If for the solution curve y=f(x) of the differential equation d y ~d x+(tan x) y= 2+sec x (1+2 sec x)^2, x ∈ ( -π 2, π 2 ), f ( π 3 )= √3 10, then f…

$\displaystyle \begin{aligned} & \text { If } \mathrm{e}^{\int \tan \mathrm{xdx}}=\mathrm{e}^{\ln (\sec \mathrm{x})}=\sec \mathrm{x} \\ & \therefore \mathrm{y} \cdot \sec \mathrm{x}=\int\left\{\frac{2+\sec \mathrm{x}}{(1+2 \sec \mathrm{x})^2}\right\} \sec \mathrm{xdx} \\ & =\int \frac{2 \cos \mathrm{x}+1}{(\cos \mathrm{x}+2)^2} \mathrm{dx} \text { Let } \cos \mathrm{x}=\frac{1-\mathrm{t}^2}{1+\mathrm{t}^2} \\ & =\int \frac{2\left(\frac{1-\mathrm{t}^2}{1+\mathrm{t}^2}\right)+1}{\left(\frac{1-\mathrm{t}^2}{1+\mathrm{t}^2}+2\right)^2} 2 \mathrm{dt}\end{aligned}$​ If e∫tanxdx=eln(secx)=secx∴y⋅secx=∫{(1+2secx)22+secx​}secxdx=∫(cosx+2)22cosx+1​dx Let cosx=1+t21−t2​=∫(1+t21−t2​+2)22(1+t21−t2​)+1​2dt​ $\begin{aligned}
& =\int \frac{2-2 t^2+1+t^2}{\left(1-t^2+2+2 t^2\right)^2} \times 2 \mathrm{dt} \
& =2 \int \frac{3-t^2}{\left(t^2+3\right)^2} \mathrm{dt}
\end{aligned}$Let$Let\mathrm{t}+\frac{3}{\mathrm{t}}=\mathrm{u}$$\begin{aligned}
& \left(1-\frac{3}{\mathrm{t}^2}\right) \mathrm{dt}=\mathrm{du} \
& =-2 \int \frac{\mathrm{du}}{\mathrm{u}^2}
\end{aligned}$$\begin{aligned}
& y \cdot(\sec x)=\frac{2}{u}+c \
& y \cdot \sec x=\frac{2}{t+\frac{3}{t}}+c...(I)
\end{aligned}$$\text { At } \mathrm{x}=\frac{\pi}{3}, \mathrm{t}=\tan \frac{\mathrm{x}}{2}=\frac{1}{\sqrt{3}}$$2 \cdot \frac{\sqrt{3}}{10}=\frac{2}{\frac{1}{\sqrt{3}}+3 \sqrt{3}}+\mathrm{c}$$\begin{aligned} & \text { 2. } \frac{\sqrt{3}}{10}=\frac{2 \sqrt{3}}{10}+\mathrm{c} \Rightarrow \mathrm{C}=0 \ & \text { At } \mathrm{x}=\frac{\pi}{4}, \mathrm{t}=\tan \frac{\mathrm{x}}{2}=\sqrt{2}-1 \ & \therefore \mathrm{y} \cdot \sqrt{2}=\frac{2}{\sqrt{2}-1+\frac{3}{\sqrt{2}-1}} \ & \mathrm{y} \cdot \sqrt{2}=\frac{2(\sqrt{2}-1)}{6-2 \sqrt{2}} \ & \mathrm{y}=\frac{\sqrt{2}(\sqrt{2}-1)}{2(3-\sqrt{2})}=\frac{1}{\sqrt{2}} \times \frac{2 \sqrt{2}-1}{7} \ & =\frac{4-\sqrt{2}}{14}\end{aligned}$