Question

How do you find the general solution to $\dfrac{{dy}}{{dx}} = {e^{x - y}}$?

Answer 1

Hint: The given equation is a differential equation. A differential equation is an equation which involves the derivatives of a variable (which is a dependent variable) with respect to another variable (which is an independent variable).
$\dfrac{{dy}}{{dx}} = f(x)$.
Here, $y$ is the dependent variable
$x$ is the independent variable
and $f(x)$ is a function in terms of the independent variable $x$.
A general solution of ${n^{th}}$ order differential equation can be said to be the solution that includes $n$ arbitrary constants. We can find the general solution of this differential equation by integrating both sides. The general solution of a differential equation is the relation between the x and y variable, that is obtained after the derivatives have been eliminated, where the relationship requires arbitrary constants to describe an equation's order.

Complete step by step answer:
We have to find the general solution to the equation
$\dfrac{{dy}}{{dx}} = {e^{x - y}}$
Using the law of exponents, we get:
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{{e^x}}}{{{e^y}}}$
We will use a variable separable method where we will separate the terms of a particular variable on each side of the equation. We can write the above equation as:
\[ \Rightarrow {e^y}dy = {e^x}dx\]
Now, taking integration on both the sides, we can write:
\[
  {e^y}dy = {e^x}dx \\
   \Rightarrow \int {{e^y}dy} = \int {{e^x}dx} \\
\]
We can solve the above integration as follows,
Let us first solve \[\int {{e^x}dx} \]
Let \[{e^x} = t\]
Then on differentiating both sides we get,
${e^x}dx = dt$
Thus we can write,
\[\int {{e^x}dx} = \int {dt} = t + C = {e^x} + C\]
Similarly, we can get,
\[\int {{e^y}dy} = {e^y} + C\]
Thus we get,
\[
  \int {{e^y}dy} = \int {{e^x}dx} \\
   \Rightarrow {e^y} = {e^x} + C \\
\]
where $C$ is any arbitrary constant.

Hence, the general solution to $\dfrac{{dy}}{{dx}} = {e^{x - y}}$ is given as \[{e^y} - {e^x} = C\], where $C$ is the arbitrary constant.

Note: We have used the variable separable method here to solve the question. In the variable separable method we try to separate all the terms of a particular variable on one side of the equation and then integrate both sides to find the solution. Any indefinite integration involves a constant in the solution.