Question

Evaluate: $\frac{dy}{dx}=\frac{2(y+2{)}^2}{(x+y-1{)}^2}$

Answer 1

$\frac{dy}{dx}=\frac{2(y+2{)}^2}{(x+y-1{)}^2}$

$\frac{dy}{dx}=\frac{2(y+2{)}^2}{(x-3+y+2{)}^2}$ ...(1)

Let $y+2=t$ .....$\left(3\right)$

$\Rightarrow dy=dt$

$\Rightarrow x-3=z$ ....$\left(4\right)$

$\Rightarrow dx=dz$

Put in $1$

$\frac{dt}{dz}=\frac{2t^2}{{(z+t)}^2}$

Put $t=vz$ ---$\left(2\right)$

$\Rightarrow \frac{dt}{dz}=v+z\frac{dv}{dz}$

$\Rightarrow v+z\frac{dv}{dz}=\frac{2v^2{z}^2}{{(z+vz)}^2}=\frac{2v^2{z}^2}{z^2{(1+v)}^2}$

$\Rightarrow z\frac{dv}{dz}=\frac{-v(1+v^2)}{{(1+v)}^2}$

$\Rightarrow \int \frac{(1+v^2+2v)}{v(1+v^2)}dv=-\int \frac{dz}{z}$ (Integral both side)

$\Rightarrow \int \frac{dv}{v}+\int \frac{2dv}{1+v^2}=-\int \frac{dz}{z}$

$\Rightarrow \ln v+2{\tan }^{-1}v=-\ln z+\ln c$

$\Rightarrow \ln \frac{vz}{c}=-2{\tan }^{-1}v$

Put $2$ in above equation

$\Rightarrow \ln \frac{tz}{cz}=-2{\tan }^{-1}\frac{t}{z}$

Put $3$ and $4$ in above eqution,

$\Rightarrow \ln c(y+c)=-2{\tan }^{-1}\frac{y+2}{x-3}$

$\Rightarrow e^{-2ta{n}^{-2}\frac{y+2}{x+3}}=c.(y+2)$