Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDifferential Equations5 Marks
Question
Solve the following differential equation : $\left(x^2+1\right) \frac{d y}{d x}+2 x y=\sqrt{x^2+4}$
✓
Answer
The given differential equation is
$\left(x^2+1\right) \frac{d y}{d x}+2 x y$
$=\sqrt{x^2+4}$
$\Rightarrow \frac{d y}{d x}+\frac{2 x}{x^2+1} y$
$=\frac{\sqrt{x^2+4}}{x^2+1}$
Which is a linear differential equation.
Here $ P =\frac{2 x}{x^2+1}$ and $ Q =\frac{\sqrt{x^2+4}}{x^2+1}$
Therefore integrating factor
$\text { I.F. } =e^{\int P d x}$
$ =e^{\int \frac{2 x}{1+x^2} d x}$
Let $1+x^2=t $
$\therefore 2 x \ d x=d t$
$\therefore \text { I.F. }=e^{\int \frac{d t}{t}}=e^{\log t}=t$
Putting the value of $t$
$\text { I.F. }=\left(1+x^2\right)$
Hence the required solution is
$\text { y.I.F. }=\int(I . F .)(Q) d x+C$
$\Rightarrow y \cdot\left(1+x^2\right)=\int \frac{\left(1+x^2\right) \sqrt{x^2+4}}{x^2+1} d x+C$
$\Rightarrow y \cdot\left(1+x^2\right)=\int \sqrt{(x)^2+(2)^2} d x+C$
${\left(\because \int \sqrt{x^2+a^2} d x=\frac{x}{2} \sqrt{x^2+a^2}+\frac{a^2}{2} \log \right.)}$
$\left.\left|\left(x+\sqrt{x^2+a^2}\right)\right|+C\right)$
$\Rightarrow y\left(x^2+1\right)=\frac{x}{2} \sqrt{x^2+4}+\frac{4}{2} \log$
$\left|\left(x+\sqrt{x^2+4}\right)\right|+C$
$\Rightarrow y\left(x^2+1\right)=\frac{1}{2} x \sqrt{x^2+4}+2 \log$
$\left|\left(x+\sqrt{x^2+4}\right)\right|+C$
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