SATHEE: Differential Equations 2 Question 17

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) $\frac{π}{3} - \frac{\sqrt{3}}{2}$

(b) $\frac{π}{3} - \frac{\sqrt{3}}{4}$

(d) $\frac{π}{6} - \frac{\sqrt{3}}{2}$

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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

$\begin{aligned} y \cdot (I F) & = \int Q \cdot (I F) d x + c \\ where, & F = e^{\int P d x} \end{aligned}$

(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.

$\begin{aligned} I F = e^{\int \frac{x}{x^{2} - 1} d x} = e^{\frac{1}{2} \ln | x^{2} - 1 |} = \sqrt{1 - x^{2}} \\ \Rightarrow Solution is y \sqrt{1 - x^{2}} = \int \frac{x (x^{3} + 2)}{\sqrt{1 - x^{2}}} \cdot \sqrt{1 - x^{2}} d x \\ or y \sqrt{1 - x^{2}} = \int (x^{4} + 2 x) d x = \frac{x^{5}}{5} + x^{2} + c \\ f (0) = 0 \Rightarrow c = 0 \Rightarrow f (x) \sqrt{1 - x^{2}} = \frac{x^{5}}{5} + x^{2} \\ Now, \int_{- \sqrt{3} / 2}^{\sqrt{3} / 2} f (x) d x = \int_{- \sqrt{3} / 2}^{\sqrt{3} / 2} \frac{x^{2}}{\sqrt{1 - x^{2}}} d x \\ = 2 \int_{0}^{\sqrt{3} / 2} \frac{x^{2}}{\sqrt{1 - x^{2}}} d x \\ = 2 \int_{0}^{π / 3} \frac{\sin^{2} θ}{\cos θ} \cos θ d θ [taking x = \sin θ] \\ = 2 \int_{0}^{π / 3} \sin^{2} θ d θ = \int_{0}^{π / 3} (1 - \cos 2 θ) d θ \\ = θ - {\frac{\sin 2 θ}{2}}_{0}^{π / 3} = \frac{π}{3} - \frac{\sin 2 π / 3}{2} = \frac{π}{3} - \frac{\sqrt{3}}{4} \end{aligned}$

Whenever we have linear differential equation containing inequality, we should always check for increasing or decreasing,

$i.e. for \frac{d y}{d x} + P y < 0 \Rightarrow \frac{d y}{d x} + P y > 0$

Multiply by integrating factor, i.e. $e^{\int P d x}$ and convert into total differential equation.

Here, $f^{'} (x) < 2 f (x)$ , multiplying by $e^{- \int 2 d x}$

$f^{'} (x) \cdot e^{- 2 x} - 2 e^{- 2 x} f (x) < 0 \Rightarrow \frac{d}{d x} (f (x) \cdot e^{- 2 x}) < 0$

$∴ φ (x) = f (x) e^{- 2 x}$ is decreasing for $x \in \frac{1}{2}, 1$

Thus, when $x > \frac{1}{2}$

$φ (x) < φ \frac{1}{2} \Rightarrow e^{- 2 x} f (x) < e^{- 1} \cdot f \frac{1}{2}$

$\begin{aligned} \Rightarrow f (x) < e^{2 x - 1} \cdot 1, given f \frac{1}{2} = 1 \\ \Rightarrow 0 < \int_{1 / 2}^{1} f (x) d x < \int_{1 / 2}^{1} e^{2 x - 1} d x \end{aligned}$