Question

Find the particular solution of the differential equation $({{\tan }^{-1}}y-x)dy=(1+{{y}^{2}})dx$ given that,

& x=0\,\,,\,\,y=0 \\\ & \\\ \end{aligned}$$

Answer 1

To solve the above question, we have to use the concept of solving linear equations i.e.,
If $\dfrac{dx}{dy}+Px=Q$ is a linear differential equation, where P, Q are the functions of Y the particular then the solution will be $x(I.F.)=\int{Q(I.F.)dy+c}$ where $I.F.={{e}^{\int{p(y)dy}}}$ after putting the values in this formula then we will get the final answer.

Complete step-by-step solution:
So, here have the given equation $({{\tan }^{-1}}y-x)dy=(1+{{y}^{2}})dx$
On rearranging the above equation, we will get,
$\dfrac{dx}{dy}=\dfrac{{{\tan }^{-1}}y-x}{1+{{y}^{2}}}$
After this we have to split the equations into two parts and then transfer it to the other side,
$\dfrac{dx}{dy}+\dfrac{x}{1+{{y}^{2}}}=\dfrac{{{\tan }^{-1}}y}{1+{{y}^{2}}}$
On comparing the above equation with $\dfrac{dx}{dy}+Px=Q$
Here we have the values of P and Q;

& P=\dfrac{1}{1+{{y}^{2}}} \\\ & Q=\dfrac{{{\tan }^{-1}}y}{1+{{y}^{2}}} \\\ \end{aligned}$$ Then, we have the integrating factor (I.F.) = $${{e}^{\int{p(y)dy}}}$$=$${{e}^{\int{\dfrac{1}{1+{{y}^{2}}}dy}}}$$ $$\Rightarrow I.F.={{e}^{{{\tan }^{-1}}y}}$$ Then the solution will be put the values in the below equation then we get, $$x(I.F.)=\int{Q(I.F.)dy}$$ $$\Rightarrow x({{e}^{{{\tan }^{-1}}y}})=\int{\dfrac{{{\tan }^{-1}}y}{1+{{y}^{2}}}}{{e}^{{{\tan }^{-1}}y}}dy$$ For simplifying this let, $$\,t={{\tan }^{-1}}y$$ Then we will get; $$dt=\dfrac{1}{1+{{y}^{2}}}dy$$ $$\therefore x({{e}^{{{\tan }^{-1}}y}})=\int{{{e}^{t}}t dt}$$ $$\Rightarrow x({{e}^{{{\tan }^{-1}}y}})={{e}^{{{\tan }^{-1}}y}}({{\tan }^{-1}}y-1)+C\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ \because \int{{{e}^{x}}x dx={{e}^{x}}(x-1)}+C \right]$$ Substituting the value of t in above equation then we will get; $$\begin{aligned} & \Rightarrow x({{e}^{{{\tan }^{-1}}y}})={{e}^{{{\tan }^{-1}}y}}({{\tan }^{-1}}y-1)+1...........(1) \\\ & 0={{e}^{0}}\left( {{\tan }^{-1}}(0)-1 \right)+C \\\ & \Rightarrow C=1 \\\ & \Rightarrow x{{e}^{{{\tan }^{-1}}y}}-{{e}^{{{\tan }^{-1}}y}}({{\tan }^{-1}}y-1)=1 \\\ & \Rightarrow {{e}^{{{\tan }^{-1}}y}}[x-{{\tan }^{-1}}y+1]=1 \\\ \end{aligned}$$ We have given that y=0 when x=0 $$\Rightarrow y(0)=0$$ Now, we will put the values of x=0 and y=0 in (1) then we get; $$\begin{aligned} & 0={{e}^{0}}\left( {{\tan }^{-1}}(0)-1 \right)+C \\\ & \Rightarrow C=1 \\\ \end{aligned}$$ Putting C= 1 we get; $$x({{e}^{{{\tan }^{-1}}y}})={{e}^{{{\tan }^{-1}}y}}({{\tan }^{-1}}y-1)+1$$ $$\Rightarrow {{e}^{{{\tan }^{-1}}y}}[x-{{\tan }^{-1}}y+1]=1$$ **Hence the particular solution of $$({{\tan }^{-1}}y-x)dy=(1+{{y}^{2}})dx$$ when x=0,y=0 is $${{e}^{{{\tan }^{-1}}y}}[x-{{\tan }^{-1}}y+1]=1$$** **Note:** In this context of finding the particular solution of the given equation we have a lot of methods based on the given question, many of us have doubt that which method is to be taken from a lot of methods we already have, for overcoming this problem we have to analyse all the methods keenly a day practice we can easily solve.