Solving the Differential Equation xy-1 dy xy1 dx: A Comprehensive Guide
Understanding and solving differential equations is a fundamental skill in mathematics, finding applications in various fields such as physics, engineering, and even economics. This article will guide you through the process of solving a specific type of differential equation: xy-1 dy xy1 dx. We'll explore the method, step-by-step, and provide insights into the mathematical tools used.
Standard Form and Separation of Variables
To begin with, let's convert the given differential equation into a more standard form:
xy-1 dy xy1 dx
By rearranging the terms, we can transform the equation as follows:
xy-1 dy - xy1 dx 0
This allows us to express the derivative dy/dx as:
frac{dy}{dx} frac{xy1}{xy-1}
Since this is a separable differential equation, we can separate the variables by rewriting it:
frac{dy}{dx} frac{xy1}{xy-1}
We can further manipulate this to separate the variables:
dy frac{xy1}{xy-1} dx
To solve this, let's denote z xy. Then, using the chain rule, we have:
frac{dy}{dx} frac{dz - dx}{dx} frac{z - 1}{z - 1} dx
Substituting back, we get:
dz - dx frac{z - 1}{z - 1} dx
Rearranging:
dz left(1 frac{z - 1}{z - 1}right) dx
Further simplifying:
1 frac{z - 1}{z - 1} frac{2z}{z - 1}
Thus, we have:
dz frac{2z}{z - 1} dx
We can now separate variables:
frac{dz}{2z} frac{dx}{z - 1}
Integrating both sides:
Integration
int frac{1}{2z} dz int frac{1}{z - 1} dx
This leads to:
frac{1}{2} ln z ln z - ln (z - 1) C
Exponentiating both sides, we obtain:
z^{1/2} k (z - 1)
Where:
k e^{2C} text{ is a constant.}
Substituting back z xy, we get:
sqrt{xy} k (xy - 1)
This equation represents the implicit solution to the original differential equation. Depending on the value of the constant k, you may want to manipulate this further or solve for y explicitly in terms of x.
Another Approach
Another way to solve this differential equation is as follows:
xy - 1 dy xy dy - dx
By dividing both sides by xy, we get:
frac{dy - dx}{xy} frac{dy dx}{xy}
Applying integration on both sides:
int{(dy - dx)} int{frac{dy dx}{xy}}
This results in:
y - x ln{xy} - C
Therefore:
y x ln{xy} C
Alternatively, let xy V:
Substitution Method
Let:
xy V
1 frac{dy}{dx} frac{dV}{dx}
Substituting into the original differential equation:
frac{dV}{dx} - 1 frac{V 1}{V - 1}
Further simplifying:
frac{dV}{dx} frac{V 1}{V - 1} 1
Which simplifies to:
frac{V - 1}{V} dV 2 dx
Applying integration on both sides:
int{(1 - frac{1}{V}) dV} int{2 dx}
This leads to:
V - ln{V} 2x C
But we have:
V xy
Thus:
xy - ln{xy} 2x C
Further simplification yields:
y x ln{xy} C
This solution represents the same implicit form of the original differential equation.
Conclusion
The methodology illustrated in this article highlights the importance of separation of variables and substitution in solving differential equations. Both approaches we've discussed—using the basic form and applying substitution—are valid and can be used depending on the problem at hand. The final form of the solution:
y x ln{xy} C
represents the implicit solution to the differential equation xy-1 dy xy1 dx.