JEE Main 2024 (Online) 4th April Morning ShiftQuestion 1
If the solution $y=y(x)$ of the differential equation $\left(x^{4}+2x^{3}+3x^{2}+2x+2\right)dy-\left(2x^{2}+2x+3\right)dx=0$ satisfies $y(-1)=-\dfrac{\pi}{4}$, then $y(0)$ is equal to :
Differential Equation
Solution
Concept. Factorise the quartic; the fraction splits into two inverse-tangent derivatives. Formulas used. (1) $x^4+2x^3+3x^2+2x+2=(x^2+1)(x^2+2x+2)$(2) $\int\dfrac{dx}{(x+1)^2+1}=\tan^{-1}(x+1)$Step 1 - Split. By (1), $$\frac{2x^2+2x+3}{(x^2+1)(x^2+2x+2)}=\frac{1}{x^2+1}+\frac{1}{x^2+2x+2}.$$ (Check: the two numerators add to $x^2+2x+2+x^2+1=2x^2+2x+3$.) Step 2 - Integrate. By (2), $$y=\tan^{-1}x+\tan^{-1}(x+1)+C.$$ Step 3 - Constant. $y(-1)=-\frac\pi4+0+C=-\frac\pi4$, so $C=0$. Step 4 - At $x=0$. $y(0)=0+\frac\pi4=\frac\pi4$. Answer: (A)