Q81 / 150 · MHT CET 2023 (13 May Shift 1) · Mathematics

The particular solution of differential equation $\mathrm{e}^{\frac{d y}{d x}}=(x+1), y(0)=3$ is

  • A
    $y=x \log x-x+2$
  • Correct answer
    B
    $y=(x+1) \log (x+1)-x+3$
  • C
    $y=(x+1) \log (x+1)+x-3$
  • D
    $y=x \log x+x-2$

Solution

$$ \begin{aligned} & \mathrm{e}^{\frac{\mathrm{d}}{\mathrm{d} x}}=(x+1) \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\log (x+1) \end{aligned} $$ Integrating on both sides, we get $$ \begin{aligned} \int \mathrm{d} y & =\int \log (x+1) \mathrm{d} x+\mathrm{c} \\ \Rightarrow y & =x \log (x+1)-\int \frac{x}{x+1} \mathrm{~d} x+\mathrm{c} \\ & =x \log (x+1)-\int \frac{x+1-1}{x+1} \mathrm{~d} x+\mathrm{c} \\ & =x \log (x+1)-\int\left(1-\frac{1}{x+1}\right) \mathrm{d} x+\mathrm{c} \\ y & =x \log (x+1)-x+\log (x+1)+\mathrm{c} . \end{aligned} $$ Since $y(0)=3$, i.e., $y=3$ when $x=0$ $$ \begin{aligned} & 3=0+\mathrm{c} \Rightarrow \mathrm{c}=3 \\ & y=x \log (x+1)+\log (x+1)-x+3 \\ & y=(x+1) \log (x+1)-x+3 \end{aligned} $$

Hence, the correct answer is option: 'B'.

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