Question 19 of 96intermediate🔧 ApplyShort Answer2 marks

In each of the Exercises 1 to 10 verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation: xy=log⁡y+Cxy = \log y + C : y′=y21−xy(xy≠1)y' = \frac{y^2}{1-xy} (xy \neq 1)

Correct Answer

Given function: xy=log⁡y+Cxy = \log y + C

Differentiating both sides with respect to xx:

Left side: Using product rule,

ddx(xy)=x⋅dydx+y⋅1=xy′+y\frac{d}{dx}(xy) = x \cdot \frac{dy}{dx} + y \cdot 1 = xy' + y

Right side: Using chain rule,

ddx(log⁡y+C)=1y⋅dydx=y′y\frac{d}{dx}(\log y + C) = \frac{1}{y} \cdot \frac{dy}{dx} = \frac{y'}{y}

Equating both sides:

xy′+y=y′yxy' + y = \frac{y'}{y}

Multiplying by yy:

xy⋅y′+y2=y′xy \cdot y' + y^2 = y'

Rearranging:

y2=y′−xy⋅y′=y′(1−xy)y^2 = y' - xy \cdot y' = y'(1 - xy)

Therefore, y′=y21−xyy' = \frac{y^2}{1 - xy}

This matches the given differential equation. Hence, xy=log⁡y+Cxy = \log y + C is a solution of the differential equation y′=y21−xyy' = \frac{y^2}{1-xy} (where xy≠1xy \neq 1).

Exercise: EXERCISE 9.2 | Q: 7 | (Page 7)
For More Understanding

Explanation

This is Exercise 9.2, Question 7 from NCERT Class 12 Mathematics. The verification requires implicit differentiation of the given function. The key steps involve applying the product rule on the left side (xy) and the chain rule on the right side (log y). After differentiation, algebraic manipulation yields the required differential equation, confirming the solution.

Solution Steps

  1. Step 1: Differentiate the given function xy=log⁡y+Cxy = \log y + C implicitly with respect to xx.

  2. Step 2: Apply product rule on LHS: ddx(xy)=xy′+y\frac{d}{dx}(xy) = xy' + y

  3. Step 3: Apply chain rule on RHS: ddx(log⁡y)=y′y\frac{d}{dx}(\log y) = \frac{y'}{y}

  4. Step 4: Equate and rearrange to get y′=y21−xyy' = \frac{y^2}{1-xy}

  5. Step 5: Conclude that the function satisfies the differential equation.

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