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

JEE Main · 2023 · 11 Apr 2023 · Shift 2 · Differential Equations

Let

y = y (x)

be the solution of the differential equations

\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

Answer: (C)

This is a first-order linear ODE. The integrating factor is:

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

Substitute

t = 1 + x^{- 5}

d t = - 5 x^{- 6} d x

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

Multiplying the ODE by I.F.:

y \cdot \frac{x ^{5}}{1 + x ^{5}} = \int \frac{x ^{5}}{1 + x ^{5}} \cdot \frac{( 1 + x ^{5} ) ^{2}}{x ^{7}} d x = \int (x^{3} + x^{- 2}) d x = \frac{x ^{4}}{4} - \frac{1}{x} + c

Using

y (1) = 2

2 \cdot \frac{1}{2} = \frac{1}{4} - 1 + c \Rightarrow c = \frac{7}{4}

y \cdot \frac{x ^{5}}{1 + x ^{5}} = \frac{x ^{4}}{4} - \frac{1}{x} + \frac{7}{4}

.

At

x = 2

y \cdot \frac{32}{33} = \frac{16}{4} - \frac{1}{2} + \frac{7}{4} = 4 - 0.5 + 1.75 = 5.25 = \frac{21}{4}

y = \frac{21}{4} \cdot \frac{33}{32} = \frac{693}{128}

JEE Main · 2023 · 11 Apr 2023 · Shift 2 · Single correct option