Let y=y(x) be the solution of the differential equation dy dx= ( tanx)+y sinx( secx- sinx tanx),x∈ (0, π 2) satisfying the condition y( π 4)=2. Then,…

Given: $\frac{dy}{dx}=\frac{\mathrm{tan}x+y}{\mathrm{sin}x(\mathrm{sec}x-\mathrm{sin}x\mathrm{tan}x)}$dxdy​=sinx(secx−sinxtanx)tanx+y​

$$\Rightarrow \frac{dy}{dx}=\frac{\mathrm{tan}x+y}{\mathrm{sin}x(\frac{1}{\mathrm{cos}x}-\frac{{\mathrm{sin}}^{2}x}{\mathrm{cos}x})}$$⇒dxdy​=sinx(cosx1​−cosxsin2x​)tanx+y​

$$\Rightarrow \frac{dy}{dx}=\frac{\mathrm{tan}x+y}{\mathrm{sin}x(\frac{{\mathrm{cos}}^{2}x}{\mathrm{cos}x})}$$⇒dxdy​=sinx(cosxcos2x​)tanx+y​

$$\Rightarrow \frac{dy}{dx}=\frac{\mathrm{tan}x+y}{\mathrm{sin}x\mathrm{cos}x}$$⇒dxdy​=sinxcosxtanx+y​

$$\Rightarrow \frac{dy}{dx}-\frac{2y}{\mathrm{sin}2x}={\mathrm{sec}}^{2}x$$⇒dxdy​−sin2x2y​=sec2x

$$\Rightarrow \mathrm{IF}={e}^{-2\int \mathrm{cosec}2xdx}$$⇒IF=e−2∫cosec2xdx

$$\Rightarrow \mathrm{IF}={e}^{-\mathrm{log}|\mathrm{cosec}2x-\mathrm{cot}2x|}$$⇒IF=e−log∣cosec2x−cot2x∣

$$\Rightarrow \mathrm{IF}=\frac{1}{\mathrm{cosec}2x-\mathrm{cot}2x}=\mathrm{cosec}2x+\mathrm{cot}2x$$⇒IF=cosec2x−cot2x1​=cosec2x+cot2x

$$\Rightarrow \mathrm{IF}=\frac{1}{\mathrm{sin}2x}+\frac{\mathrm{cos}2x}{\mathrm{sin}2x}$$⇒IF=sin2x1​+sin2xcos2x​

$$\Rightarrow \mathrm{IF}=\frac{2{\mathrm{cos}}^{2}x}{2\mathrm{sin}x\mathrm{cos}x}$$⇒IF=2sinxcosx2cos2x​

$$\Rightarrow \mathrm{IF}=\mathrm{cot}x$$⇒IF=cotx

$$\Rightarrow y\times \mathrm{cot}x=\int \mathrm{cot}x\times {\mathrm{sec}}^{2}xdx$$⇒y×cotx=∫cotx×sec2xdx

$$\Rightarrow y\times \mathrm{cot}x=2\int \mathrm{cosec}2xdx$$⇒y×cotx=2∫cosec2xdx

$$\Rightarrow y\mathrm{cot}x=\mathrm{log}|\mathrm{cosec}2x-\mathrm{cot}2x|+c$$⇒ycotx=log∣cosec2x−cot2x∣+c

Now, using $y(\frac{\pi }{4})=2$y(4π​)=2 we get,

$$\Rightarrow 2\mathrm{cot}\frac{\pi }{4}=\mathrm{log}|\mathrm{cosec}\frac{\pi }{2}-\mathrm{cot}\frac{\pi }{2}|+c$$⇒2cot4π​=log∣cosec2π​−cot2π​∣+c

$$\Rightarrow c=2$$⇒c=2

$$\Rightarrow y\mathrm{cot}x=\mathrm{log}|\mathrm{cosec}2x-\mathrm{cot}2x|+2$$⇒ycotx=log∣cosec2x−cot2x∣+2

Now, finding $y(\frac{\pi }{3})$y(3π​) we get,

$$\Rightarrow y\mathrm{cot}\frac{\pi }{3}=\mathrm{log}|\mathrm{cosec}\frac{2\pi }{3}-\mathrm{cot}\frac{2\pi }{3}|+2$$⇒ycot3π​=log∣cosec32π​−cot32π​∣+2

$$\Rightarrow y\times \frac{1}{\sqrt{3}}=\mathrm{log}|\frac{2}{\sqrt{3}}+\frac{1}{\sqrt{3}}|+2$$⇒y×3​1​=log∣3​2​+3​1​∣+2

$$\Rightarrow y\times \frac{1}{\sqrt{3}}=\frac{1}{2}\mathrm{log}3+2$$⇒y×3​1​=21​log3+2

$$\Rightarrow y=\frac{\sqrt{3}}{2}\mathrm{log}3+2\sqrt{3}$$⇒y=23​​log3+23​

$$\Rightarrow y=\sqrt{3}(\mathrm{log}\sqrt{3}+2)$$⇒y=3​(log3​+2)