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If ey(x+1)=1e^y(x + 1) = 1 and d2ydx2=k(dydx)2\frac{d^2y}{dx^2} = k(\frac{dy}{dx})^2, then k is equal to

Solution

✅ Correct Option: 2

Given ey(x+1)=1e^y(x + 1) = 1

ey=1x+1e^y = \frac{1}{x+1}

Taking natural logarithm on both sides:

y=−ln⁡(x+1)y = -\ln(x+1)


Differentiating with respect to xx:

dydx=−1x+1\frac{dy}{dx} = -\frac{1}{x+1}


Differentiating again with respect to xx:

d2ydx2=1(x+1)2\frac{d^2y}{dx^2} = \frac{1}{(x+1)^2}


Given that d2ydx2=k(dydx)2\frac{d^2y}{dx^2} = k\left(\frac{dy}{dx}\right)^2

1(x+1)2=k(−1x+1)2\frac{1}{(x+1)^2} = k\left(-\frac{1}{x+1}\right)^2

1(x+1)2=k⋅1(x+1)2\frac{1}{(x+1)^2} = k \cdot \frac{1}{(x+1)^2}

Therefore, k=1k = 1

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