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 :
Given differential equation is \[(x+y)dx+xdy=0\] \[\Rightarrow \] \[xdx=-(x+y)dy\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{-(x+y)}{x}\] It is a homogeneous differential-equation, So, putting\[y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}\]we get \[v+x\frac{dv}{dx}=-\frac{x+vx}{x}=-\frac{1+v}{1}\] \[\Rightarrow \] \[x\frac{dv}{dx}=-1-2v\] \[\Rightarrow \] \[\int{\frac{dv}{1+2v}}=-\int{\frac{dx}{x}}\] \[\Rightarrow \] \[\log (1+2v)=-\log x+\log {{C}_{1}}\] \[\Rightarrow \] \[\log \left( 1+\frac{2y}{x} \right)=2\log \frac{{{C}_{1}}}{x}\] \[\Rightarrow \] \[\frac{x+2y}{x}={{\left( \frac{{{C}_{1}}}{x} \right)}^{2}}\] \[\Rightarrow \] \[{{x}^{2}}+2xy=C\] (where\[C=C_{1}^{2})\]
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