Solving the Differential Equation dy/dx - 1 e^(x - y) Using Substitution and Separation Methods

Introduction

Differential equations are a fundamental tool in mathematical modeling, widely used in various fields such as physics, engineering, and economics. This article focuses on solving a specific first-order nonlinear ordinary differential equation, namely, dy/dx - 1 e^(x - y). We will employ the method of substitution and separation of variables to derive the general solution. This guide is aligned with Google's indexing standards for comprehensive and well-formatted articles.

Step-by-Step Solution

Given the differential equation frac{dy}{dx} - 1 e^{x - y}, we can rearrange it to isolate frac{dy}{dx}:

frac{dy}{dx} e^{x - y} 1.

This is a first-order nonlinear ordinary differential equation. To simplify it, we can rewrite it as:

frac{dy}{dx} e^x e^{-y} 1.

Let's use the substitution ( v e^{-y} ), which implies ( y -ln v ). Then, we find frac{dy}{dx}:

frac{dy}{dx} -frac{1}{v} frac{dv}{dx}.

Substituting this into our equation, we get:

-frac{1}{v} frac{dv}{dx} e^x v 1.

Multiplying both sides by -v results in:

frac{dv}{dx} -v e^x v - v.

This can be rearranged to:

frac{dv}{dx} -e^x v^2 - v.

This is a separable differential equation. We can separate the variables:

frac{1}{v e^x v 1} dv -dx.

Now we integrate both sides. The left-hand side can be split using partial fractions:

frac{1}{v e^x v 1} frac{A}{v} - frac{B e^x}{v e^x 1}.

To find A and B, we solve:

1 A e^x v B v - A e^x - B.

Setting ( v 0 ) gives ( A 1 ). Now substituting ( A 1 ) back in gives:

1 e^x B v 1 - A e^x.

Setting the coefficient of ( v ) to zero, we find ( B -e^x ). Thus we have:

frac{1}{v e^x v 1} frac{1}{v} - frac{e^x}{v e^x 1}.

Integrating both sides, we get:

int left(frac{1}{v} - frac{e^x}{v e^x 1}right) dv -int dx.

The left side integrates to:

ln v - ln (v e^x 1) -x C,

where ( C ) is the constant of integration. Combining the logarithms, we get:

ln frac{v}{v e^x 1} -x C.

Exponentiating gives:

frac{v}{v e^x 1} Ce^{-x}.

Substituting back ( v e^{-y} ), we obtain:

frac{e^{-y}}{e^x e^{-y} 1} Ce^{-x}.

Cross-multiplying leads to:

e^{-y} Ce^{-x} e^x e^{-y} Ce^{-x}.

This simplifies to find ( y ) in terms of ( x ) and the constant ( C ).

Finally, this gives us a relationship between ( y ) and ( x ) which can be further manipulated to express ( y ) explicitly if needed.

Conclusion

In summary, although the exact solution can be complex, the general solution involves using substitutions and integrating to find a relationship between ( y ) and ( x ).

Key Terms: Differential Equation, Substitution Method, Separation of Variables, Exponential Function

For more information on differential equations and solving techniques, you can explore further reading on online resources or textbooks. This guide is designed to help you understand the step-by-step process of solving this specific type of differential equation, which is a common problem in mathematical modeling and various scientific fields.