Question

The general solution of the differential equation \[\left( 1+{{y}^{2}} \right)dx+\left( 1+{{x}^{2}} \right)dy=0\] is
a) \[x-y=C\left( 1-xy \right)\]
b) \[x-y=C\left( 1+xy \right)\]
c) \[x+y=C\left( 1-xy \right)\]
d) \[x+y=C\left( 1+xy \right)\]

Answer 1

Hint: Here we need to find the general solution of the given differential equation by interchanging the terms. We know that the integration formula, i.e., \[\int{{{\tan }^{-1}}\left( x \right)}.dx=\dfrac{1}{1+{{x}^{2}}}+c\], where c is the integration constant. Using this we can solve the given problem.

Step by step solution:
Given
\[\left( 1+{{y}^{2}} \right)dx+\left( 1+{{x}^{2}} \right)dy=0\]
Dividing the whole equation by \[\left( 1+{{y}^{2}} \right)\left( 1+{{x}^{2}} \right)\], We have
\[\Rightarrow \dfrac{\left( 1+{{y}^{2}} \right)dx+\left( 1+{{x}^{2}} \right)dy}{\left( 1+{{y}^{2}} \right)\left( 1+{{x}^{2}} \right)}=\dfrac{0}{\left( 1+{{y}^{2}} \right)\left( 1+{{x}^{2}} \right)}\]
\[\Rightarrow \dfrac{\left( 1+{{y}^{2}} \right)dx}{\left( 1+{{y}^{2}} \right)\left( 1+{{x}^{2}} \right)}+\dfrac{\left( 1+{{x}^{2}} \right)dy}{\left( 1+{{y}^{2}} \right)\left( 1+{{x}^{2}} \right)}=0\]
\[\Rightarrow \dfrac{dx}{\left( 1+{{x}^{2}} \right)}+\dfrac{dy}{\left( 1+{{y}^{2}} \right)}=0\]
Integrating on both sides we have
\[\Rightarrow {{\tan }^{-1}}x+{{\tan }^{-1}}y=c\], Where c is the integration constant.
We know that \[{{\tan }^{-1}}(x)+{{\tan }^{-1}}(y)={{\tan }^{-1}}\left( \dfrac{x-y}{1+xy} \right)\], applying this we have
\[\Rightarrow {{\tan }^{-1}}\left( \dfrac{x-y}{1+xy} \right)=c\]
Or
\[\Rightarrow \left( \dfrac{x-y}{1+xy} \right)=\tan c\]
Again, we know that \[\tan c\] is a constant, hence we take \[\tan c=C\]. Then we have,
\[\Rightarrow \left( \dfrac{x-y}{1+xy} \right)=C\]
\[\Rightarrow \left( x-y \right)=C\left( 1+xy \right)\]

Hence, option (c) is correct.

Note: We know that we will have an integration constant in indefinite integration, whereas in definite integral we will not have an integration constant because of the lower and upper limit of the integral. We also know that the integration of zero or any constant is not equal to zero. But differentiation of a constant is zero.