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Differential Equations — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDifferential Equations5 Marks
Question
Solve the following differential equation $\frac{\text{dy}}{\text{dx}}=\frac{1+\text{y}^2}{\text{y}^3}$
✓

Answer

We have
$\frac{\text{dy}}{\text{dx}}=\frac{1+\text{y}^2}{\text{y}^3}$
$\Rightarrow\frac{\text{y}^3}{1+\text{y}^2}$
$\Rightarrow\text{dx}=\frac{\text{y}^3}{1+\text{y}^2}\ \text{dy}$
Integrating both sides, we get
$\int\text{dx}=\int\frac{\text{y}^3}{1+\text{y}^2}\text{dy}$
$\Rightarrow\text{x}=\int\frac{\text{y}+\text{y}^3-\text{y}}{1+\text{y}^2}\text{dy}$
$\Rightarrow\text{x}=\int\frac{(1+\text{y}^2)\text{y}-\text{y}}{1+\text{y}^2}\text{dy}$
$\Rightarrow\text{x}=\int\text{y dy}-\int\frac{\text{y}}{1+\text{y}^2}\ \text{dy}$
$\Rightarrow\text{x}=\frac{\text{y}^2}{2}-\int\frac{\text{y}}{1+\text{y}^2}\ \text{dy}$
Putting $1 + y^2 = t$ we get
$2y dy = dt$
$\therefore\text{x}=\frac{\text{y}^2}{2}-\frac{1}{2}\int\frac{1}{\text{t}}\text{dt}$
$\Rightarrow\text{x}=\frac{\text{y}^2}{2}-\frac{1}{2}\log|\text{t}|+\text{C}$
$\Rightarrow\text{x}=\frac{\text{y}^2}{2}-\frac{1}{2}\log|1+\text{y}^2|+\text{C}$
Hence, $\text{x}=\frac{\text{y}^2}{2}-\frac{1}{2}\log|1+\text{y}^2|+\text{C}$ is the required solution.

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