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) $\frac{1}{16}$

Show Answer

Answer:

Correct Answer: 7. (d)

Solution:

  1. 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 $\sqrt{a}$ ]

$$ \begin{aligned} \quad \sqrt{a} y(1) & =\sqrt{a} \frac{\tan ^{-1}(1)}{1+1}=\sqrt{a} \frac{\frac{\pi}{4}}{2}=\frac{\sqrt{a} \pi}{8}=\frac{\pi}{32} \text { (given) } \\ \therefore \quad \sqrt{a} & =\frac{1}{4} \Rightarrow a=\frac{1}{16} \end{aligned} $$



NCERT Chapter Video Solution

Dual Pane