# Which of the following differential equation has

Question:

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. $\frac{d^{2} y}{d x^{2}}+x \frac{d y}{d x}+x y=0$

Solution:

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.