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NCERT Exemplar · Q71

Q.The differential equation ydydx+x=cy\frac{dy}{dx}+x=c represents:
(A) Family of hyperbolas
(B) Family of parabolas
(C) Family of ellipses
(D) Family of circles

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The given differential equation ydydx+x=cy\frac{dy}{dx}+x=c can be rewritten as y dy+x dx=c dxy\,dy + x\,dx = c\,dx. Integrating gives x2+y2=2cx+kx^2 + y^2 = 2cx + k, which is the equation of a family of circles. The correct option is (D).

We are asked: what family of curves does ydydx+x=cy\frac{dy}{dx}+x=c represent? The key is to recognise that this is a first-order differential equation that can be solved by separating variables — but more importantly, the structure hints at a relation between xx and yy that is symmetric and quadratic.

Let’s rewrite it cleanly:

ydydx+x=cy\frac{dy}{dx} + x = c

Bring xx to the other side:

ydydx=c−xy\frac{dy}{dx} = c - x

Now multiply both sides by dxdx (treating dy/dxdy/dx as a ratio, which is valid here):

y dy=(c−x) dxy\,dy = (c - x)\,dx

This is a separable differential equation. We can integrate both sides directly.

Tip

The form y dy+x dx=c dxy\,dy + x\,dx = c\,dx is a dead giveaway that after integration we get x2+y2x^2 + y^2 terms — the hallmark of a circle.

Step 1: Integrate both sides

∫y dy=∫(c−x) dx\int y\,dy = \int (c - x)\,dx

y22=cx−x22+C\frac{y^2}{2} = cx - \frac{x^2}{2} + C

where CC is the constant of integration.

Step 2: Rearrange into a recognisable form

Multiply through by 2:

y2=2cx−x2+2Cy^2 = 2cx - x^2 + 2C

Bring all terms to one side:

x2+y2−2cx=2Cx^2 + y^2 - 2cx = 2C

Step 3: Complete the square in xx

We have x2−2cxx^2 - 2cx. Add and subtract c2c^2:

(x2−2cx+c2)+y2=2C+c2(x^2 - 2cx + c^2) + y^2 = 2C + c^2

(x−c)2+y2=c2+2C(x - c)^2 + y^2 = c^2 + 2C

Let R2=c2+2CR^2 = c^2 + 2C (which is a constant, since cc is fixed and CC is arbitrary). Then:

(x−c)2+y2=R2(x - c)^2 + y^2 = R^2 …

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