Question

If $\frac{dy}{dx}=ta{n}^2\left(x+y\right)$, then $sin2\left(x+y\right)=$

• A. $x-y+c$
• B. $2(x-y)+c$
• C. $x+y+c$
• D. $2(x+y)+c$

Answer 1

Answer: (B)

Substitute $x+y=z$
$\Rightarrow \frac{dy}{dx}=\frac{dz}{dx}-1$
So, the given equation becomes
$\frac{dz}{dx}=se{c}^{2}z$
$\Rightarrow dx=co{s}^{2}zdz=\frac{1+cos2z}{2}dz$
$\Rightarrow x=\frac{1}{2}\left(z+\frac{sin2z}{2}\right)+c_1$
$\Rightarrow 2x=x+y+\frac{sin2\left(x+y\right)}{2}+c_2$
$\Rightarrow 2\left(x-y\right)=sin2\left(x+y\right)-c$
$\Rightarrow sin2\left(x+y\right)=2\left(x-y\right)+c$