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Solving the Differential Equation (1+x^2)dy + (xy + x^3 + x)dx = 0
This article will explore the solution to the given differential equation:
(1 + x^2)dy + (xy + x^3 + x)dx = 0
This is a first-order differential equation that can be solved using the method of exact differential equations.
Identifying an Exact Differential Equation
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
Let's apply this to our equation:
- M(x,y) = xy + x^3 + x
- N(x,y) = 1 + x^2
Calculating the partial derivatives:
- ∂M/∂y = x
- ∂N/∂x = 2x
Since ∂M/∂y ≠ ∂N/∂x, the given differential equation is not exact in its current form.
Making the Equation Exact
To make the equation exact, we can find an integrating factor, denoted by μ(x,y), which satisfies the following condition:
(∂(μM)/∂y) = (∂(μN)/∂x)
In this case, since the given equation only involves x, we can use an integrating factor that depends only on x: μ(x).
The condition for finding the integrating factor becomes:
(μN)' = (μM)'
Expanding this, we get:
μ'N + μN' = μ'M + μM'
Since N and M are functions of x and y, we can rewrite this as:
μ'(N - M) = μ(M' - N')
Simplifying and rearranging, we get:
μ'/μ = (M' - N') / (N - M)
The right side of this equation should only involve x, which is the case in our equation. Let's calculate the right side:
(M' - N') / (N - M) = (y + 3x^2 + 1 - 2x) / (1 + x^2 - xy - x^3 - x)
This simplifies to:
(M' - N') / (N - M) = -1/x
Therefore, we have:
μ'/μ = -1/x
Integrating both sides with respect to x, we get:
ln(μ) = -ln(x)
Solving for μ, we obtain:
μ(x) = 1/x
Solving the Exact Differential Equation
Now, we multiply the original differential equation by the integrating factor μ(x):
(1/x)(1 + x^2)dy + (1/x)(xy + x^3 + x)dx = 0
This simplifies to:
(1/x + x)dy + (y + x^2 + 1)dx = 0
Now, the equation is exact. We can proceed to find a solution.
Since the equation is exact, we know there exists a function F(x,y) such that:
dF = (∂F/∂x)dx + (∂F/∂y)dy = 0
Comparing this to our exact equation, we have:
- ∂F/∂x = y + x^2 + 1
- ∂F/∂y = 1/x + x
Integrating the first equation with respect to x, we get:
F(x,y) = xy + (1/3)x^3 + x + g(y)
where g(y) is an arbitrary function of y.
Differentiating F(x,y) with respect to y, we get:
∂F/∂y = x + g'(y)
Comparing this with the second equation, we get:
x + g'(y) = 1/x + x
Therefore, g'(y) = 1/x. However, since g(y) should only depend on y, this implies that g(y) is a constant.
Finally, the solution to the given differential equation is given by:
F(x,y) = xy + (1/3)x^3 + x + C = 0
where C is an arbitrary constant.
This implicit equation represents the family of solutions to the given differential equation.