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Question Number 13025 by ARJUN SUBBA last updated on 11/May/17
Commented by prakash jain last updated on 13/May/17
$$\mathrm{A}\:\mathrm{should}\:\mathrm{be}\:−\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Answered by prakash jain last updated on 12/May/17
![A should be −(1/2) (2/(√5))ln (((√5)+1+2v)/((√5)−1−2v))=ln x+(1/2)ln (2−2v−2v^2 )+ln c_1 (2/(√5))ln (((1+(√5))x+2y)/(((√5)−1)x−2y))=((2ln x+ln (((2x^2 −2xy−2y^2 ))/x^2 )+ln c_1 )/2) (2/(√5))ln (((1+(√5))x+2y)/(((√5)−1)x−2y))=((ln x^2 +ln (((2x^2 −2xy−2y^2 ))/x^2 )+ln c_1 )/2) (2/(√5))ln (((1+(√5))x+2y)/(((√5)−1)x−2y))=((ln x^2 .(((2x^2 −2xy−2y^2 ))/x^2 )+ln c_1 )/2) (2/(√5))ln (((1+(√5))x+2y)/(((√5)−1)x−2y))=((ln(2x^2 −2xy−2y^2 ) +ln c_1 )/2) (2/(√5))ln (((1+(√5))x+2y)/(((√5)−1)x−2y))=((ln(x^2 −xy−y^2 ) +ln 2+ln c_1 )/2) ln 2+ln c_1 =const=ln c (2/(√5))ln (((1+(√5))x+2y)/(((√5)−1)x−2y))=((ln(x^2 −xy−y^2 )∙c)/2) ln (((1+(√5))x+2y)/(((√5)−1)x−2y))=((√5)/2)∙((ln[(x^2 −xy−y^2 )∙c])/2) (((1+(√5))x+2y)/(((√5)−1)x−2y))=(x^2 −xy−y^2 )^((√5)/4) ∙c_2 (x^2 −xy−y^2 )^((√5)/4) (((1+(√5))x+2y)/(((√5)−1)x−2y))=c](Q13065.png)
$$\mathrm{A}\:\mathrm{should}\:\mathrm{be}\:−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{\mathrm{2}}{\sqrt{\mathrm{5}}}\mathrm{ln}\:\frac{\sqrt{\mathrm{5}}+\mathrm{1}+\mathrm{2}{v}}{\sqrt{\mathrm{5}}−\mathrm{1}−\mathrm{2}{v}}=\mathrm{ln}\:{x}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left(\mathrm{2}−\mathrm{2}{v}−\mathrm{2}{v}^{\mathrm{2}} \right)+\mathrm{ln}\:\mathrm{c}_{\mathrm{1}} \\ $$$$\frac{\mathrm{2}}{\sqrt{\mathrm{5}}}\mathrm{ln}\:\frac{\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)\mathrm{x}+\mathrm{2}{y}}{\left(\sqrt{\mathrm{5}}−\mathrm{1}\right){x}−\mathrm{2}{y}}=\frac{\mathrm{2ln}\:\mathrm{x}+\mathrm{ln}\:\frac{\left(\mathrm{2x}^{\mathrm{2}} −\mathrm{2xy}−\mathrm{2y}^{\mathrm{2}} \right)}{\mathrm{x}^{\mathrm{2}} }+\mathrm{ln}\:\mathrm{c}_{\mathrm{1}} }{\mathrm{2}} \\ $$$$\frac{\mathrm{2}}{\sqrt{\mathrm{5}}}\mathrm{ln}\:\frac{\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)\mathrm{x}+\mathrm{2}{y}}{\left(\sqrt{\mathrm{5}}−\mathrm{1}\right){x}−\mathrm{2}{y}}=\frac{\mathrm{ln}\:\mathrm{x}^{\mathrm{2}} +\mathrm{ln}\:\frac{\left(\mathrm{2x}^{\mathrm{2}} −\mathrm{2xy}−\mathrm{2y}^{\mathrm{2}} \right)}{\mathrm{x}^{\mathrm{2}} }+\mathrm{ln}\:\mathrm{c}_{\mathrm{1}} }{\mathrm{2}} \\ $$$$\frac{\mathrm{2}}{\sqrt{\mathrm{5}}}\mathrm{ln}\:\frac{\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)\mathrm{x}+\mathrm{2}{y}}{\left(\sqrt{\mathrm{5}}−\mathrm{1}\right){x}−\mathrm{2}{y}}=\frac{\mathrm{ln}\:\mathrm{x}^{\mathrm{2}} .