrm{3}\mid \\ $$$$\therefore\:{ln}\frac{{x}+{y}}{{x}+\mathrm{3}}+{ln}\mid\mathrm{1}−{ln}\frac{{x}+{y}}{{x}+\mathrm{3}}\mid={B}−{ln}\mid{x}+\mathrm{3}\mid \\ $$$${ln}\left({x}+{y}\right)−{ln}\left({x}+\mathrm{3}\right)+{ln}\mid\mathrm{1}+{ln}\left({x}+\mathrm{3}\right)−{ln}\left({x}+{y}\right)\mid={B}−{ln}\mid{x}+\mathrm{3}\mid \\ $$$${For}\:{ln}\left({x}+\mathrm{3}\right)\in\mathbb{R},\:{x}+\mathrm{3}>\mathrm{0}\Rightarrow\mid{x}+\mathrm{3}\mid={x}+\mathrm{3} \\ $$$$\therefore\:{ln}\left({x}+{y}\right)−{ln}\left({x}+\mathrm{3}\right)+{ln}\mid\mathrm{1}+{ln}\left({x}+\mathrm{3}\right)−{ln}\left({x}+{y}\right)\mid={B}−{ln}\left({x}+\mathrm{3}\right) \\ $$$$ \\ $$$${ln}\left({x}+{y}\right)+{ln}\mid\mathrm{1}+{ln}\left({x}+\mathrm{3}\right)−{ln}\left({x}+{y}\right)\mid+{C}=\mathrm{0} \\ $$$${C}={constant} \\ $$
Commented by peter james last updated on 17/Aug/16
Thank you so much sir for the help.
$$\:\boldsymbol{{T}}{hank}\:{you}\:{so}\:{much}\:{sir}\:{for}\:{the}\:{help}. \\ $$

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