A) \[(3,0)\]
B) \[(\sqrt{3,}0)\]
C) \[(-1,\,\,2)\]
D) \[\left( -\sqrt{2},1 \right)\]
Correct Answer: B
Solution :
\[\frac{dy}{dx}=\frac{{{x}^{2}}-2y}{x}\] (Given) |
\[\frac{dy}{dx}+2\frac{y}{x}=x\] |
I.F \[={{e}^{\int{\frac{2}{x}dx}}}={{x}^{2}}\] |
\[\therefore \] \[y.{{x}^{2}}=\int{x.{{x}^{2}}}dx+C\] |
\[y.{{x}^{2}}=\frac{{{x}^{4}}}{y}+C\] |
Hence, b passes through\[(1,-2)\] |
\[\Rightarrow \] \[C=-\frac{9}{4}\] |
\[\therefore \] \[y{{x}^{2}}=\frac{{{x}^{4}}}{4}-\frac{9}{4}\] |
Checking options, only option [B] satisfy |
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