QJEEM24Ma1e46: JEE Main 2024 Mathematics - Class 12 JEEM24Ma1e46 Solution: If the solution of the differential equation satisfies , then…

$\begin{aligned}& \left(x^4+2 x^3+3 x^2+2 x+2\right) d y-\left(2 x^2+2 x+3\right) d x=0 \\& \int d y=\int\left(\frac{2 x^2+2 x+3}{x^4+2 x^3+3 x^2+2 x+2} d x\right. \\& \int d y=\int \frac{1}{x^2+1} d x+\int \frac{}{x^2+2 x+2} d x \\& y=\tan ^{-1}(x)+\tan ^{-1}(1+x)+C \\& y(-1)=\tan ^{-1}(-1)+\tan ^{-1}(1-1)+C \\& y(-1)=-\frac{\pi}{4}+C=\left(\frac{-\pi}{4}\right)-\{\text { given }\} \\& \Rightarrow C=0 \\& \text { So, } y(x)=\tan ^{-1}(x)+\tan ^{-1}(1+x) \\& y(0)=\tan ^{-1}(0)+\tan ^{-1}(1+0) \\& y(0)=\frac{\pi}{4}\end{aligned}$