Differential Equations 2 Question 7
7. Let $y=y(x)$ be the solution of the differential equation, $\left(x^{2}+1\right)^{2} \frac{d y}{d x}+2 x\left(x^{2}+1\right) y=1$ such that $y(0)=0$. If $\sqrt{a} y(1)=\frac{\pi}{32}$, then the value of ’ $a$ ’ is $\quad$ (2019 Main, 8 April I)
(a) $\frac{1}{4}$
(b) $\frac{1}{2}$
(c) 1
(d) $\frac{1}{16}$
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Answer:
Correct Answer: 7. (d)
Solution:
- Given differential equation is
$$ \begin{aligned} & \left(x^{2}+1\right)^{2} \frac{d y}{d x}+2 x\left(x^{2}+1\right) y=1 \\ \Rightarrow \quad & \quad \frac{d y}{d x}+\frac{2 x}{1+x^{2}} \quad y=\frac{1}{\left(1+x^{2}\right)^{2}} \end{aligned} $$
$$ \text { [dividing each term by }\left(1+x^{2}\right)^{2} \text { ] } $$
This is a linear differential equation of the form
$$ \frac{d y}{d x}+P \cdot y=Q $$
Here, $P=\frac{2 x}{\left(1+x^{2}\right)}$ and $Q=\frac{1}{\left(1+x^{2}\right)^{2}}$
$\therefore$ Integrating Factor (IF) $=e^{\int \frac{2 x}{1+x^{2}} d x}$
$$ =e^{\ln \left(1+x^{2}\right)}=\left(1+x^{2}\right) $$
and required solution of differential Eq. (i) is given by
$$ \begin{array}{rlrl} & y \cdot(IF) & =\int Q(IF) d x+C \\ \Rightarrow & y\left(1+x^{2}\right) & =\int \frac{1}{\left(1+x^{2}\right)^{2}}\left(1+x^{2}\right) d x+C & \\ \Rightarrow & y\left(1+x^{2}\right) & =\int \frac{d x}{1+x^{2}}+C \\ \Rightarrow & y\left(1+x^{2}\right) & =\tan ^{-1}(x)+C \\ \because & y(0) & =0 \\ \therefore & & C & \\ \therefore & y\left(1+x^{2}\right) & =\tan ^{-1} x \\ \Rightarrow & y & =\frac{\tan ^{-1} x}{1+x^{2}} \\ \Rightarrow & & \\ & & & \\ & & & \end{array} $$
Now, $\quad$ at $x=1$
[multiplying both sides by $\sqrt{a}$ ]
$$ \begin{aligned} \quad \sqrt{a} y(1) & =\sqrt{a} \frac{\tan ^{-1}(1)}{1+1}=\sqrt{a} \frac{\frac{\pi}{4}}{2}=\frac{\sqrt{a} \pi}{8}=\frac{\pi}{32} \text { (given) } \\ \therefore \quad \sqrt{a} & =\frac{1}{4} \Rightarrow a=\frac{1}{16} \end{aligned} $$
