Let y = y (x) be the solution of the differential equation d y d x + 5 x ( x 5 + 1 ) y = ( x 5 + 1 ) 2 x 7 , x 0 . If y ( 1 ) = 2 , then y ( 2 ) is equal to [2023]

Question

Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} + \frac{5}{x (x^{5} + 1)} y = \frac{{(x^{5} + 1)}^{2}}{x^{7}}, x > 0 .$ If $y (1) = 2$ , then $y (2)$ is equal to [2023]

Smart Achievers · Accepted Answer

(3)

$I.F. = e^{\int \frac{5 d x}{x (x^{5} + 1)}} = e^{\int \frac{5 x^{- 6}}{x^{- 5} + 1} d x}$

Put $1 + x^{- 5} = t \Rightarrow - 5 x^{- 6} d x = d t$

$∴$ $I.F. = e^{\int \frac{- d t}{t}} = \frac{1}{t} = \frac{x^{5}}{1 + x^{5}}$

$Solution is given by \frac{y x^{5}}{1 + x^{5}} = \int \frac{x^{5}}{(1 + x^{5})} \times \frac{{(1 + x^{5})}^{2}}{x^{7}} d x + C$

$= \int x^{3} d x + \int x^{- 2} d x + C$ $∴$ $\frac{y x^{5}}{1 + x^{5}} = \frac{x^{4}}{4} - \frac{1}{x} + C$

At $x = 1, y = 2$

$\frac{2}{2} = \frac{1}{4} - 1 + C \Rightarrow C = \frac{7}{4}$ $∴$ $\frac{y x^{5}}{1 + x^{5}} = \frac{x^{4}}{4} - \frac{1}{x} + \frac{7}{4}$

$At x = 2$

$y (\frac{32}{33}) = \frac{21}{4} = \frac{693}{128}$

Solution

Q.
Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} + \frac{5}{x (x^{5} + 1)} y = \frac{{(x^{5} + 1)}^{2}}{x^{7}}, x > 0 .$ If $y (1) = 2$ , then $y (2)$ is equal to [2023]

1 $\frac{637}{128}$

2 $\frac{679}{128}$

3 $\frac{693}{128}$

4 $\frac{697}{128}$

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Solution

Q. Let y=y(x) be the solution of the differential equation dydx+5x(x5+1)y=(x5+1)2x7, x>0. If y(1)=2, then y(2) is equal to [2023]

1 637128

2 679128

3 693128

4 697128