A) \[{{x}^{2}}=cy(y+2x)\]
B) \[xy=c(y+2x)\]
C) \[{{x}^{2}}y=c(y+2x)\]
D) None of these
Correct Answer: C
Solution :
| The given differential equation is \[\frac{dy}{dx}=-\frac{y(x+y)}{{{x}^{2}}},\] which is homogeneous |
| Put \[y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}\]and we get \[v+x\frac{dv}{dx}=-\frac{vx(x+vx)}{{{x}^{2}}}=-v(1+v)\] |
| \[\Rightarrow \,\,x\frac{dv}{dx}=-2v-{{v}^{2}}\Rightarrow \frac{dv}{v(v+2)}=-\frac{dx}{x}\] |
| On integrating, we get \[\int{\frac{dv}{v(v+2)}=\int{-\frac{dx}{x}+a,}}\] a is an arbitrary constant |
| \[\Rightarrow \,\frac{1}{2}\int{\left( \frac{1}{v}-\frac{1}{v+2} \right)}dv=-\int{\frac{dx}{x}+a}\] |
| \[\Rightarrow \,\frac{1}{2}\,\,[\ell nv-\ell n(v+a)]=-\ell nx+a\] |
| \[\Rightarrow \,\frac{1}{2}\,\,\ell n\frac{v}{v+2}=-\ell n\,x+a\] |
| \[\Rightarrow \,\,\ell n\sqrt{\frac{y}{y+2x}}+\ell n\,\,\,x=a\] \[\left[ \text{Resubstiuting}\,\,v=\frac{y}{x} \right]\] |
| \[\Rightarrow \,\,\ell n\left[ x\sqrt{\frac{y}{y+2x}} \right]=a\Rightarrow x\sqrt{\frac{y}{y+2x}}={{e}^{a}}=b\] (a new constant) |
| \[\Rightarrow \,\,{{x}^{2}}y=(y+2x),\] where \[c={{b}^{2}}\]another constant |
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