Question

The solution of $(1+y^2)dx({\tan }^{-1}y-x)dy$ is

• A. $x{e}^{{\tan }^{-1}y}=({\tan }^{-1}y-1)e^{{\tan }^{-1}y}+c$
• B. $x{e}^{{\tan }^{-1}y}=({\tan }^{-1}y-1)e^{{\tan }^{-1}y}-x+c$
• C. $x=({\tan }^{-1}y-1)e^{{\tan }^{-1}y}-y+c$
• D. $x=({\tan }^{-1}y-1)e^{{\tan }^{-1}y}-x+y+c$

Answer 1

Answer: (A)

$x{e}^{{\tan }^{-1}y}=({\tan }^{-1}y-1)e^{{\tan }^{-1}y}+c$

$(ta{n}^{-1}y-x)dy=(1+y^2)dx$

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

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

$\frac{dx}{dy}+\frac{x}{1+y^2}=\frac{ta{n}^{-1}y}{1+y^2}$

compare it with $\frac{dx}{dy}+px=Q$

$p=\frac{1}{1+y^2}\int fdy$

Integrating factor =e $|Q=\frac{{\tan }^{-1}y}{1+y^2}$

$=\int \frac{1}{1+y^2}dy=ta{n}^{-1}y$

So I.F=$e^{tan-1y}$

standard differential $e{q}^n$

$x.If=\int QIf+c$

$X.e^{ta{n}^{-1}y}=\int \frac{ta{n}^{-1}y}{1+y^2}{e}^{ta{n}^{-1}y}+c$.......(i)

$\int \frac{ta{n}^{-1}y}{1+y^2}{e}^{ta{n}^{-1}y}.dy$

put $ta{n}^{-1}y=d$

$\left(\frac{1}{1+y^2}\right)dy=dt\Rightarrow dy=dt(1+y^2)$

$\int \frac{t}{(1+y^2)}{e}^t.dt(1+y^2)=\int t.e^{t}dt$

$\Rightarrow t\int e^t-\int e^t.1=t{e}^t-e^t=e^t(t-1)$

Put $t=ta{n}^{-1}y$

we get,

$e^{ta{n}^{-1}y}(ta{n}^{-1}y-1)$

Putting in $e{q}^n$ (i)

$x.e^{ta{n}^{-1}y}=e^{ta{n}^{-1y}}(ta{n}^{-1}y-1)+c$

$x=(ta{n}^{-1}y-1)+c{e}^{-ta{n}^{-1}y}$