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If x=∫0ydt1+9t2x = \int_0^y \frac{dt}{\sqrt{1+9t^2}} and d2ydx2=λy\frac{d^2y}{dx^2} = \lambda y, then, λ\lambda is equal to

Solution

✅ Correct Option: 3

dxdy=11+9y2\frac{dx}{dy} = \frac{1}{\sqrt{1+9y^2}}, so dydx=1+9y2\frac{dy}{dx} = \sqrt{1+9y^2}.

d2ydx2=18y21+9y2⋅dydx=9y1+9y2⋅1+9y2=9y\frac{d^2y}{dx^2} = \frac{18y}{2\sqrt{1+9y^2}} \cdot \frac{dy}{dx} = \frac{9y}{\sqrt{1+9y^2}} \cdot \sqrt{1+9y^2} = 9y.

Hence λ=9\lambda = 9.

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