\begin{array}{1 1}(A)\;\frac{d^2y}{dx^2}-x^2\frac{dy}{dx}+xy=x\qquad(B)\;\frac{d^2y}{dx^2}+x\frac{dy}{dx}+xy=x\\(C)\;\frac{d^2y}{dx^2}-x^2\frac{dy}{dx}+xy=0\qquad(D)\;\frac{d^2y}{dx^2}+x\frac{dy}{dx}+xy=0\end{array}

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- Derivative of $y$ is $y'$ and derivative of $y'$ is $y''$

Step 1:

Given : $y = x$

differentiating this we get

$y' = 1$

Again differentiating this we get,

$y'' = 0$

Step 2:

Substituting for $y'$ and $y''$ and $y$ in the given options we see in the option C

LHS is $y'' - x^2 y' + xy$

On substituting for $y''$ and $y$ we get

$0 - x^2.1 + x^2 = 0$

Hence option C satisfies as the solution

Hence option C is the correct answer.

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