A) \[{{x}^{2}}+{{y}^{2}}=c\]
B) \[2{{x}^{2}}-{{y}^{2}}=c\]
C) \[{{x}^{2}}+2xy=c\]
D) \[{{y}^{2}}+2xy=c\]
Correct Answer: C
Solution :
\[(x+y)dx+xdy=0\] Þ \[xdy=-(x+y)dx\] Þ \[\frac{dy}{dx}=-\frac{x+y}{x}\] It is homogenous equation, hence put \[y=vx\] and \[\frac{dy}{dx}=v+x\frac{dv}{dx},\] we get \[v+x\frac{dv}{dx}=-\frac{x+vx}{x}=-\frac{1+v}{1}\] Þ \[x\frac{dv}{dx}=-1-2v\]Þ \[\int_{{}}^{{}}{\frac{dv}{1+2v}}=-\int_{{}}^{{}}{\frac{dx}{x}}\] Þ \[\frac{1}{2}\log (1+2v)=-\log x+\log c\]Þ \[\log \left( 1+2\frac{y}{x} \right)=2\log \frac{c}{x}\] Þ \[\frac{x+2y}{x}={{\left( \frac{c}{x} \right)}^{2}}\]Þ\[{{x}^{2}}+2xy=c\].
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