TS EAMCET 11-Sep-2020 Shift 2 Solved Paper

Given differential equation, $\sin x \;\frac{dy}{dx}+y\cos x=e^{2x}, x\in(0,\pi)$ or $\;\; \;\frac{dy}{dx}+y\cot x=\frac{e^{2x}}{\sin x}$ This is linear differential equation and general form is given by $\;\frac{dy}{dx}+P(x)y=Q(x)$ Here $P(x)=\cot x, Q(x)=\;\frac{e^{2x}}{\sin x}$ Integrating factor, IF $=e^{\int P(x)dx}=e^{\int \cot x dx}=e^{\ln\sin x}$ $IF=\sin x$ Solution of linear differential equation $y(IF)=\int Q(x) IF dx$ $y(\sin x)=\int\;\frac{e^{2x}}{\sin x} \cdot \sin x dx$ $y\sin x=\int e^{2x} dx \Rightarrow y\sin x=\;\frac{e^{2x}}{2}+C \;\; ...(i)$ \{where, $C=$ integrating constant $$< br> Given that,$y(\pi/2)=0$< br> < div class="hscrollenable">$\therefore \;\; 0 \times \sin\frac{\pi}{2}=\frac{e^{2\times \pi/2}}{2}+C \Rightarrow C=-\frac{e^{\pi}}{2}$< br> </div>Putting the value of$c$in Eq. (i), we get < br>$y\sin x=\;\frac{e^{2x}}{2}-\;\frac{e^{\pi}}{2}$< br> < div class="hscrollenable">$\text{At}, x=\;\frac{\pi}{6}, y\left(\;\frac{\pi}{6}\right)\sin\;\frac{\pi}{6}=\;\frac{e^{2\times \pi/6}}{2}-\;\frac{e^{\pi}}{2}$< br>$y\left(\;\frac{\pi}{6}\right) \times \;\frac{1}{2}=\;\frac{1}{2}(e^{\pi/3}-e^{n}) \Rightarrow y\left(\;\frac{\pi}{6}\right)=e^{\pi/3}-e^{\pi}$