Question

solve the given differential equation. (2x + y + 1)y = 1

Explanation:

Step1: Rewrite the differential equation

Given $(2x + y+1)y'=1$, we can rewrite it as $y'=\frac{1}{2x + y + 1}$, or $\frac{dy}{dx}=\frac{1}{2x + y + 1}$. Let $u = 2x + y+1$, then $\frac{du}{dx}=2+\frac{dy}{dx}$. So, $\frac{dy}{dx}=\frac{du}{dx}-2$.
The differential - equation becomes $\frac{du}{dx}-2=\frac{1}{u}$.

Step2: Rearrange the equation

Rearrange $\frac{du}{dx}-2=\frac{1}{u}$ to get $\frac{du}{dx}=\frac{1 + 2u}{u}$. Then, separate the variables: $\frac{u}{2u + 1}du=dx$.
We can rewrite $\frac{u}{2u + 1}$ as $\frac{\frac{1}{2}(2u + 1)-\frac{1}{2}}{2u + 1}=\frac{1}{2}-\frac{1}{2(2u + 1)}$.
So the equation is $(\frac{1}{2}-\frac{1}{2(2u + 1)})du=dx$.

Step3: Integrate both sides

Integrate the left - hand side: $\int(\frac{1}{2}-\frac{1}{2(2u + 1)})du=\frac{1}{2}u-\frac{1}{4}\ln|2u + 1|+C_1$.
Integrate the right - hand side: $\int dx=x + C_2$.
So, $\frac{1}{2}u-\frac{1}{4}\ln|2u + 1|=x + C$.

Step4: Substitute back $u = 2x + y+1$

Substitute $u = 2x + y+1$ into the equation:
$\frac{1}{2}(2x + y+1)-\frac{1}{4}\ln|2(2x + y+1)+1|=x + C$.
Expand the left - hand side: $x+\frac{1}{2}y+\frac{1}{2}-\frac{1}{4}\ln|4x + 2y+3|=x + C$.
Simplify to get $\frac{1}{2}y+\frac{1}{2}-\frac{1}{4}\ln|4x + 2y+3|=C$.
Multiply through by 4 to obtain $2y + 2-\ln|4x + 2y+3|=C_3$.

Answer:

$2y + 2-\ln|4x + 2y+3|=C$ (where $C$ is an arbitrary constant)