Question

The solution of the differential equation $\frac{dy}{dx}=\sin (x+y)\tan (x+y)-1$ is

• A. $\csc \left(x+y\right)+\tan (x+y)=C$
• B. $x+\csc \left(x+y\right)=C$
• C. $x+\tan (x+y)=C$
• D. $x+\sec \left(x+y\right)=C$

Answer 1

The correct option is B

$x+\csc \left(x+y\right)=C$

Explanation for the correct option

Given: $\frac{dy}{dx}=\sin (x+y)\tan (x+y)-1$

Let $x+y=z$

differentiating both sides

$$\begin{gathered} \Rightarrow 1+\frac{dy}{dx}=\frac{dz}{dx} \\ \Rightarrow \frac{dy}{dx}=\frac{dz}{dx}-1 \end{gathered}$$

$$\begin{gathered} \Rightarrow \frac{dz}{dx}-1=\sin \left(z\right)\tan \left(z\right)-1 \\ \Rightarrow \frac{dz}{dx}=\sin \left(z\right)\tan \left(z\right) \\ \Rightarrow \frac{dz}{dx}=\sin \left(z\right)·\frac{\sin \left(z\right)}{\cos \left(z\right)} \\ \Rightarrow \frac{\cos \left(z\right)dz}{{\sin }^2\left(z\right)}=dx \end{gathered}$$

Let $\sin \left(z\right)=t$

$\cos \left(z\right)dz=dt$

$$\begin{gathered} \Rightarrow \int \frac{dt}{t^2}=\int dx \\ \Rightarrow \frac{t^{-2+1}}{-2+1}=x \\ \Rightarrow -\frac{1}{t}=x \end{gathered}$$

put $t=\sin \left(z\right)$

$-\frac{1}{\sin \left(z\right)}=x$

put $z=x+y$

$\Rightarrow -\csc \left(x+y\right)=x+C$

Hence, option B is correct.