If for the solution curve y=f(x) of the differential equation (d y)/(d x)+( x) y=(2+ x)/((…

Question

If for the solution curve $y=f(x)$ of the differential equation $\frac{d y}{d x}+(	an x) y=\frac{2+\sec x}{(1+2 \sec x)^2}$, $x \in\left(\frac{-\pi}{2}, \frac{\pi}{2}ight), f\left(\frac{\pi}{3}ight)=\frac{\sqrt{3}}{10}$

Accepted Answer

$\begin{gathered} {e}^{\int 	an xdx}={e}^{\ln (\sec x)}=\sec x \ 	herefore y\cdot \sec x=\int {\frac{2+\sec x}{(1+2\sec x{)}^{2}}}\sec xdx \end{gathered}$

$\begin{gathered} =\int \frac{2\cos x+1}{(\cos x+2{)}^{2}}dx\cos x=\frac{1-{t}^{2}}{1+{t}^{2}} \ =\int \frac{2(\frac{1-{t}^{2}}{1+{t}^{2}})+1}{{(\frac{1-{t}^{2}}{1+{t}^{2}}+2)}^{2}}2dt \ =\int \frac{2-2{t}^{2}+1+{t}^{2}}{{(1-{t}^{2}+2+2{t}^{2})}^{2}}	imes 2dt \ =2\int \frac{3-{t}^{2}}{{({t}^{2}+3)}^{2}}dt \ t+\frac{3}{t}=u \ (1-\frac{3}{{t}^{2}})dt=du \end{gathered}$

$=-2\int \frac{du}{{u}^{2}}$

$\begin{gathered} y\cdot (\sec x)=\frac{2}{u}+c \ y\cdot \sec x=\frac{2}{t+\frac{3}{t}}+c\, \end{gathered}$

$\begin{gathered} x=\frac{\pi }{3},t=	an \frac{x}{2}=\frac{1}{\sqrt{3}} \ \frac{\sqrt{3}}{10}=\frac{2}{\frac{1}{\sqrt{3}}+3\sqrt{3}}+c \ \frac{\sqrt{3}}{10}=\frac{2\sqrt{3}}{10}+c\Rightarrow C=0 \ x=\frac{\pi }{4},t=	an \frac{x}{2}=\sqrt{2}-1 \end{gathered}$

$\begin{gathered} 	herefore y\cdot \sqrt{2}=\frac{2}{\sqrt{2}-1+\frac{3}{\sqrt{2}-1}} \ y\cdot \sqrt{2}=\frac{2(\sqrt{2}-1)}{6-2\sqrt{2}} \ y=\frac{\sqrt{2}(\sqrt{2}-1)}{2(3-\sqrt{2})}=\frac{1}{\sqrt{2}}	imes \frac{2\sqrt{2}-1}{7} \ =\frac{4-\sqrt{2}}{14} \end{gathered}$

Solutions · 1