JEE Main 2025 (Online) 7th April Evening ShiftQuestion 1
Let $y=y(x)$ be the solution of the differential equation $\left(x^{2}+1\right)y'-2xy=\left(x^{4}+2x^{2}+1\right)\cos x$, $y(0)=1$. Then $\int_{-3}^{3}y(x)\,dx$ is :
Differential Equation
Solution
Concept. Divide by $(x^2+1)^2$: the left side is $\left(\frac{y}{x^2+1}\right)'$. Formulas used. (1) $\left(\dfrac{y}{x^2+1}\right)'=\dfrac{(x^2+1)y'-2xy}{(x^2+1)^2}$(2) $\int_{-a}^{a}(\text{odd})\,dx=0$Step 1 - Integrate. $x^4+2x^2+1=(x^2+1)^2$, so by (1) $$\left(\frac{y}{x^2+1}\right)'=\cos x\ \Longrightarrow\ y=(x^2+1)(\sin x+C).$$ Step 2 - Constant. $y(0)=1$ gives $C=1$. Step 3 - The integral. $(x^2+1)\sin x$ is odd, so by (2) $$\int_{-3}^{3}y\,dx=\int_{-3}^{3}(x^2+1)\,dx=2(9+3)=24.$$ Answer: (B)