Question

Solve the following: $ \left( 1+{{y}^{2}} \right)dx=\left( {{\tan }^{-1}}y-x \right)dy $ .

Answer 1

Hint: We are asked to find the solution of $ \left( 1+{{y}^{2}} \right)dx=\left( {{\tan }^{-1}}y-x \right)dy $ . We use the knowledge of linear differential equations to solve this; we simplify our equation into linear form $ \dfrac{dx}{dy}+P\left( y \right)x=\theta \left( y \right) $ . Once we have linear form then find the integrating factor using $ I.F={{e}^{\int{P\left( y \right)dy}}} $ . Once we get the integrity factor we will go for finding a solution, for which we use $ x.IF=\int{Q\left( y \right)}.I.F+C $ . We simplify and get our required solution.

Complete step by step solution:
We are given $ \left( 1+{{y}^{2}} \right)dx=\left( {{\tan }^{-1}}y-x \right)dy $ $ \left( 1+{{y}^{2}} \right)dx=\left( {{\tan }^{-1}}y-x \right)dy $
We have to solve this and find the solution to this equation.
To solve this we need the knowledge of linear differential equations.
We know equation $ \dfrac{dx}{dy}+P\left( y \right)x=\theta \left( y \right) $ $ \dfrac{dx}{dy}+P\left( y \right)x=\theta \left( y \right) $
Such forms are called a linear differential equation.
Solution of this form is given as - $ x.IF=\int{Q\left( y \right)}.I.F+C $ $ x.IF=\int{Q\left( y \right)}.I.F+C $
Where,
I.F is called integrating factor and is given as $ I.F={{e}^{\int{P\left( y \right)dy}}} $
We are given our equation as -
 $ \left( 1+{{y}^{2}} \right)dx=\left( {{\tan }^{-1}}y-x \right)dy $
Simplifying, we get
 $ \dfrac{dx}{dy}=\dfrac{{{\tan }^{-1}}\theta }{1+{{y}^{2}}}-\dfrac{x}{1+{{y}^{2}}} $
Now, simplifying further, we get
 $ \dfrac{dx}{dy}+\dfrac{x}{1+{{y}^{2}}}=\dfrac{{{\tan }^{-1}}y}{1+{{y}^{2}}} $
We see closely that this statement is the same as a linear differential equation.
Hence, we have $ P\left( y \right)=\dfrac{1}{1+{{y}^{2}}} $ and we have
 $ Q\left( y \right)=\dfrac{{{\tan }^{-1}}y}{1+{{y}^{2}}} $
To find the solution to our problem, we will find an integrating factor.
As we know $ I.F={{e}^{\int{P\left( y \right)dy}}} $
We have the function $ P\left( y \right)=\dfrac{1}{1+{{y}^{2}}} $ , so we can write $ I.F={{e}^{\int{\dfrac{1}{1+{{y}^{2}}}dy}}} $ .
As we know that $ \int{\dfrac{1}{1+{{y}^{2}}}dy}={{\tan }^{-1}}y $ , so we get $ I.F={{e}^{\int{\dfrac{1}{1+{{y}^{2}}}dy}}}={{e}^{{{\tan }^{-1}}y}} $
Now, we know the solution is given as –
 $ x.I.F=\int{Q\left( y \right).I.F+C} $
Substituting $ I.F={{e}^{{{\tan }^{-1}}y}} $ , we get
 $ x.{{e}^{{{\tan }^{-1}}y}}=\int{\dfrac{{{\tan }^{-1}}y{{e}^{{{\tan }^{-1}}y}}}{1+{{y}^{2}}}+C} $ …………………………….. (i)
Now we have to solve $ \int{\dfrac{{{\tan }^{-1}}y{{e}^{{{\tan }^{-1}}y}}}{1+{{y}^{2}}}dy} $
We get $ {{\tan }^{-1}}y=t $
So, differentiating both sides we get
 $ \dfrac{1}{1+{{y}^{2}}}dy=at $
So, our integral become
 $ \int{\dfrac{{{\tan }^{-1}}y{{e}^{{{\tan }^{-1}}y}}}{1+{{y}^{2}}}dy=\int{t{{e}^{t}}at}} $
Now we use the ILATE rule. Here we have algebraic and exponential function, so as per ILATE we will choose ‘t’ as first function and $ {{e}^{t}} $ as second function. So, we get-
 $ \int{\dfrac{{{\tan }^{-1}}y{{e}^{{{\tan }^{-1}}y}}}{1+{{y}^{2}}}dy=\int{t{{e}^{t}}at}} $
Solving, we get
 $ =t{{e}^{t}}-\int{i.{{e}^{t}}at} $
Simplifying, we get –
 $ \begin{align}
  & =t{{e}^{t}}-{{e}^{t}} \\
 & =\left( t-1 \right){{e}^{t}} \\
\end{align} $
So, we get –
\[\int{\dfrac{{{\tan }^{-1}}y{{e}^{{{\tan }^{-1}}y}}}{1+{{y}^{2}}}dy=\left( {{\tan }^{-1}}y-1 \right){{e}^{{{\tan }^{-1}}y}}}\]
Using this in equation (i), we get –
 $ x.{{e}^{{{\tan }^{-1}}y}}=\left( {{\tan }^{-1}}y-1 \right){{e}^{{{\tan }^{-1}}y}}+C $
Dividing by $ {{e}^{{{\tan }^{-1}}y}} $ , we get –
 $ x=\left( {{\tan }^{-1}}y-1 \right)+\dfrac{C}{{{e}^{{{\tan }^{-1}}y}}} $
So, solution is $ x=\left( {{\tan }^{-1}}y-1 \right)+C.{{e}^{{{\tan }^{-1}}y}} $ .

Note:
 While solving these integrals, we need to be very clear about choosing the first function and the second function. We need to be more clear on choosing the term $ \dfrac{dx}{dy} $ or $ \dfrac{dy}{dx} $ to form a linear equation. When we substitute our function as ‘t’. So, we have to re-substitute the original value at the last.