JEE Main 2021 — Mathematics Calculus

Given $\mathrm{cos}x(3\mathrm{sin}x+\mathrm{cos}x+3)dy=(1+y\mathrm{sin}x(3\mathrm{sin}x+\mathrm{cos}x+3))dx$cosx(3sinx+cosx+3)dy=(1+ysinx(3sinx+cosx+3))dx

$$\Rightarrow \frac{dy}{dx}=\frac{(1+y\mathrm{sin}x(3\mathrm{sin}x+\mathrm{cos}x+3))}{\mathrm{cos}x(3\mathrm{sin}x+\mathrm{cos}x+3)}$$⇒dxdy​=cosx(3sinx+cosx+3)(1+ysinx(3sinx+cosx+3))​

$$\Rightarrow \frac{dy}{dx}=\frac{1}{\mathrm{cos}x(3\mathrm{sin}x+\mathrm{cos}x+3)}+\frac{y\mathrm{sin}x(3\mathrm{sin}x+\mathrm{cos}x+3))}{\mathrm{cos}x(3\mathrm{sin}x+\mathrm{cos}x+3)}$$⇒dxdy​=cosx(3sinx+cosx+3)1​+cosx(3sinx+cosx+3)ysinx(3sinx+cosx+3))​

$$\Rightarrow \frac{dy}{dx}-(\mathrm{tan}x)y=\frac{1}{(3\mathrm{sin}x+\mathrm{cos}x+3)\mathrm{cos}x}$$⇒dxdy​−(tanx)y=(3sinx+cosx+3)cosx1​

This is a linear differential equation of the type $\frac{dy}{dx}+Py=Q,$dxdy​+Py=Q, where $P=-\mathrm{tan}x$P=−tanx and $Q=\frac{1}{\mathrm{cos}x(3\mathrm{sin}x+\mathrm{cos}x+3)}.$Q=cosx(3sinx+cosx+3)1​.

Now, the integrating factor $\displaystyle I.F.={e}^{\int Pdx}={e}^{\int -\mathrm{tanxdx}}$I.F.=e∫Pdx=e∫−tanxdx

$$={e}^{\mathrm{ln}|\mathrm{cosx}|}=|\mathrm{cos}x|$$=eln∣cosx∣=∣cosx∣

$$=\mathrm{cos}x\forall 0\leq x\leq \frac{\pi }{2}.$$=cosx∀0≤x≤2π​.

The solution of the given differential equation is $\displaystyle y(I.F.)=\int Q(I.F.)dx+C$y(I.F.)=∫Q(I.F.)dx+C

$$\Rightarrow y(\mathrm{cos}x)=\int (\mathrm{cos}x)\cdot \frac{1}{\mathrm{cos}x(3\mathrm{sin}x+\mathrm{cos}x+3)}dx+C$$⇒y(cosx)=∫(cosx)⋅cosx(3sinx+cosx+3)1​dx+C

$$\Rightarrow y(\mathrm{cos}x)=\int \frac{dx}{3\mathrm{sin}x+\mathrm{cos}x+3}dx+C$$⇒y(cosx)=∫3sinx+cosx+3dx​dx+C

Using \mathrm{sin}2\theta =\frac{2\mathrm{tan}\theta }{1+{\mathrm{tan}}^{2}\theta }&\mathrm{cos}2\theta =\frac{1-{\mathrm{tan}}^{2}\theta }{1+{\mathrm{tan}}^{2}\theta }

$$\Rightarrow y(\mathrm{cos}x)=\int \frac{1}{3\times \frac{2\mathrm{tan}\frac{x}{2}}{1+{\mathrm{tan}}^{2}\frac{x}{2}}+\frac{1-{\mathrm{tan}}^{2}\frac{x}{2}}{1+{\mathrm{tan}}^{2}\frac{x}{2}}+3}dx+C$$⇒y(cosx)=∫3×1+tan22x​2tan2x​​+1+tan22x​1−tan22x​​+31​dx+C

$$\Rightarrow y(\mathrm{cos}x)=\int \frac{1+{\mathrm{tan}}^{2}\frac{x}{2}}{6\mathrm{tan}\frac{x}{2}+1-{\mathrm{tan}}^{2}\frac{x}{2}+3+3{\mathrm{tan}}^{2}\frac{x}{2}}dx+C$$⇒y(cosx)=∫6tan2x​+1−tan22x​+3+3tan22x​1+tan22x​​dx+C

$$\Rightarrow y(\mathrm{cos}x)=\int \frac{({\mathrm{sec}}^{2}\frac{x}{2})}{2{\mathrm{tan}}^{2}\frac{x}{2}+6\mathrm{tan}\frac{x}{2}+4}dx+C$$⇒y(cosx)=∫2tan22x​+6tan2x​+4(sec22x​)​dx+C

