Q60 / 100 · BITSAT 2024 (Memory Based Paper 2) · Mathematics

If $\frac{d y}{d x}-y \log _e 2=2^{\sin x}(\cos x-1) \log _e 2$, then $y$

  • Correct answer
    A
    $2^{\sin x}+c 2^x$
  • B
    $2^{\cos x}+c 2^x$
  • C
    $2^{\sin x}+c 2^{-x}$
  • D
    $2^{\cos x}+c 2^{-x}$

Solution

$\frac{d y}{d x}-y \log _e 2=2^{\sin x}(\cos x-1) \log _e 2$ This is linear differential equation $\text { I.F. }=e^{-\log _e 2 \int d x}=e^{-x \log _e 2}=2^{-x}$ then general Solution is $y 2^{-x}=\int 2^{-x} 2^{\sin x}(\cos x-1) \log _e 2 d x+c$ Now let $\sin x-x=t \Rightarrow(\cos x-1) d x=d t$ $\begin{aligned} & \therefore y 2^{-x}=\log _e 2 \int 2^t d t+c \\ & \therefore y 2^{-x}=2^t+c \\ & \therefore y=2^{x+t}+c 2^x \\ & \therefore y=2^{\sin x}+c 2^x \end{aligned}$

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

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