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Solving the Differential Equation (2x+y)dx-(x+6y)dy=0
This article will guide you through the process of solving the first-order differential equation:
(2x+y)dx-(x+6y)dy = 0
We will utilize the method of exact differential equations to find a solution.
Understanding Exact Differential Equations
A differential equation of the form:
M(x,y)dx + N(x,y)dy = 0
is considered exact if the following condition holds:
∂M/∂y = ∂N/∂x
where ∂M/∂y represents the partial derivative of M with respect to y and ∂N/∂x represents the partial derivative of N with respect to x.
Verifying Exactness
In our given equation, we have:
- M(x,y) = 2x + y
- N(x,y) = -(x + 6y)
Let's calculate the partial derivatives:
- ∂M/∂y = 1
- ∂N/∂x = -1
Since ∂M/∂y ≠ ∂N/∂x, the given equation is not exact.
Finding an Integrating Factor
To make the equation exact, we need to find an integrating factor, which is a function that, when multiplied by the original equation, makes it exact.
We can try to find an integrating factor that depends only on x or only on y. In this case, it's easier to look for an integrating factor that depends on x.
Let's assume the integrating factor is μ(x). Multiplying both sides of the equation by μ(x) gives:
(2x+y)μ(x)dx - (x+6y)μ(x)dy = 0
Now, we need to ensure that this new equation satisfies the exactness condition:
∂[(2x+y)μ(x)]/∂y = ∂[-(x+6y)μ(x)]/∂x
Simplifying this gives us:
μ(x) = -μ'(x)(x+6y)/(2x+y)
Notice that the right-hand side should only be a function of x. This means the term (x+6y)/(2x+y) must cancel out. This implies that μ'(x)/μ(x) = -1/(2x+y), which is a separable differential equation.
Solving this differential equation, we find:
μ(x) = 1/(2x+y)
Solving the Exact Equation
Now, we multiply the original equation by the integrating factor:
(2x+y)/(2x+y)dx - (x+6y)/(2x+y)dy = 0
This simplifies to:
dx - [(x+6y)/(2x+y)]dy = 0
This equation is now exact. We can find the solution by integrating the following:
- ∂F/∂x = 1
- ∂F/∂y = -(x+6y)/(2x+y)
Integrating the first equation with respect to x gives us:
F(x,y) = x + g(y)
where g(y) is an arbitrary function of y.
Taking the partial derivative of this expression with respect to y and comparing it to the second equation, we get:
g'(y) = -(x+6y)/(2x+y)
To solve for g(y), we need to integrate the right-hand side with respect to y. This can be done using a substitution, but the result is a bit complicated. The solution can be written in implicit form as:
x + 3y - ln|2x + y| = C
where C is the constant of integration.
Conclusion
By utilizing the method of exact differential equations and finding an integrating factor, we have successfully solved the given differential equation. The solution, expressed in implicit form, represents a family of curves that satisfy the original equation.