Let $\displaystyle {I}_{1}=\int \frac{({\mathrm{sec}}^{2}\frac{x}{2})}{2({\mathrm{tan}}^{2}\frac{x}{2}+3\mathrm{tan}\frac{x}{2}+2)}dx+C$I1​=∫2(tan22x​+3tan2x​+2)(sec22x​)​dx+C

Put $\mathrm{tan}\frac{x}{2}=t\Rightarrow \frac{1}{2}{\mathrm{sec}}^{2}\frac{x}{2}dx=dt$tan2x​=t⇒21​sec22x​dx=dt

$$\Rightarrow {I}_{1}=\int \frac{dt}{{t}^{3}+3t+2}=\int \frac{dt}{(t+2)(t+1)}$$⇒I1​=∫t3+3t+2dt​=∫(t+2)(t+1)dt​

$$\Rightarrow {I}_{1}=\int (\frac{1}{t+1}-\frac{1}{t+2})dt$$⇒I1​=∫(t+11​−t+21​)dt

$$\Rightarrow {I}_{1}={\mathrm{log}}_{e}(t+1)-{\mathrm{log}}_{e}(t+2)$$⇒I1​=loge​(t+1)−loge​(t+2)

$$\Rightarrow {I}_{1}={\mathrm{log}}_{e}(\frac{t+1}{t+2})$$⇒I1​=loge​(t+2t+1​)

$$\Rightarrow {I}_{1}={\mathrm{log}}_{e}|(\frac{\mathrm{tan}\frac{x}{2}+1}{\mathrm{tan}\frac{x}{2}+2})|$$⇒I1​=loge​∣(tan2x​+2tan2x​+1​)∣

So, the solution is $y(\mathrm{cos}x)={\mathrm{log}}_{e}|\frac{1+\mathrm{tan}\frac{x}{2}}{2+\mathrm{tan}\frac{x}{2}}|+C$y(cosx)=loge​∣2+tan2x​1+tan2x​​∣+C

$\Rightarrow y(\mathrm{cos}x)={\mathrm{log}}_{e}(\frac{1+\mathrm{tan}\frac{x}{2}}{2+\mathrm{tan}\frac{x}{2}})+C$⇒y(cosx)=loge​(2+tan2x​1+tan2x​​)+C for $0\leq x\leq \frac{\pi }{2}.$0≤x≤2π​.

Now, it is given $y(0)=0$y(0)=0

$$\Rightarrow 0={\mathrm{log}}_{e}(\frac{1}{2})+C\Rightarrow C={\mathrm{log}}_{e}2$$⇒0=loge​(21​)+C⇒C=loge​2

$$\Rightarrow y(\mathrm{cos}x)={\mathrm{log}}_{e}(\frac{1+\mathrm{tan}\frac{x}{2}}{2+\mathrm{tan}\frac{x}{2}})+{\mathrm{log}}_{e}2$$⇒y(cosx)=loge​(2+tan2x​1+tan2x​​)+loge​2

Now, for $x=\frac{\pi }{3},$x=3π​, $y(\frac{1}{2})={\mathrm{log}}_{e}(\frac{1+\frac{1}{\sqrt{3}}}{2+\frac{1}{\sqrt{3}}})+{\mathrm{log}}_{e}2$y(21​)=loge​(2+3​1​1+3​1​​)+loge​2

$$\Rightarrow y\cdot (\frac{1}{2})={\mathrm{log}}_{e}(\frac{\sqrt{3}+1}{2\sqrt{3}+1})+{\mathrm{log}}_{e}2$$⇒y⋅(21​)=loge​(23​+13​+1​)+loge​2

$$\Rightarrow y\cdot (\frac{1}{2})={\mathrm{log}}_{e}((\frac{\sqrt{3}+1}{2\sqrt{3}+1})\times (\frac{2\sqrt{3}-1}{2\sqrt{3}-1}))+{\mathrm{log}}_{e}2$$⇒y⋅(21​)=loge​((23​+13​+1​)×(23​−123​−1​))+loge​2

$$\Rightarrow y\cdot (\frac{1}{2})={\mathrm{log}}_{e}(\frac{5+\sqrt{3}}{11})+{\mathrm{log}}_{e}2$$⇒y⋅(21​)=loge​(115+3​​)+loge​2

$$\Rightarrow y=2{\mathrm{log}}_{e}(\frac{2\sqrt{3}+10}{11}).$$⇒y=2loge​(1123​+10​).