Answer to Question #258384 in Differential Equations for Zee

Question

Use linear substitution to solve the following first-order differential equation

𝑑𝑦/𝑑𝑥=(2𝑥+𝑦)/(2𝑥+𝑦+1)

Answer 1

$\frac{d y}{d x}=\frac{2 x+y}{2 x+y+1} \ ...(1) \\\text{By substitution } \\\text{let}\ v=2 x+y \\\text { then } \frac{d v}{d x} =2+\frac{d y}{d x} \\ \text { so } \frac{d y}{d x} =\frac{d v}{d x}-2 \\\therefore \text{From (1) we get} \\\frac{d v}{d x}-2=\frac{v}{v+2} \\\frac{d v}{d x}=\frac{v}{(v+1)}+2 =\frac{v+2(v+1)}{(v+1)} \\ \frac{d v}{d x}=\frac{3 v+2}{(v+1)} \\\therefore \quad \frac{(v+1)}{(3 v+2)} d v=d x \\\text{Integrating both sides} \\\int \frac{(v+1)}{(3 v+2)} d v=\int d x+c \\\frac{1}{3} \int\left[\frac{3 v+2+1}{3 v+2}\right] d v=x+c \\\Rightarrow \frac{1}{3} \int\left[1+\frac{1}{(3 v+2)}\right] d v=x+c \\\Rightarrow \quad \int 1 d v+\frac{1}{3} \int \frac{d v}{\left(v+\frac{2}{3}\right)}=3 x+c \\\Rightarrow \quad v+\frac{1}{3} \ln|v+\frac{2}{3}|=3 x+c \\\text{Putting}\ v=2 x+y\ \text{we have} \\2 x+y+\frac{1}{3} \ln \left|2 x+y+\frac{2}{3}\right|=3 x+c \\ \\\Rightarrow \quad y+\frac{1}{3} \ln \left|2 x+y+\frac{2}{3}\right|= x+C \\ \\\Rightarrow \quad y+\frac{1}{3} \ln \left|2 x+y+\frac{2}{3}\right|-x=C$