Question

The general solution of $\frac{dy}{dx}\sqrt{1+x+y}=x+y-1,$ is
(where $c$ is constant of integration and $\log$ has natural base ${}^{\prime }{e}^{\prime }$)

• A. $2[\sqrt{1+x+y}+\frac{1}{3}\log |\sqrt{1+x+y}-1|-\frac{4}{3}\log |\sqrt{1+x+y}+2|]=x+c$
• B. $3[\sqrt{1+x+y}+\frac{1}{2}\log |\sqrt{1+x+y}-2|-\frac{4}{3}\log |\sqrt{1+x+y}+5|]=x+c$
• C. $2[\sqrt{x+y-1}+\frac{1}{3}\log |\sqrt{x+y-1}-1|+\frac{4}{3}\log |\sqrt{1+x+y}+2|]=x+c$
• D. $3[\sqrt{x+y-1}+\frac{1}{2}\log |\sqrt{x+y-1}-2|-\frac{4}{3}\log |\sqrt{1+x+y}+5|]=x+c$

Answer 1

The correct option is A $2[\sqrt{1+x+y}+\frac{1}{3}\log |\sqrt{1+x+y}-1|-\frac{4}{3}\log |\sqrt{1+x+y}+2|]=x+c$

Putting $\sqrt{1+x+y}=v,$ we get
$\Rightarrow x+y-1=v^2-2$
$1+\frac{dy}{dx}=2v\frac{dv}{dx}$
Then the given equation transformed to
$(2v\frac{dv}{dx}-1)v=v^2-2$
$\Rightarrow \frac{dv}{dx}=\frac{v^2+v-2}{2v^2}$
$\Rightarrow \int \frac{2v^2}{v^2+v-2}dv=\int dx$
$\Rightarrow 2\int [1+\frac{1}{3(v-1)}-\frac{4}{3(v+2)}]dv=\int dx$
$\Rightarrow 2[v+\frac{1}{3}\log |v-1|-\frac{4}{3}\log |v+2|]=x+c$ where $v=\sqrt{1+x+y}$
$\therefore 2[\sqrt{1+x+y}+\frac{1}{3}\log |\sqrt{1+x+y}-1|-\frac{4}{3}\log |\sqrt{1+x+y}+2|]=x+c$