JEE Main 2024 — Mathematics Calculus

$$\begin{aligned} & \frac{d y}{d x}+\frac{y}{1+x^2}=\frac{e^{\tan ^{-1} x}}{1+x^2} \\ & \text { I.F. }=e^{\int \frac{1}{1+x^2} d x}=e^{\tan ^{-1} x} \\ & y \cdot e^{\tan ^{-1} x}=\int\left(\frac{e^{\tan ^{-1} x}}{1+x^2}\right) e^{\tan ^{-1} x} \cdot d x \\ & \text { Let } \tan ^{-1} x=z \quad \therefore \frac{d x}{1+x^2}=d z \\ & \therefore y \cdot e^z=\int e^{2 z} d z=\frac{e^{2 z}}{2}+C \\ & y \cdot e^{\tan ^{-1} x}=\frac{e^{2 \tan ^{-1} x}}{2}+C \\ & \Rightarrow y=\frac{e^{\tan ^{-1} x}}{2}+\frac{C}{e^{\tan ^{-1} x}}\end{aligned}$$​dxdy​+1+x2y​=1+x2etan−1x​ I.F. =e∫1+x21​dx=etan−1xy⋅etan−1x=∫(1+x2etan−1x​)etan−1x⋅dx Let tan−1x=z∴1+x2dx​=dz∴y⋅ez=∫e2zdz=2e2z​+Cy⋅etan−1x=2e2tan−1x​+C⇒y=2etan−1x​+etan−1xC​​

$$\begin{aligned} & \because \mathrm{y}(1)=0 \Rightarrow 0=\frac{\mathrm{e}^{\pi / 4}}{2}+\frac{\mathrm{C}}{\mathrm{e}^{\pi / 4}} \Rightarrow \mathrm{C}=\frac{-\mathrm{e}^{\pi / 2}}{2} \\ & \therefore \mathrm{y}=\frac{\mathrm{e}^{\tan ^{-1} \mathrm{x}}}{2}-\frac{\mathrm{e}^{\pi / 2}}{2 \mathrm{e}^{\tan ^{-1} \mathrm{x}}} \\ & \therefore \mathrm{y}(0)=\frac{1-\mathrm{e}^{\pi / 2}}{2}\end{aligned}$$​∵y(1)=0⇒0=2eπ/4​+eπ/4C​⇒C=2−eπ/2​∴y=2etan−1x​−2etan−1xeπ/2​∴y(0)=21−eπ/2​​