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Miscellaneous Exercise 6(I) · Q98

Q.x2+y2=a2x^2+y^2=a^2 is a solution of... (A) d2ydx2+dydx−y=0\dfrac{d^2y}{dx^2}+\dfrac{dy}{dx}-y=0 (B) y=x1+(dydx)2+a2yy=x\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}+a^2y (C) y=xdydx+a1+(dydx)2y=x\dfrac{dy}{dx}+a\sqrt{1+\left(\dfrac{dy}{dx}\right)^2} (D) d2ydx2=(x+1)dydx\dfrac{d^2y}{dx^2}=(x+1)\dfrac{dy}{dx}

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x2+y2=a2x^2+y^2=a^2 gives dydx=−xy\dfrac{dy}{dx}=-\dfrac{x}{y} on differentiating. Testing option (C): xdydx+a1+(dydx)2=x(−xy)+a1+x2y2=−x2y+a⋅x2+y2y=−x2y+a⋅ay=a2−x2y=y2y=yx\dfrac{dy}{dx}+a\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}=x\left(-\dfrac{x}{y}\right)+a\sqrt{1+\dfrac{x^2}{y^2}}=-\dfrac{x^2}{y}+a\cdot\dfrac{\sqrt{x^2+y^2}}{y}=-\dfrac{x^2}{y}+a\cdot\dfrac{a}{y}=\dfrac{a^2-x^2}{y}=\dfrac{y^2}{y}=y, which matches the left side yy exactly.

✓Final answer

Option (C)

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