36. Which of the following differential equations has y = x as one of its particular solutions: (A)d2ydx2−x2dydx+xy=x(B)d2ydx2+xdydx+xy=x(C)d2ydx2−x2dydx+xy=0(D)d2ydx2+xdydx+xy=0

The given equation of curve is y = x . Differentiating with respect to x, we get: dydx=1 .......... (1) Again, differentiating with respect to x, we get: d2ydx2=0.......... (2) Now, on substituting the values of y, d2ydx2 and dydx from equation (1) and (2) in each of the given alternatives, we find that only the differential equation given in alternative C is correct d2ydx2−x2dydx+xy=0−x2.1+x.x=−x2+x2=0 Therefore, option (C) is correct.

36. Which of the following differential equations has y = x as one of its particular solutions:

$\begin{array}{l} (A) \frac{d^{2} y}{d x^{2}} - x^{2} \frac{d y}{d x} + x y = x \\ (B) \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} + x y = x \\ (C) \frac{d^{2} y}{d x^{2}} - x^{2} \frac{d y}{d x} + x y = 0 \\ (D) \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} + x y = 0 \end{array}$

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Vishal Baghel | Contributor-Level 10

10 months ago

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 C

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ln|y| = ln (lnx)²) + C.
y = A (lnx)².
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