JEE Main 2021 (Online) 20th July Evening ShiftQuestion 1
Let $y=y(x)$ satisfy the equation $\dfrac{dy}{dx}-|A|=0$, for all $x>0$, where $A=\begin{bmatrix}y&\sin x&1\\0&-1&1\\2&0&\dfrac{1}{x}\end{bmatrix}$. If $y(\pi)=\pi+2$, then the value of $y\left(\dfrac{\pi}{2}\right)$ is :
Differential Equation
Solution
Concept. Expand the determinant; the equation becomes $(xy)'=$ a function of $x$. Formulas used. (1) Expansion along the first row(2) $(xy)'=y+xy'$Step 1 - Determinant. By (1), $$|A|=y\left(-\frac1x-0\right)-\sin x\,(0-2)+1\,(0+2)=-\frac yx+2\sin x+2.$$ Step 2 - The equation. $y'=-\frac yx+2\sin x+2$, so $xy'+y=2x\sin x+2x$. By (2), $$xy=\int(2x\sin x+2x)\,dx=-2x\cos x+2\sin x+x^2+c.$$ Step 3 - Use $y(\pi)=\pi+2$. $\pi^2+2\pi=2\pi+0+\pi^2+c$, so $c=0$. Step 4 - At $x=\frac\pi2$. $$y\left(\frac\pi2\right)=\frac{2+\frac{\pi^2}{4}}{\frac\pi2}=\frac4\pi+\frac\pi2.$$ Answer: (A)