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प्रश्न
If y = a + bx2, a, b arbitrary constants, then
विकल्प
\[\frac{d^2 y}{d x^2} = 2xy\]
\[x\frac{d^2 y}{d x^2} = y_1\]
\[x\frac{d^2 y}{d x^2} - \frac{dy}{dx} + y = 0\]
\[x\frac{d^2 y}{d x^2} = 2 xy\]
MCQ
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उत्तर
(b) \[x\frac{d^2 y}{d x^2} = y_1\]
Here,
\[y = a + b x^2 \]
\[ \Rightarrow y_1 = 2bx\]
\[ \Rightarrow y_2 = 2b\]
\[\text { Multiplying by x on both sides we get,} \]
\[ x y_2 = 2bx = y_1 \]
\[ \Rightarrow x\frac{d^2 y}{d x^2} = y_1\]
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