Question

Solve:

$\frac{dy}{dx}=\frac{x+y+1}{2x+2y+3}$

• A. $x-y+\frac{4}{3}=c{e}^{(x+2y)}$
• B. $x+y-\frac{4}{3}=c{e}^{(x-y)}$
• C. $x+y+\frac{4}{3}=c{e}^{3(x-2y)}$
• D. $x+y+\frac{4}{3}=c{e}^{3(x+y)}$

Answer 1

Answer: (C)

$x+y+\frac{4}{3}=c{e}^{3(x-2y)}$

$\frac{dy}{dx}=\frac{x+y+1}{2x+2y+3}$
which is not a homogeneous differential eqn.
Here, $\frac{a_1}{a_2}=\frac{1}{2};\frac{b_1}{b_2}=\frac{1}{2}$$\Rightarrow \frac{a_1}{a_2}=\frac{b_1}{b_2}$
$\frac{dy}{dx}=\frac{x+y+1}{2(x+y)+3}$ .....(1)
Put $x+y=v$
$1+\frac{dy}{dx}=\frac{dv}{dx}$$\Rightarrow \frac{dy}{dx}=\frac{dv}{dx}-1$
So, eqn (1) becomes,
$\frac{dv}{dx}-1=\frac{v+1}{2v+3}$$\Rightarrow \frac{dv}{dx}=\frac{3v+4}{2v+3}$
$\Rightarrow \frac{2v+3}{3v+4}dv=dx$
Integrating both sides, we get
$\Rightarrow \frac{1}{3}\int \frac{2(3v+4)+1}{3v+4}dv=\int dx$
$2v+\log |3v+4|=3x+\log c$
$\Rightarrow \log |\frac{3x+3y+4}{c}|=x-2y$