Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDIFFERENTIAL EQUATIONS5 Marks
Question
Solve the following differential equations:$\text{y}(1-\text{x}^2)\frac{\text{dy}}{\text{dx}}=\text{x}(1+\text{y}^2)$
✓
Answer
We have,
$\text{y}(1-\text{x}^2)\frac{\text{dy}}{\text{dx}}=\text{x}(1+\text{y}^2)$
$\Rightarrow\frac{\text{y}}{1+\text{y}^2}\text{dy}=\frac{\text{x}}{1-\text{x}^2}\text{dx}$
Integrating both sides,
$\int\frac{\text{y}}{1+\text{y}^2}\text{dy}=\int\frac{\text{x}}{1-\text{x}^2}\text{dx}$
Substituting $1+\text{y}^2=\text{t}$ and $1-\text{x}^2=\text{u}$
$2\text{ydy = dt}$ and $-2\text{x dx = du}$
$\therefore\frac{1}{2}\int\frac{1}{\text{t}}=\frac{-1}{2}\int\frac{1}{\text{u}}\text{du}$
$\Rightarrow\frac{1}2{}\log|\text{t}|=-\frac{1}{2}\log|\text{u}|+\log\text{C}$
$\Rightarrow\frac{1}{2}|1+\text{y}^2|=-\frac{1}{2}\log|1-\text{x}^2|+\log\text{C}$
$\Rightarrow\frac{1}{2}\big[\log|1+\text{y}^2|+\log|1-\text{x}^2|\big]=\log\text{C}$
$\Rightarrow\log(|1+\text{y}^2||1-\text{x}^2|)=2\log\text{C}$
$\Rightarrow(1+\text{y}^2)(1-\text{x}^2)=\text{C}^2$
$\Rightarrow(1+\text{y}^2)(1-\text{x}^2)=\text{C}_1,$ where $\text{C}_1=\text{C}^2$
Hence, $(1+\text{y}^2)(1-\text{x}^2)=\text{C}_1$ is the required solution.
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