A) \[\frac{{{d}^{3}}y}{d{{x}^{3}}}=0\]
B) \[{{x}^{2}}\frac{{{d}^{2}}y}{d{{x}^{2}}}-2x\frac{dy}{dx}+2y=0\]
C) \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=0\]
D) \[{{x}^{2}}\frac{{{d}^{2}}y}{d{{x}^{2}}}+y=0\]
E) \[2\frac{{{d}^{2}}y}{d{{x}^{2}}}+\frac{dy}{dx}=0\]
Correct Answer: B
Solution :
\[y=A{{x}^{2}}+Bx\] ?..(i) On differentiating, we get \[\frac{dy}{dx}=2Ax+B\] \[\Rightarrow \] \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=2A\] \[\therefore \] \[\frac{dy}{dx}-\frac{{{d}^{2}}y}{d{{x}^{2}}}x=B\] From (i) \[y=\frac{1}{2}\frac{{{d}^{2}}y}{d{{x}^{2}}}{{x}^{2}}+x\left[ \frac{dy}{dx}-\frac{{{d}^{2}}y}{d{{x}^{2}}}x \right]\] \[\Rightarrow \] \[y=\frac{1}{2}{{x}^{2}}\frac{{{d}^{2}}y}{d{{x}^{2}}}-{{x}^{2}}\frac{{{d}^{2}}y}{d{{x}^{2}}}+x\frac{dy}{dx}\] \[\Rightarrow \] \[y=-\frac{1}{2}{{x}^{2}}\frac{{{d}^{2}}y}{d{{x}^{2}}}+x\frac{dy}{dx}\] \[\Rightarrow \] \[{{x}^{2}}\frac{{{d}^{2}}y}{d{{x}^{2}}}-2x\frac{dy}{dx}+2y=0\]
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