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NCERT Exemplar · Q31

Q.Find the equation of a curve passing through origin if the slope of the tangent to the curve at any point (x, y)(x,\,y) is equal to the square of the difference of the abscissa and ordinate of the point.

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Modelling gives dydx=(x−y)2\frac{dy}{dx}=(x-y)^2; with v=x−yv=x-y it separates to 1+v1−v=e2x\frac{1+v}{1-v}=e^{2x}, so y=x−e2x−1e2x+1=x−tanh⁡xy=x-\frac{e^{2x}-1}{e^{2x}+1}=x-\tanh x.

Translate the words

'Abscissa' is xx, 'ordinate' is yy, and the tangent slope is dydx\frac{dy}{dx}. 'Slope equals the square of their difference' means

dydx=(x−y)2.\frac{dy}{dx}=(x-y)^2.

A substitution to separate it

The right side depends only on x−yx-y, so let v=x−yv=x-y. Then dvdx=1−dydx\frac{dv}{dx}=1-\frac{dy}{dx}, and the equation becomes

1−dvdx=v2⇒dvdx=1−v2,1-\frac{dv}{dx}=v^2\quad\Rightarrow\quad\frac{dv}{dx}=1-v^2,

which is separable.

Integrate

∫dv1−v2=∫dx.\int\frac{dv}{1-v^2}=\int dx.

With 11−v2=12(11−v+11+v)\dfrac{1}{1-v^2}=\dfrac12\Big(\dfrac{1}{1-v}+\dfrac{1}{1+v}\Big),

12log⁡∣1+v1−v∣=x+C.\frac12\log\left|\frac{1+v}{1-v}\right|=x+C.

Use the origin

At (0,0)(0,0), v=0v=0, so 12log⁡1=0+C⇒C=0\frac12\log 1=0+C\Rightarrow C=0. Then …

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