Question
Solve the following differential equations:$(1+\text{x})(1+\text{y}^2)\text{dx}+(1+\text{y})(1+\text{x}^2)\text{dy}=0$
✓
Answer
We have,
$(1+\text{x})(1+\text{y}^2)\text{dx}+(1+\text{y})(1+\text{x}^2)\text{dy}=0$
$\Rightarrow(1+\text{x})(1+\text{y}^2)\text{dx}=-(1+\text{y})(1+\text{x}^2)\text{dy}$
$\Rightarrow\frac{1+\text{x}}{1+\text{x}^2}\text{dx}=-\frac{1+\text{y}}{1+\text{y}^2}\text{dy}$
Integrating both sides, we get
$\Rightarrow\frac{1+\text{x}}{1+\text{x}^2}\text{dx}=-\frac{1+\text{y}}{1+\text{y}^2}\text{dy}$
$\Rightarrow\int\frac{1}{1+\text{x}^2}\text{dx}+\int\frac{\text{x}}{1+\text{x}^2}\text{dx}=-\int\frac{1}{1+\text{y}^2}\text{dy}-\int\frac{\text{y}}{1+\text{y}^2}\text{dy}$
Substituting $1+\text{x}^2=\text{t}$ in the second integral of LHS and $1+\text{y}^2=\text{u}$ in the second integral of RHS, we get
$2\text{x dx = dt}$ and $2\text{y dy = du}$
$\therefore\int \frac{1}{1+\text{x}^2}\text{dx}+\frac{1}{2}\int\text{dt}=-\int\frac{1}{1+\text{y}^2}\text{dy}-\frac{1}{2}\int\frac{1}{\text{u}}\text{du}$
$\Rightarrow\tan^{-1}\text{x}+\frac{1}{2}\log|\text{t}|=-\tan^{-1}\text{y}-\frac{1}{2}\log|\text{u}|+\text{C}$
$\Rightarrow\tan^{-1}\text{x}+\frac{1}{2}\log|1+\text{x}^2|=-\tan^{-1}\text{y}-\frac{1}{2}\log|1+\text{y}^2|+\text{C}$
$\Rightarrow\tan^{-1}\text{x}+\tan^{-1}\text{y}+\frac{1}{2}\log|1+\text{x}^2|+\frac{1}{2}\log|1+\text{y}^2|=\text{C}$
$\Rightarrow\tan^{-1}\text{x}+\tan^{-1}\text{y}+\frac{1}{2}\log\big|(1+\text{x}^2)(1+\text{y}^2)\big|=\text{C}$
Hence, $\tan^{-1}\text{x}+\tan^{-1}\text{y}+\frac{1}{2}\log\big|(1+\text{x}^2)(1+\text{y}^2)\big|=\text{C}$ is the required solution.
$(1+\text{x})(1+\text{y}^2)\text{dx}+(1+\text{y})(1+\text{x}^2)\text{dy}=0$
$\Rightarrow(1+\text{x})(1+\text{y}^2)\text{dx}=-(1+\text{y})(1+\text{x}^2)\text{dy}$
$\Rightarrow\frac{1+\text{x}}{1+\text{x}^2}\text{dx}=-\frac{1+\text{y}}{1+\text{y}^2}\text{dy}$
Integrating both sides, we get
$\Rightarrow\frac{1+\text{x}}{1+\text{x}^2}\text{dx}=-\frac{1+\text{y}}{1+\text{y}^2}\text{dy}$
$\Rightarrow\int\frac{1}{1+\text{x}^2}\text{dx}+\int\frac{\text{x}}{1+\text{x}^2}\text{dx}=-\int\frac{1}{1+\text{y}^2}\text{dy}-\int\frac{\text{y}}{1+\text{y}^2}\text{dy}$
Substituting $1+\text{x}^2=\text{t}$ in the second integral of LHS and $1+\text{y}^2=\text{u}$ in the second integral of RHS, we get
$2\text{x dx = dt}$ and $2\text{y dy = du}$
$\therefore\int \frac{1}{1+\text{x}^2}\text{dx}+\frac{1}{2}\int\text{dt}=-\int\frac{1}{1+\text{y}^2}\text{dy}-\frac{1}{2}\int\frac{1}{\text{u}}\text{du}$
$\Rightarrow\tan^{-1}\text{x}+\frac{1}{2}\log|\text{t}|=-\tan^{-1}\text{y}-\frac{1}{2}\log|\text{u}|+\text{C}$
$\Rightarrow\tan^{-1}\text{x}+\frac{1}{2}\log|1+\text{x}^2|=-\tan^{-1}\text{y}-\frac{1}{2}\log|1+\text{y}^2|+\text{C}$
$\Rightarrow\tan^{-1}\text{x}+\tan^{-1}\text{y}+\frac{1}{2}\log|1+\text{x}^2|+\frac{1}{2}\log|1+\text{y}^2|=\text{C}$
$\Rightarrow\tan^{-1}\text{x}+\tan^{-1}\text{y}+\frac{1}{2}\log\big|(1+\text{x}^2)(1+\text{y}^2)\big|=\text{C}$
Hence, $\tan^{-1}\text{x}+\tan^{-1}\text{y}+\frac{1}{2}\log\big|(1+\text{x}^2)(1+\text{y}^2)\big|=\text{C}$ is the required solution.
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