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.