Question

Solve: $\frac{dy}{dx}+2y\tan x=\sin x,$ given that $y=0$, when $x=\frac{\pi }{3}$. Show that maximum value of $y$ is $\frac{1}{8}$.

Answer 1

$\frac{dy}{dx}=2y\tan x=\sin x$

The given differential equation is of the form

$\frac{dy}{dx}+Py=Q$

Comparing the two equations,

$P=2y\tan x\&Q=\sin x$

Now, we will find the integrating factor

If=$e^{\int Pdx}=e^{\int 2\tan xdx}=e^{2\int \tan xdx}=e^{2\left[logsecx\right]}=e^{log\left({sec}^{2}x\right)}={sec}^{2}x$

$\therefore$ Solution to the differential equation

$\Rightarrow y\left(IF\right)=\int (Q\times IF)dx+C\Rightarrow y{sec}^{2}x=\int \sin x{sec}^{2}xdx+C\Rightarrow y{sec}^{2}x=\int \frac{sinx}{{cos}^{2}x}dx+C=\int \frac{sinx}{cosx}\times \frac{1}{cosx}dx+C\Rightarrow y{sec}^{2}x=\int \tan x\sec xdx+C\Rightarrow y{sec}^{2}x=\sec x+C\Rightarrow \Rightarrow y=\frac{1}{secx}+\frac{C}{{sec}^{2}x}=cosx+\left({cos}^{2}x\right)Cwhenx=\frac{\pi }{3},y=0\Rightarrow y=0=cos\left(\frac{\pi }{3}\right)+{cos}^2\left(\frac{\pi }{3}\right)C\Rightarrow \frac{1}{2}+\left(\frac{1}{4}\right)C=0\Rightarrow C=-2\therefore y=cosx-2{cos}^{2}x\frac{y}{2}=-{cos}^{2}x+\frac{1}{2}cosx=-({cos}^{2}x-\frac{1}{2}cosx)=-[{(cosx-\frac{1}{4})}^2\cdot {\left(\frac{1}{4}\right)}^2]\Rightarrow \frac{y}{2}=-[{(cosx-\frac{1}{4})}^2\cdot {\left(\frac{1}{4}\right)}^2]=-{(cosx-\frac{1}{4})}^2+\frac{1}{16}$

${(cosx-\frac{1}{4})}^2$ will always be positive & therefore $-{(cosx-\frac{1}{4})}^2\le 0$

Hence $\frac{y}{2}$ will be max. value of ${\left(\frac{1}{4}\right)}^2=\frac{1}{16}$

So the max. value of $y=\frac{1}{16}\times 2=\frac{1}{8}$