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Question: How do you find the general solution to \(\dfrac{{dy}}{{dx}} = {e^{x - y}}\)?...

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

Explanation

Solution

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).
dydx=f(x)\dfrac{{dy}}{{dx}} = f(x).
Here, yy is the dependent variable
xx is the independent variable
and f(x)f(x) is a function in terms of the independent variable xx.
A general solution of nth{n^{th}} order differential equation can be said to be the solution that includes nn 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
dydx=exy\dfrac{{dy}}{{dx}} = {e^{x - y}}
Using the law of exponents, we get:
dydx=exey\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:
eydy=exdx\Rightarrow {e^y}dy = {e^x}dx
Now, taking integration on both the sides, we can write:

eydy=exdx eydy=exdx  {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 exdx\int {{e^x}dx}
Let ex=t{e^x} = t
Then on differentiating both sides we get,
exdx=dt{e^x}dx = dt
Thus we can write,
exdx=dt=t+C=ex+C\int {{e^x}dx} = \int {dt} = t + C = {e^x} + C
Similarly, we can get,
eydy=ey+C\int {{e^y}dy} = {e^y} + C
Thus we get,

eydy=exdx ey=ex+C  \int {{e^y}dy} = \int {{e^x}dx} \\\ \Rightarrow {e^y} = {e^x} + C \\\

where CC is any arbitrary constant.

Hence, the general solution to dydx=exy\dfrac{{dy}}{{dx}} = {e^{x - y}} is given as eyex=C{e^y} - {e^x} = C, where CC 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.