CUET Preparation Today
If $y = \sqrt{2024x+2025}$, then which of the following is correct? |
$y\frac{d^2y}{dx^2}-2\frac{dy}{dx}=0$ $y\frac{d^2y}{dx^2}-2(\frac{dy}{dx})^2=0$ $y\frac{d^2y}{dx^2}+x(\frac{dy}{dx})^2=0$ $y\frac{d^2y}{dx^2}+(\frac{dy}{dx})^2=0$ |
$y\frac{d^2y}{dx^2}+(\frac{dy}{dx})^2=0$ |
The correct answer is Option (4) → $y\frac{d^2y}{dx^2}+(\frac{dy}{dx})^2=0$ Given $y=\sqrt{2024x+2025}$ $y=(2024x+2025)^{\frac{1}{2}}$ $\frac{dy}{dx}=\frac{1}{2}(2024x+2025)^{-\frac{1}{2}}\cdot2024$ $\frac{dy}{dx}=\frac{1012}{\sqrt{2024x+2025}}$ $\frac{d^2y}{dx^2}=1012\cdot\left(-\frac{1}{2}\right)(2024x+2025)^{-\frac{3}{2}}\cdot2024$ $\frac{d^2y}{dx^2}=-\frac{1012\cdot2024}{2(2024x+2025)^{\frac{3}{2}}}$ $\frac{d^2y}{dx^2}=-\frac{1012\cdot1012}{(2024x+2025)^{\frac{3}{2}}}$ $y=\sqrt{2024x+2025}$, $\frac{dy}{dx}=\frac{1012}{y}$ $\frac{d^2y}{dx^2}=-\frac{1012^2}{y^3}$ Now compute $y\frac{d^2y}{dx^2}+\left(\frac{dy}{dx}\right)^2$ $=y\left(-\frac{1012^2}{y^3}\right)+\left(\frac{1012}{y}\right)^2$ $=-\frac{1012^2}{y^2}+\frac{1012^2}{y^2}$ $=0$ Hence the correct relation is $y\frac{d^2y}{dx^2}+\left(\frac{dy}{dx}\right)^2=0$ |
