Question

If y = a + bx², a, b arbitrary constants, then

 

पर्याय

• \[\frac{d^2 y}{d x^2} = 2xy\] 

• \[x\frac{d^2 y}{d x^2} = y_1\]

• \[x\frac{d^2 y}{d x^2} - \frac{dy}{dx} + y = 0\]

• \[x\frac{d^2 y}{d x^2} = 2 xy\]

Answer 1

(b)  \[x\frac{d^2 y}{d x^2} = y_1\]

Here,

\[y = a + b x^2 \]

\[ \Rightarrow y_1 = 2bx\]

\[ \Rightarrow y_2 = 2b\]

\[\text { Multiplying by x on both sides we get,} \]

\[ x y_2 = 2bx = y_1 \]

\[ \Rightarrow x\frac{d^2 y}{d x^2} = y_1\]