If   $\frac{dy}{dx}+\frac{2^{x-y}\left(2^y-1\right)}{2^x-1}=0,x,y>0,y\left(1\right)=1,$ then y(2) is equal to:

I = ∫ (e? (x²+1)/ (x+1)² dx = f (x)e? + c
I = ∫ (e? (x²-1+1+1)/ (x+1)² dx
I = ∫e? [ (x-1)/ (x+1) + 2/ (x+1)² ] dx
for x = 1
f' (1) = 12/24 - 12/16 = 3/4

$L.R.=\frac{\left|3\times 2+4x-3-5\right|}{\sqrt[]{3^2+4^2}}=\frac{11}{5}$              

Equation of family of parabolas

${\left(x-h\right)}^2=\frac{11}{5}\left(y-k\right)$              

Differentiate 2 (x – h) = $\frac{11}{5}\frac{dy}{dx}$  

Again differentiate 2 = $\frac{11}{5}\frac{d^{2}y}{d{x}^2}$  

$11\frac{d^{2}y}{d{x}^2}=10$              

$2x^2\frac{dy}{dx}-2xy+3y^2=0$

$\frac{dy}{dx}=\frac{y}{x}-\frac{3}{2}{\left(\frac{y}{x}\right)}^2$

Put y = vx

  $\frac{dy}{dx}=v+x\frac{dv}{dx}$

$Point\left(e,\frac{e}{3}\right)$

$\Rightarrow \frac{-e}{\left(\frac{e}{3}\right)}+\frac{3}{2}\mathcal{l}ne=C$

$C=\frac{3}{2}-3=\frac{-3}{2}$

$\frac{3}{2}\mathcal{l}n\left|x\right|\frac{-x}{y}+\frac{3}{2}=0$

$x=1\Rightarrow y=\frac{2}{3}$              

$b_n={\displaystyle \underset{0}{\overset{\pi /2}{\int }}\frac{co{s}^{2}nx}{sinx}dx}$

$={\displaystyle \underset{0}{\overset{\pi /2}{\int }}\frac{1+cos2nx}{2sinx}dx}$

$b_n-b_{n-1}={\displaystyle \underset{0}{\overset{\pi /2}{\int }}\frac{cos\left(2nx\right)-cos\left(2\left(n-1\right)x\right)}{2sinx}}dx$

$={\frac{cos\left(2n-1\right)x}{\left(2n-1\right)}]}_0^{\pi /2}=\frac{1}{2n-1}\left[0-1\right]=\frac{-1}{2n-1}$

$b_{n-1}-b_n=\frac{1}{2n-1}$

$b_1-b_2=\frac{1}{3}\left(A\right)-\frac{1}{5},......$

$b_2-b_3=\frac{1}{3}\left(B\right)-5,-7,-9,.....$

$b_3-b_4=\frac{1}{7}\left(C\right)$

$\frac{dy}{dx}-y=x,$  linear differential equation.

IF = e^-x

$y.e^{-x}={\displaystyle \int x{e}^{-x}dx=-e^{-x}\left(x+1\right)+C}$                

$y=-\left(x+1\right)+C.e^x$                

$y_2\left(x\right)=-\left(x+1\right)+2e^x$                

y₂ > y₁, no solution.