\frac{\left(\mathrm{2x}^{\mathrm{2}} −\mathrm{2xy}−\mathrm{2y}^{\mathrm{2}} \right)}{\mathrm{x}^{\mathrm{2}} }+\mathrm{ln}\:\mathrm{c}_{\mathrm{1}} }{\mathrm{2}} \\ $$$$\frac{\mathrm{2}}{\sqrt{\mathrm{5}}}\mathrm{ln}\:\frac{\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)\mathrm{x}+\mathrm{2}{y}}{\left(\sqrt{\mathrm{5}}−\mathrm{1}\right){x}−\mathrm{2}{y}}=\frac{\mathrm{ln}\left(\mathrm{2x}^{\mathrm{2}} −\mathrm{2xy}−\mathrm{2y}^{\mathrm{2}} \right)\:+\mathrm{ln}\:\mathrm{c}_{\mathrm{1}} }{\mathrm{2}} \\ $$$$\frac{\mathrm{2}}{\sqrt{\mathrm{5}}}\mathrm{ln}\:\frac{\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)\mathrm{x}+\mathrm{2}{y}}{\left(\sqrt{\mathrm{5}}−\mathrm{1}\right){x}−\mathrm{2}{y}}=\frac{\mathrm{ln}\left(\mathrm{x}^{\mathrm{2}} −\mathrm{xy}−\mathrm{y}^{\mathrm{2}} \right)\:+\mathrm{ln}\:\mathrm{2}+\mathrm{ln}\:\mathrm{c}_{\mathrm{1}} }{\mathrm{2}} \\ $$$$\mathrm{ln}\:\mathrm{2}+\mathrm{ln}\:{c}_{\mathrm{1}} \:={const}=\mathrm{ln}\:{c} \\ $$$$\frac{\mathrm{2}}{\sqrt{\mathrm{5}}}\mathrm{ln}\:\frac{\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)\mathrm{x}+\mathrm{2}{y}}{\left(\sqrt{\mathrm{5}}−\mathrm{1}\right){x}−\mathrm{2}{y}}=\frac{\mathrm{ln}\left(\mathrm{x}^{\mathrm{2}} −\mathrm{xy}−\mathrm{y}^{\mathrm{2}} \right)\centerdot{c}}{\mathrm{2}} \\ $$$$\mathrm{ln}\:\frac{\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)\mathrm{x}+\mathrm{2}{y}}{\left(\sqrt{\mathrm{5}}−\mathrm{1}\right){x}−\mathrm{2}{y}}=\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\centerdot\frac{\mathrm{ln}\left[\left(\mathrm{x}^{\mathrm{2}} −\mathrm{xy}−\mathrm{y}^{\mathrm{2}} \right)\centerdot{c}\right]}{\mathrm{2}} \\ $$$$\:\frac{\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)\mathrm{x}+\mathrm{2}{y}}{\left(\sqrt{\mathrm{5}}−\mathrm{1}\right){x}−\mathrm{2}{y}}=\left(\mathrm{x}^{\mathrm{2}} −\mathrm{xy}−\mathrm{y}^{\mathrm{2}} \right)^{\sqrt{\mathrm{5}}/\mathrm{4}} \centerdot{c}_{\mathrm{2}} \\ $$$$\:\left(\mathrm{x}^{\mathrm{2}} −\mathrm{xy}−\mathrm{y}^{\mathrm{2}} \right)^{\sqrt{\mathrm{5}}/\mathrm{4}} \frac{\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)\mathrm{x}+\mathrm{2}{y}}{\left(\sqrt{\mathrm{5}}−\mathrm{1}\right){x}−\mathrm{2}{y}}={c} \\ $$
Commented by prakash jain last updated on 13/May/17
$$\mathrm{it}\:\mathrm{does}\:\mathrm{not}\:\mathrm{match}\:\mathrm{with}\:\mathrm{your}\:\mathrm{final} \\ $$$$\mathrm{amswer}.\:\mathrm{Most}\:\mathrm{likely}\:\mathrm{some}\:\mathrm{typo} \\ $$$$\mathrm{error}\:\mathrm{in}\:\mathrm{ur}\:\mathrm{final}\:\mathrm{answer}. \\ $$