Which of the following differential equation has $y=x$ as one of its particular solution?
A. $\frac{d^{2} y}{d x^{2}}-x^{2} \frac{d y}{d x}+x y=x$
B. $\frac{d^{2} y}{d x^{2}}+x \frac{d y}{d x}+x y=x$
C. $\frac{d^{2} y}{d x^{2}}-x^{2} \frac{d y}{d x}+x y=0$
D. $\frac{d^{2} y}{d x^{2}}+x \frac{d y}{d x}+x y=0$
The given equation of curve is y = x.
Differentiating with respect to x, we get:
$\frac{d y}{d x}=1$ ...(1)
Again, differentiating with respect to x, we get:
$\frac{d^{2} y}{d x^{2}}=0$ ...(2)
Now, on substituting the values of $y, \frac{d^{2} y}{d x^{2}}$, and $\frac{d y}{d x}$ from equation (1) and (2) in each of the given alternatives, we find that only the differential equation given in alternative $\boldsymbol{C}$ is correct.
$\frac{d^{2} y}{d x^{2}}-x^{2} \frac{d y}{d x}+x y=0-x^{2} \cdot 1+x \cdot x$
$=-x^{2}+x^{2}$
$=0$
Hence, the correct answer is C.