Let y = y(x) be the solution of the differential equation , –1 < x < 1, y(0) = 0. If , m and n are co-prime numbers, then m + n is equal to __________. [2024]

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Solution

Q.
Let y = y(x) be the solution of the differential equation $(1 - x^{2}) d y = [x y + (x^{3} + 2) \sqrt{3 (1 - x^{2})}] d x$ , –1 < x < 1, y(0) = 0. If $y (\frac{1}{2}) = \frac{m}{n}$ , m and n are co-prime numbers, then m + n is equal to __________. [2024]

Ans.
(97)

We have, $(1 - x^{2}) d y = [x y + (x^{3} + 2) \sqrt{3 (1 - x^{2})}] d x$

$\Rightarrow \frac{d y}{d x} = \frac{x}{(1 - x^{2})} y + \frac{(x^{3} + 2) \sqrt{3 - 3 x^{2}}}{(1 - x^{2})}$

$\Rightarrow \frac{d y}{d x} - \frac{x}{1 - x^{2}} y = \frac{(x^{3} + 2) \sqrt{3 - 3 x^{2}}}{1 - x^{2}}$

$I . F . = e^{\int \frac{- x}{1 - x^{2}} d x}$

Put $1 - x^{2} = t \Rightarrow - x d x = \frac{d t}{2}$

$\Rightarrow I . F . = e^{\int \frac{d t}{2 t}} = e^{\frac{1}{2} \ln t} = e^{\frac{1}{2} \ln (1 - x^{2})}$

$= e^{\ln {(1 - x^{2})}^{\frac{1}{2}}} = {(1 - x^{2})}^{\frac{1}{2}} = \sqrt{1 - x^{2}}$

$∴ y \sqrt{(1 - x^{2})} = \int \frac{(x^{3} + 2) \sqrt{3 (1 - x^{2})}}{(1 - x^{2})} \sqrt{1 - x^{2}} d x$

$\Rightarrow y \sqrt{1 - x^{2}} = \sqrt{3} \int (x^{3} + 2) d x$

$\Rightarrow y \sqrt{1 - x^{2}} = \sqrt{3} [\frac{x^{4}}{4} + 2 x] + c$

Since, y(0) = 0, so $0 = c \Rightarrow y = \sqrt{\frac{3}{1 - x^{2}}} (\frac{x^{4}}{4} + 2 x)$

Now, $y (\frac{1}{2}) = \sqrt{\frac{3}{1 - \frac{1}{4}}} (\frac{{(\frac{1}{2})}^{4}}{4} + 2 \times \frac{1}{2}) = 2 (\frac{1}{64} + 1)$

$= 2 (\frac{65}{64}) = \frac{65}{32} = \frac{m}{n}$

$∴$ m = 65, n = 32 $\Rightarrow$ m + n = 65 +32 = 97.

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