Question

Find the particular solution of the differential equation $(ta{n}^{-1}y-x)dy=(1+y^2)dx$, given that when $x=0,y=0$.

Answer 1

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

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

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

The above equation is in the form of $\frac{dy}{dx}+P\left(y\right)x+Q\left(y\right)$

Here, $P\left(y\right)=\frac{1}{1+y^2},Q\left(y\right)=\frac{ta{n}^{-1}y}{1+y^2}$

Integrating factor = $e^{\frac{1}{1+y^2}}=e^{ta{n}^{-1}y}$

So, $x{e}^{ta{n}^{-1}y}=\int e^{ta{n}^{-1}y}\left(\frac{ta{n}^{-1}y-x}{1+y^2}\right)dy$

Put $ta{n}^{-1}y=t=\frac{dy}{1+y^2}=dt$

$x{e}^{ta{n}^{-1}y}=\int e^{t}tdt\Rightarrow x{e}^{ta{n}^{-1}y}=t\int e^{t}dt-\int (\frac{d\left(t\right)}{dt}\int e^{t}dt)dt$

$x{e}^{ta{n}^{-1}y}=t{e}^t-e^t+C\Rightarrow x{e}^{ta{n}^{-1}y}=(ta{n}^{-1}y-1)e^{ta{n}^{-1}}y+C$

Given that $x=0,y=0,$ we get

$0.e^{ta{n}^{-1}0}=(ta{n}^{-1}0-1)e^{ta{n}^{-1}}0+C$

$C=1$

So, the required solution is: $x=ta{n}^{-1}y+e^{-ta{n}^{-1}y}-1$