Q69 / 150 · MHT CET 2023 (12 May Shift 1) · Mathematics

The differential equation $\cos (x+y) \mathrm{d} y=\mathrm{d} x$ has the general solution given by

  • A
    $y=\sin (x+y)+\mathrm{c}$, where $\mathrm{c}$ is a constant.
  • B
    $y=\tan (x+y)+\mathrm{c}$, where c is a constant
  • Correct answer
    C
    $y=\tan \left(\frac{x+y}{2}\right)+\mathrm{c}$, where $\mathrm{c}$ is a constant
  • D
    $y=\frac{1}{2} \tan (x+y)+\mathrm{c}$, where $\mathrm{c}$ is a constant

Solution

$$ \begin{aligned} & \cos (x+y) \mathrm{d} y=\mathrm{d} x \\ \therefore \quad & \frac{\mathrm{d} x}{\mathrm{~d} y}=\cos (x+y) \end{aligned} $$ Put $x+y=\mathrm{u}$ Differentiating w.r.t. $y$, we get $\frac{\mathrm{d} x}{\mathrm{~d} y}+1=\frac{\mathrm{du}}{\mathrm{d} y}$ $$ \therefore \quad \frac{\mathrm{d} x}{\mathrm{~d} y}=\frac{\mathrm{du}}{\mathrm{d} y}-1 $$ Substituting (ii) and (iii) in (i), we get $$ \begin{aligned} & \frac{\mathrm{du}}{\mathrm{d} y}-1=\cos \mathrm{u} \\ & \therefore \quad \frac{\mathrm{du}}{1+\operatorname{cosu}}=\mathrm{d} y \\ & \therefore \quad \frac{\mathrm{du}}{2 \cos ^2\left(\frac{\mathrm{u}}{2}\right)}=\mathrm{d} y \end{aligned} $$ Integrating on both sides, we get $$ \begin{aligned} & \quad \frac{1}{2} \int \sec ^2\left(\frac{\mathrm{u}}{2}\right) \mathrm{du}=\int \mathrm{d} y \\ & \therefore \quad y=\tan \left(\frac{x+y}{2}\right)+\mathrm{c} \end{aligned} $$

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

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