Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDIFFERENTIAL EQUATIONS5 Marks
Question
Solve the differential equation: $\frac{\text{dy}}{\text{dx}}-\frac{2\text{x}}{1+\text{x}^2}\text{y}=\text{x}^2+2$
✓
Answer
The given differential equation is
$\frac{\text{dy}}{\text{dx}}-\frac{2\text{x}}{1+\text{x}^2}\text{y}=\text{x}^2+2$
This equation is of the form $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q},$
where $\text{P}=\frac{-2\text{x}}{1+\text{x}^2}$ and $\text{Q}=\text{x}^2+2$
Now, $\text{l.F}=\text{e}^{\int\frac{2\text{x}}{1+\text{x}^2}\text{dx}}$
$=\text{e}^{-\log\Big(\frac{1}{1+\text{x}^2}\Big)}=\frac{1}{1+\text{x}^2}$
The general solution of the given differential equation is
$\text{y}\times\text{l.F}.=\int(\text{Q}\times\text{l.F.})\text{dx}+\text{C},$ where C is an aribatry contant
$\Rightarrow\frac{\text{y}}{1+\text{x}^2}=\int\frac{\text{x}^2+2}{1+\text{x}^2}\text{dx}+\text{C}$
$=\int\Big(1+\frac{1}{\text{x}^2+1}\Big)\text{dx}+\text{C}$
$=\int\text{dx}+\int\frac{1}{\text{x}^2+1}\text{dx}+\text{C}$
$=\text{x}+\tan^{-1}\text{x}+\text{C}$
$\text{y}=(1+\text{x}^2)(\text{x}+\tan^{-1})\text{x}+\text{C}$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.