If y = y(x) is the solution of the differential equation, dy/dx + 2ytanx = sinx, y(π/3) = 0, then the maximum value of the function y(x) over R is equal to :

Let $l={\displaystyle \int \frac{2e^x+3e^{-x}}{4e^x+7e^{-x}}}dx={\displaystyle \int \frac{2e^{2x}+3}{4e^{2x}+7}dx}$  

$={\displaystyle \int \frac{2e^{2x}}{4e^{2x}+7}dx+{\displaystyle \int \frac{3e^{-2x}}{4+7e^{-2x}}dx}}$               

Put $4e^{2x}+7=t4+7e^{-2x}=\lambda$  

$8e^{2x}dx=dt-14e^{-2x}dx=d\lambda$              

$=\frac{1}{4}{\displaystyle \int \frac{dt}{t}-\frac{3}{14}{\displaystyle \int \frac{d\lambda }{\lambda }=\frac{1}{4}lnt-\frac{3}{14}ln\lambda +c}}$

$u=\frac{13}{2},v=\frac{1}{2}$              

->so, u + v = 7

$\left(2x-10y^3\right)dy+ydx=0$              

$\frac{dx}{dy}+\frac{2x}{y}-10y^2=0$              

$\frac{dx}{dy}+\frac{2}{y}x=10y^2\to$ Linear differential equation

$P=\frac{2}{y},Q=10y^2$              

$Nowputx=2,y=\beta then2{\beta }^2=2{\beta }^5-2$              

or ${\beta }^5-{\beta }^2-1=0$  

So B will be roots of $y^5-y^2-1=0$  

$\frac{dy}{dx}+\frac{e^y-2x}{2x^2}=0$  

$e^{-y}\frac{dy}{dx}-\frac{e^{-y}}{x}=-\frac{-1}{2x^2}$              

Put $e^{-y}=t\Rightarrow -e^{-y}\frac{dy}{dx}=\frac{dt}{dx}$  

$\frac{dt}{dx}+\frac{t}{x}=\frac{1}{2x^2}$              

$\frac{dt}{dx}+\frac{t}{x}=\frac{1}{2x^2}$

 I. F. $=e^{{\displaystyle \int \frac{dx}{x}}}=e^{lnx}=x$  

Soln. tx = ${\displaystyle \int \frac{x}{2x^2}dx=\frac{1}{2}lnx+c}$  

$x=1\Rightarrow e^{-y}=\frac{1}{2}\Rightarrow y=ln2$