Question

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

Answer 1

$\text{Given that}(1+y^2)dx=(ta{n}^{-1}y-x)dy\Rightarrow \frac{dx}{dy}=\frac{ta{n}^{-1}y}{1+y^2}-\frac{x}{1+y^2}\Rightarrow \frac{dx}{dy}+\frac{x}{1+y^2}=\frac{ta{n}^{-1}y}{1+y^2}\text{The differential equation is in the form}\frac{dx}{dy}+P\left(y\right)x=Q\left(y\right)\text{where}P\left(y\right)=\frac{1}{1+y^2}\text{and}Q\left(y\right)=\frac{ta{n}^{-1}y}{1+y^2}\text{So, Integration factor, (I.F.)}=e^{\int \frac{1}{1+y^2}dy}=e^{ta{n}^{-1}y}\therefore \text{solution is given as :}x\times e^{ta{n}^{-1}y}=\int e^{ta{n}^{-1}y}\times \frac{ta{n}^{-1}y}{1+y^2}dy+C\text{Put}ta{n}^{-1}y=t\Rightarrow \frac{1}{1+y^2}dy=dt.\therefore x{e}^{ta{n}^{-1}y}=\int e^{t}tdt+C\Rightarrow x{e}^{ta{n}^{-1}y}=t\int e^{t}dt-\int \left(\frac{d}{dt}\right(t)\int e^{t}dt)+C\Rightarrow x{e}^{ta{n}^{-1}y}=t{e}^t-e^t+C\Rightarrow x{e}^{ta{n}^{-1}y}=e^{ta{n}^{-1}y}(ta{n}^{-1}y-1)+Cor,x=ta{n}^{-1}y-1+C{e}^{-ta{n}^{-1}y}\text{is the required solution}$