Differential Equations 2 Question 17
17. The function $y=f(x)$ is the solution of the differential equation $\frac{d y}{d x}+\frac{x y}{x^{2}-1}=\frac{x^{4}+2 x}{\sqrt{1-x^{2}}}$ in $(-1,1)$ satisfying $f(0)=0$. Then, $\int _{-\frac{\sqrt{3}}{2}}^{\sqrt{3} / 2} f(x) d x$ is
(2014 Adv.)
(a) $\frac{\pi}{3}-\frac{\sqrt{3}}{2}$
(b) $\frac{\pi}{3}-\frac{\sqrt{3}}{4}$
(c) $\frac{\pi}{6}-\frac{\sqrt{3}}{4}$
(d) $\frac{\pi}{6}-\frac{\sqrt{3}}{2}$
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Answer:
Correct Answer: 17. (b)
Solution:
- PLAN (i) Solution of the differential equation $\frac{d y}{d x}+P y=Q$ is
$$ \begin{aligned} y \cdot(IF) & =\int Q \cdot(IF) d x+c \\ \text { where, } \quad & \quad \mathbb{F}=e^{\int P d x} \end{aligned} $$
(ii) $\int _{-a}^{a} f(x) d x=2 \int _0^{a} f(x) d x$, if $f(-x)=f(x)$
Given differential equation
$$ \frac{d y}{d x}+\frac{x}{x^{2}-1} y=\frac{x^{4}+2 x}{\sqrt{1-x^{2}}} $$
This is a linear differential equation.
$$ \begin{aligned} & IF=e^{\int \frac{x}{x^{2}-1} d x}=e^{\frac{1}{2} \ln \left|x^{2}-1\right|}=\sqrt{1-x^{2}} \\ & \Rightarrow \text { Solution is } y \sqrt{1-x^{2}}=\int \frac{x\left(x^{3}+2\right)}{\sqrt{1-x^{2}}} \cdot \sqrt{1-x^{2}} d x \\ & \text { or } \quad y \sqrt{1-x^{2}}=\int\left(x^{4}+2 x\right) d x=\frac{x^{5}}{5}+x^{2}+c \\ & f(0)=0 \Rightarrow c=0 \Rightarrow f(x) \sqrt{1-x^{2}}=\frac{x^{5}}{5}+x^{2} \\ & \text { Now, } \int _{-\sqrt{3} / 2}^{\sqrt{3} / 2} f(x) d x=\int _{-\sqrt{3} / 2}^{\sqrt{3} / 2} \frac{x^{2}}{\sqrt{1-x^{2}}} d x \\ & =2 \int _0^{\sqrt{3} / 2} \frac{x^{2}}{\sqrt{1-x^{2}}} d x \\ & \left.=2 \int _0^{\pi / 3} \frac{\sin ^{2} \theta}{\cos \theta} \cos \theta d \theta \quad \text { [taking } x=\sin \theta\right] \\ & =2 \int _0^{\pi / 3} \sin ^{2} \theta d \theta=\int _0^{\pi / 3}(1-\cos 2 \theta) d \theta \\ & =\theta-\frac{\sin 2 \theta}{2} _0^{\pi / 3}=\frac{\pi}{3}-\frac{\sin 2 \pi / 3}{2}=\frac{\pi}{3}-\frac{\sqrt{3}}{4} \end{aligned} $$
Whenever we have linear differential equation containing inequality, we should always check for increasing or decreasing,
$$ \text { i.e. for } \frac{d y}{d x}+P y<0 \Rightarrow \frac{d y}{d x}+P y>0 $$
Multiply by integrating factor, i.e. $e^{\int P d x}$ and convert into total differential equation.
Here, $f^{\prime}(x)<2 f(x)$, multiplying by $e^{-\int 2 d x}$
$$ f^{\prime}(x) \cdot e^{-2 x}-2 e^{-2 x} f(x)<0 \Rightarrow \frac{d}{d x}\left(f(x) \cdot e^{-2 x}\right)<0 $$
$\therefore \varphi(x)=f(x) e^{-2 x}$ is decreasing for $x \in \frac{1}{2}, 1$
Thus, when $x>\frac{1}{2}$
$$ \varphi(x)<\varphi \frac{1}{2} \Rightarrow e^{-2 x} f(x)<e^{-1} \cdot f \frac{1}{2} $$
$$ \begin{aligned} & \Rightarrow \quad f(x)<e^{2 x-1} \cdot 1, \text { given } f \frac{1}{2}=1 \\ & \Rightarrow \quad 0<\int _{1 / 2}^{1} f(x) d x<\int _{1 / 2}^{1} e^{2 x-1} d x \end{aligned} $$
$$ \begin{aligned} & \Rightarrow \quad 0<\int _{1 / 2}^{1} f(x) d x<\frac{e-1}{2} \end{aligned} $$
