The general solution of the differential equation $\mathrm{e}^{y-x} \frac{\mathrm{~d} y}{\mathrm{~d}…

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

Solution

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

Asked in: MHT CET 2024 (10 May Shift 2)

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