The differential equation $\cos (x+y) \mathrm{d} y=\mathrm{d} x$ has the general solution given by
The differential equation $\cos (x+y) \mathrm{d} y=\mathrm{d} x$ has the general solution given by
- $y=\sin (x+y)+\mathrm{c}$, where $\mathrm{c}$ is a constant.
- $y=\tan (x+y)+\mathrm{c}$, where c is a constant
- $y=\tan \left(\frac{x+y}{2}\right)+\mathrm{c}$, where $\mathrm{c}$ is a constant
- $y=\frac{1}{2} \tan (x+y)+\mathrm{c}$, 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
$\begin{aligned}
& \quad \frac{1}{2} \int \sec ^2\left(\frac{\mathrm{u}}{2}\right) \mathrm{du}=\int \mathrm{d} y \\
& \therefore \quad y=\tan \left(\frac{x+y}{2}\right)+\mathrm{c}
\end{aligned}$
Asked in: MHT CET 2023 (12 May Shift 1)
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