A) \[{{y}^{2}}+2xy-{{x}^{2}}=c\]
B) \[{{y}^{2}}+2xy+{{x}^{2}}=c\]
C) \[{{y}^{2}}-2xy-{{x}^{2}}=c\]
D) \[{{y}^{2}}-2xy+{{x}^{2}}=c\]
Correct Answer: A
Solution :
Given \[\frac{dy}{dx}=\frac{x-y}{x+y}\]. Put \[y=vx\] Þ \[\frac{dy}{dx}=v+x\,\frac{dv}{dx}\] \\[v+x\,\frac{dv}{dx}=\frac{x-vx}{x+vx}\] Þ \[v+x\,\frac{dv}{dx}=\frac{1-v}{1+v}\] Þ \[\frac{1+v}{2-{{(1+v)}^{2}}}dv=\frac{dx}{x}\] Integrating both sides, \[\int{\frac{1+v}{2-{{(1+v)}^{2}}}}\,dv=\int{\frac{dx}{x}}\] Put \[{{(1+v)}^{2}}=t\Rightarrow 2(1+v)dv=dt\] Þ \[\frac{1}{2}\int_{{}}^{{}}{\frac{dt}{2-t}}=\int_{{}}^{{}}{\frac{dx}{x}}\] Þ \[-\frac{1}{2}\log (2-t)=\log xc\] Þ \[-\frac{1}{2}\log [2-{{(1+v)}^{2}}]=\log xc\] Þ \[-\frac{1}{2}\log [-{{v}^{2}}-2v+1]=\log xc\] Þ \[\log \frac{1}{\sqrt{1-2v-{{v}^{2}}}}=\log xc\] Þ \[{{x}^{2}}{{c}^{2}}(1-2v-{{v}^{2}})=1\] Þ \[{{y}^{2}}+2xy-{{x}^{2}}={{c}_{1}}\].
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