Question

Find the particular solution of the differential equation \[({{\tan }^{-1}}y-x)dy=(1+{{y}^{2}})dx\] given that,
\[\begin{align}
  & x=0\,\,,\,\,y=0 \\
 & \\
\end{align}\]

Answer 1

Hint: 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;
\[\begin{align}
  & P=\dfrac{1}{1+{{y}^{2}}} \\
 & Q=\dfrac{{{\tan }^{-1}}y}{1+{{y}^{2}}} \\
\end{align}\]
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;
\[\begin{align}
  & \Rightarrow x({{e}^{{{\tan }^{-1}}y}})={{e}^{{{\tan }^{-1}}y}}({{\tan }^{-1}}y-1)+1...........(1) \\
 & 0={{e}^{0}}\left( {{\tan }^{-1}}(0)-1 \right)+C \\
 & \Rightarrow C=1 \\
 & \Rightarrow x{{e}^{{{\tan }^{-1}}y}}-{{e}^{{{\tan }^{-1}}y}}({{\tan }^{-1}}y-1)=1 \\
 & \Rightarrow {{e}^{{{\tan }^{-1}}y}}[x-{{\tan }^{-1}}y+1]=1 \\
\end{align}\]
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;
\[\begin{align}
  & 0={{e}^{0}}\left( {{\tan }^{-1}}(0)-1 \right)+C \\
 & \Rightarrow C=1 \\
\end{align}\]
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.