Question

$\frac{dy}{dx}=\frac{x+y+1}{2x+2y+3}$, its solution is $x+y+\frac{k}{3}=c{e}^{3(x-2y)}$, what is $k$.

• A. 1
• B. 2
• C. 5
• D. 4

Answer 1

Answer: (D)

4

$\frac{dy}{dx}=\frac{x+y+1}{2x+2y+3}$
Put $x+y=v$
$1+\frac{dy}{dx}=\frac{dv}{dx}$
$\Rightarrow \frac{dv}{dx}-1=\frac{v+1}{2v+3}$
$\Rightarrow \frac{dv}{dx}=\frac{3v+4}{2v+3}$
$\Rightarrow \frac{(2v+3)dv}{3v+4}=dx$
Integrating both sides,
$\int \frac{(2v+3)dv}{3v+4}=\int dx$
$\Rightarrow \frac{1}{3}\int \frac{2(3v+4)dv}{3v+4}+\frac{1}{3}\int \frac{1dv}{3v+4}=x+\log c$
$\Rightarrow \frac{2}{3}v+\frac{1}{9}\log 3v+4=x+\log c$
$\Rightarrow \frac{1}{9}\log (3x+3y+4)-\log c=\frac{1}{3}(x-2y)$
$\Rightarrow \log \frac{(x+y+\frac{4}{3})}{c}=3(x-2y)$
$x+y+\frac{4}{3}=c{e}^{3(x-2y)}$
Comparing with given solution, we get
$k=4$