The solution of $\mathrm{e}^{y-x} \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{y(\sin x+\cos x)}{(1+y \log y)}$ is

The solution of $\mathrm{e}^{y-x} \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{y(\sin x+\cos x)}{(1+y \log y)}$ is
  1. $\frac{\mathrm{e}^y}{y}=\mathrm{e}^x \sin x+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
  2. $\mathrm{e}^y \log y=\mathrm{e}^x \cos x+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
  3. $\mathrm{e}^y \log y=\mathrm{e}^x \sin x+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
  4. $\mathrm{e}^y y=\mathrm{e}^x \sin x+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.

Solution

$\begin{array}{ll} & \mathrm{e}^{y-x} \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{y(\sin x+\cos x)}{(1+y \log y)} \\ \therefore \quad & \mathrm{e}^y \frac{(1+y \log y)}{y} \mathrm{~d} y=\mathrm{e}^x(\sin x+\cos x) \mathrm{d} x \\ \therefore \quad & \int \mathrm{e}^y\left(\log y+\frac{1}{y}\right) \mathrm{d} y=\int \mathrm{e}^x(\sin x+\cos x) \mathrm{d} x \\ & \Rightarrow \mathrm{e}^y \log y=\mathrm{e}^x \sin x+\mathrm{c} \\ & \ldots\left[\because \int \mathrm{e}^x\left(\mathrm{f}(x)+\mathrm{f}^{\prime}(x)\right) \mathrm{d} x=\mathrm{e}^x \mathrm{f}(x)+\mathrm{c}\right]\end{array}$

Asked in: MHT CET 2023 (12 May Shift 2)

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