Q. 11 - Calculus

Question

In the process of solving the differential equation $\frac{dy}{dx}=1-y$ by separation of variables, we obtain the equation $-\ln |1-y|=x+C$. After solving for $y$, this equation becomes $y=1-A{e}^{-x}$. Given that $y>1$, how is A related to $C$?

Step-by-Step Solution

Answer

Ans:   $A$ and $C$ are related by the relation $A=-e^{-C}$

Step 1. Given information.

given equations,

     $\frac{dy}{dx}=1-y$   ,   $-\ln |1-y|=x+C$   ,    $y=1-A{e}^{-x}$

Step 2. The differential equation d y d x = 1 - y is solved by applying the variable separable method. After separating the variables and integrating both sides yields

   $\begin{array}{r}-\ln |1-y|=x+C\\ \ln |1-y|=-x-C\\ |1-y|=e^{-x-C}\end{array}$

Note that $y>1$, and the exponential function is always positive, So the above result leads to the conclusion

      $\begin{array}{r}y-1=e^{-x-c}\\ y=1+e^{-x-c}\\ =1+e^{-x}\cdot e^{-c}\end{array}$

 

Step 3.

In the above result, since $e^{-C}$ is a constant, replace it with another constant $A$, so that the solution is written in compact form as $y=1-A{e}^{-x}$. Thus, the two constants $A$ and $C$ are related by the relation $A=-e^{-C}$