Question

For the differential equations , find the general solution:$\frac {dy}{dx}+y=1, \ \ (y≠1)$

Answer 1

The given differential equation is:

$\frac {dy}{dx}+y=1$

$⇒dy+y dx=dx$

$⇒dy=(1-y)dx$

Separating the variables, we get:

$\frac {dy}{1-y}=dx$

Now, integrating both sides, we get:

$∫\frac {dy}{1-y}=∫dx$

$⇒log\ (1-y)=x+log\ C$

$⇒-log\ C-log\ (1-y)=x$

$⇒log\ C(1-y)=-x$

$⇒C(1-y)=e^{-x}$

$⇒1-y=\frac {1}{C}e^{-x}$

$⇒y=1+Ae^{-x }$ $(where \ A=-\frac 1C)$

This is the required general solution of the given differential equation.