Question

Find the general situation of $si{n}^{-1}\left(dy/dx\right)=x+y$ using variable separable method.

• A. $\left(tan\frac{1}{2}\right(x+y)+1)(x+1)+2+c=0$
• B. $(x+y)(x+1)+2+c=0$
• C. $(sin\frac{(x+y)}{2}+2)+c=0$
• D. $\left(tan\right(x+y)+1)(x+1)+c=0$

Answer 1

The correct option is A $\left(tan\frac{1}{2}\right(x+y)+1)(x+1)+2+c=0$

Look guys there are some differential equations which on the face look like can’t be solved using variable separable. One has to make proper substitutions to reduce it into a form where variables can be separated.
In this question we will put x + y = v
x+y = v
$1+\frac{dy}{dx}=\frac{dv}{dx}$
$\frac{dv}{dx}-1=\frac{dy}{dx}$ ...(1)
From the given equation $si{n}^{-1}\left(dy/dx\right)=x+y$
$\frac{dy}{dx}=sin(x+y)$
$=sin\left(v\right)$ ...(2)
So from (1) and (2)
$\frac{dv}{dx}-1=sinv$
$\Rightarrow \frac{dv}{[1+sinv]}=dx$
$\Rightarrow \frac{dv}{[sinv/2+cosv/2{]}^2}=dx$
Taking $co{s}^2\left(v/2\right)$ common from the denominator.
$\Rightarrow \frac{se{c}^{2}v/2}{(1+tanv/2{)}^2}dv=dx$
Now put 1+tanv/2=t
$\frac{1}{2}se{c}^{2}v/2dv=dt$
$\Rightarrow \frac{2}{t^2}dt=dx$
Integrating we will get
$\Rightarrow \frac{-2}{t}=x+cor\frac{-2}{1+tanv/2}=(x+c)\Rightarrow (1+tanv/2)(x+c)+2=0\Rightarrow (1+tan\frac{(x+y)}{2})(x+c)+2=0$