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Differential Equations — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations4 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
$
\begin{array}{l}
\left(x^2+1\right) \frac{d y}{d x}+2 x y=\sqrt{x^2+4} \\
\Rightarrow \quad \frac{d y}{d x}+\frac{2 x}{x^2+1} y=\frac{\sqrt{x^2+4}}{x^2+1}
\end{array}
$
Which is a linear differential equation.
Here $\quad P =\frac{2 x}{x^2+1} \quad$ and $\quad Q =\frac{\sqrt{x^2+4}}{x^2+1}$
Therefore integrating factor
$
\begin{aligned}
\text { I.F. } & =e^{\int P d x} \\
& =e^{\int \frac{2 x}{1+x^2} d x}
\end{aligned}
$
Let $1+x^2=t \therefore 2 x d x=d t$
$
\therefore \quad \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
$
\begin{array}{l}
\text { y.I.F. }=\int(I . F .)(Q) d x+C \\
\Rightarrow \quad 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 \quad 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 \quad 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 \quad 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
\end{array}
$

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