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QJEEM23Ma3653: JEE Main 2023 Mathematics - Class 12 JEEM23Ma3653 Solution: Let be the solution of the differential equation . If , then is …

Let y=y(x)y=y(x) be the solution of the differential equation \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)=2y(1)=2 , then y(2)y(2) is equal to :

Multiple Choice Options

  • A 693128\frac{693}{128}
  • B 697128\frac{697}{128}
  • C 637128\frac{637}{128}
  • D 679128\frac{679}{128}

Step-by-Step Verified Solution

I.F =e∫5dxx(x5+1)=e∫5x−6dx(x−5+1)=\mathrm{e}^{\int \frac{5 \mathrm{dx}}{\mathrm{x}\left(\mathrm{x}^5+1\right)}}=\mathrm{e}^{\int \frac{5 \mathrm{x}^{-6} \mathrm{dx}}{\left(\mathrm{x}^{-5}+1\right)}}

Put, 1+x−5=t⇒−5x−6dx=dt1+\mathrm{x}^{-5}=\mathrm{t} \Rightarrow-5 \mathrm{x}^{-6} \mathrm{dx}=\mathrm{dt}

∴\therefore e∫−dtt=e−ln⁡t=1t=x51+x5e^{\int-\frac{d t}{t}}=e^{-\ln t}=\frac{1}{t}=\frac{x^5}{1+x^5}

y⋅x51+x5amp;=∫x5(1+x5)×(1+x5)2x7dxamp;=∫x3dx+∫x−2dx\begin{aligned}y \cdot \frac{x^5}{1+x^5} & =\int \frac{x^5}{\left(1+x^5\right)} \times \frac{\left(1+x^5\right)^2}{x^7} d x \\\\& =\int x^3 d x+\int x^{-2} d x\end{aligned}

y⋅x51+x5=x44−1x+cy \cdot \frac{x^5}{1+x^5}=\frac{x^4}{4}-\frac{1}{x}+c

 Given that: x=1⇒y=2\text { Given that: } x=1 \Rightarrow y=2

amp;2⋅12=14−1+camp;c=74amp;y⋅x51+x5=x44−1x+74amp; Now put, x=2amp;y⋅(3233)=214amp;y=693128\begin{aligned}& 2 \cdot \frac{1}{2}=\frac{1}{4}-1+\mathrm{c} \\\\& \mathrm{c}=\frac{7}{4} \\\\& \mathrm{y} \cdot \frac{\mathrm{x}^5}{1+\mathrm{x}^5}=\frac{\mathrm{x}^4}{4}-\frac{1}{\mathrm{x}}+\frac{7}{4} \\\\& \text { Now put, } \mathrm{x}=2 \\\\& \mathrm{y} \cdot\left(\frac{32}{33}\right)=\frac{21}{4} \\\\& \mathrm{y}=\frac{693}{128}\end{aligned}

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