Question

Which of the following differential equations has y=x as one of its particular solution?

• A. $\frac{d^2y}{dx^2}-x^2\frac{dy}{dx}+xy=x$
• B. $\frac{d^2y}{dx^2}+x\frac{dy}{dx}+xy=x$
• C. $\frac{d^2y}{dx^2}-x^2\frac{dy}{dx}+xy=0$
• D. $\frac{d^2y}{dx^2}-x\frac{dy}{dx}+xy=0$

Answer 1

Answer: (C)

$\frac{d^2y}{dx^2}-x^2\frac{dy}{dx}+xy=0$

Answer 2

The correct option is(C): $\frac{d^2y}{dx^2}-x^2\frac{dy}{dx}+xy=0$

The given equation of curve is y=x.

Differentiating with respect to x,we get:

$\frac{dy}{dx}$=1...(1)

Again, differentiating with respect to x,we get:

$\frac{d^2y}{dx^2}=0$...(2)

Now, on substituting the values of y, $\frac{d^2y}{dx^2}$,and $\frac{dy}{dx}$ from equation (1) and (2) in each of

the given alternatives, we find that only the differential equation given in alternative C is correct.

$\frac{d^2y}{dx^2}-x^2\frac{dy}{dx}+xy=0-x^2.1+x.x$

$=-x^2+x^2$

=0

Hence,the correct answer is C.