UPSC NDA Math Model Paper 4

Given: $\left(\frac{dy}{dx}\right)^{2}-x\left(\frac{dy}{dx}\right)+y=0$ ......... (1) These types of questions will be solved by option. Checking Option 1: $y = x - 1$ Differentiating with respect to x, we get $\Rightarrow \frac{d y}{d x} = 1 - 0$ $\Rightarrow \frac{dy}{dx} = 1$ Put the value of $\frac{d y}{d x} = 1$ in equation 1st, $1^{2} - (x \times 1) + y = 0$ $\Rightarrow 1 - x + y = 0$ $\Rightarrow 1 - x + x - 1 = 0$ $(\because y = x - 1)$ $\Rightarrow 0 = 0$ Hence y = x - 1 is the solution for the differential equation. Checking Option 2: $4 y = x^{2}$Differentiating with respect to $x$, we get $\Rightarrow 4 \frac{dy}{dx} = 2 x$ $\Rightarrow \frac{d y}{d x} = \frac{x}{2}$ Put the value of $\frac{dy}{dx} = \frac{x}{2}$ in equation $1\text{st}$ $\left(\frac{x}{2}\right)^{2} - x \left(\frac{x}{2}\right) + y = 0$ $\Rightarrow \frac{x^{2}}{4} - \frac{x^{2}}{2} + y = 0$ $\Rightarrow -x^{2} + 4 y = 0$ $\Rightarrow -x^{2} + x^{2} = 0$ $(\because 4y = x^{2})$ $\Rightarrow 0 = 0$ Hence $4y = x^{2}$ is the solution for the differential equation. Checking Option 3: y = x Differentiating with respect to x, we get Put the value of $\frac{dy}{dx} = 1$ in equation $1\text{st}$, $1^{2} - (x \times 1) + y = 0$ $\Rightarrow 1 - x + y = 0$ $\Rightarrow 1 - x + x = 0 \;\;\; (\because y = x)$ $\Rightarrow 1 \neq 0$ Hence y = x is the not a solution for the differential equation. Checking Option 4: y = -x - 1 Differentiating with respect to x, we get $\Rightarrow \frac{dy}{dx} = -1 - 0$ $\Rightarrow \frac{dy}{dx} = -1$ Put the value of $\frac{dy}{dx} = -1$ in equation 1 st, $-1^{2} - (x \times -1) + y = 0$ $\Rightarrow 1 + x + y = 0$ $\Rightarrow 1 + x - x - 1 = 0 \;\;\;\; (\because y = -x - 1)$ $\Rightarrow 0 = 0$ Hence y = -x - 1 is the solution for the differential equation. So, option 3 is correct answer.