Answer
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Hint: We will solve this question by differentiating the equation with respect to $x$ and keep differentiating, until we get the values of constants A and B. When we get the values we will make some substitutions and this we give us the result.
Complete step-by-step solution -
Here we have,
$y = A{x^2} + Bx.........(1)$
Now, differentiating equation (1) with respect to $x$, we get
$\dfrac{{dy}}{{dx}} = 2Ax + B$ $.........(2)$
Again, differentiating equation (2) with respect to $x$, we get
\[
\dfrac{d}{{dx}}\left( {\dfrac{{dy}}{{dx}}} \right) = \dfrac{d}{{dx}}\left( {2Ax + B} \right) \\
\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = 2A + 0 \\
\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = 2A \\
\Rightarrow A = \dfrac{{\dfrac{{{d^2}y}}{{d{x^2}}}}}{2} \\
\]
Now, substituting the value of A in equation (2), we obtain
$
\dfrac{{dy}}{{dx}} = 2 \cdot \dfrac{{\dfrac{{{d^2}y}}{{d{x^2}}}}}{2} \cdot x + B \\
\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{{d^2}y}}{{d{x^2}}}x + B \\
\Rightarrow B = \dfrac{{dy}}{{dx}} - \dfrac{{{d^2}y}}{{d{x^2}}}x \\
$
Now, substituting the values of both A and B in equation (1), we obtain
$
y = \dfrac{{\dfrac{{{d^2}y}}{{d{x^2}}}}}{2} \cdot {x^2} + \left( {\dfrac{{dy}}{{dx}} - \dfrac{{{d^2}y}}{{d{x^2}}}x} \right)x \\
\Rightarrow 2y = \dfrac{{{d^2}y}}{{d{x^2}}}{x^2} + 2\left( {\dfrac{{dy}}{{dx}} - \dfrac{{{d^2}y}}{{d{x^2}}}x} \right)x \\
\Rightarrow 2y = \dfrac{{{d^2}y}}{{d{x^2}}}{x^2} + 2x\dfrac{{dy}}{{dx}} - 2\dfrac{{{d^2}y}}{{d{x^2}}}{x^2} \\
\Rightarrow 2y = - \dfrac{{{d^2}y}}{{d{x^2}}}{x^2} + 2x\dfrac{{dy}}{{dx}} \\
\Rightarrow {x^2}\dfrac{{{d^2}y}}{{d{x^2}}} - 2x\dfrac{{dy}}{{dx}} + 2y = 0 \\
$
Hence, option C is the correct answer.
Note: A differential equation is an equation that relates one or more functions and their derivatives. Generally, it defines a relationship between the physical quantities and their rates. These questions must be solved with full concentration as the second derivatives might be confusing.
Complete step-by-step solution -
Here we have,
$y = A{x^2} + Bx.........(1)$
Now, differentiating equation (1) with respect to $x$, we get
$\dfrac{{dy}}{{dx}} = 2Ax + B$ $.........(2)$
Again, differentiating equation (2) with respect to $x$, we get
\[
\dfrac{d}{{dx}}\left( {\dfrac{{dy}}{{dx}}} \right) = \dfrac{d}{{dx}}\left( {2Ax + B} \right) \\
\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = 2A + 0 \\
\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = 2A \\
\Rightarrow A = \dfrac{{\dfrac{{{d^2}y}}{{d{x^2}}}}}{2} \\
\]
Now, substituting the value of A in equation (2), we obtain
$
\dfrac{{dy}}{{dx}} = 2 \cdot \dfrac{{\dfrac{{{d^2}y}}{{d{x^2}}}}}{2} \cdot x + B \\
\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{{d^2}y}}{{d{x^2}}}x + B \\
\Rightarrow B = \dfrac{{dy}}{{dx}} - \dfrac{{{d^2}y}}{{d{x^2}}}x \\
$
Now, substituting the values of both A and B in equation (1), we obtain
$
y = \dfrac{{\dfrac{{{d^2}y}}{{d{x^2}}}}}{2} \cdot {x^2} + \left( {\dfrac{{dy}}{{dx}} - \dfrac{{{d^2}y}}{{d{x^2}}}x} \right)x \\
\Rightarrow 2y = \dfrac{{{d^2}y}}{{d{x^2}}}{x^2} + 2\left( {\dfrac{{dy}}{{dx}} - \dfrac{{{d^2}y}}{{d{x^2}}}x} \right)x \\
\Rightarrow 2y = \dfrac{{{d^2}y}}{{d{x^2}}}{x^2} + 2x\dfrac{{dy}}{{dx}} - 2\dfrac{{{d^2}y}}{{d{x^2}}}{x^2} \\
\Rightarrow 2y = - \dfrac{{{d^2}y}}{{d{x^2}}}{x^2} + 2x\dfrac{{dy}}{{dx}} \\
\Rightarrow {x^2}\dfrac{{{d^2}y}}{{d{x^2}}} - 2x\dfrac{{dy}}{{dx}} + 2y = 0 \\
$
Hence, option C is the correct answer.
Note: A differential equation is an equation that relates one or more functions and their derivatives. Generally, it defines a relationship between the physical quantities and their rates. These questions must be solved with full concentration as the second derivatives might be confusing.
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