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Solving the Differential Equation: (xy^3+y)dx+2(x^2y^2+x+y^4)dy=0
This article will explore the solution of the given differential equation:
(xy^3+y)dx+2(x^2y^2+x+y^4)dy=0
This equation is a first-order, non-linear, and exact differential equation. To solve it, we'll follow these steps:
1. Checking for Exactness
A differential equation of the form M(x,y)dx + N(x,y)dy = 0 is exact if and only if:
∂M/∂y = ∂N/∂x
In our case, M(x,y) = xy^3+y and N(x,y) = 2(x^2y^2+x+y^4).
Let's calculate the partial derivatives:
- ∂M/∂y = 3xy^2 + 1
- ∂N/∂x = 4xy^2 + 2
Since ∂M/∂y ≠ ∂N/∂x, the given differential equation is not exact.
2. Finding an Integrating Factor
To make the equation exact, we need to find an integrating factor. There are two common methods:
- Finding an integrating factor depending only on x: If (∂N/∂x - ∂M/∂y)/M is a function of x alone, then μ(x) = exp[∫((∂N/∂x - ∂M/∂y)/M)dx] is an integrating factor.
- Finding an integrating factor depending only on y: If (∂M/∂y - ∂N/∂x)/N is a function of y alone, then μ(y) = exp[∫((∂M/∂y - ∂N/∂x)/N)dy] is an integrating factor.
Let's try the first method:
(∂N/∂x - ∂M/∂y)/M = (4xy^2 + 2 - 3xy^2 - 1)/(xy^3+y) = (xy^2 + 1)/(xy^3+y) = 1/y
Since this is a function of y alone, we need to use the second method.
μ(y) = exp[∫(1/y)dy] = exp(ln|y|) = |y|
We can choose μ(y) = y (since y is positive in our equation).
3. Multiplying by the Integrating Factor
Multiplying the original equation by the integrating factor y:
y(xy^3+y)dx + 2y(x^2y^2+x+y^4)dy = 0
Simplifying:
(x^4y^4 + xy^2)dx + (2x^2y^3 + 2xy + 2y^5)dy = 0
Now, the equation is exact.
4. Solving the Exact Equation
Since the equation is exact, there exists a function F(x,y) such that:
∂F/∂x = x^4y^4 + xy^2 and ∂F/∂y = 2x^2y^3 + 2xy + 2y^5
Integrating the first equation with respect to x:
F(x,y) = (1/5)x^5y^4 + (1/2)x^2y^2 + g(y)
where g(y) is an arbitrary function of y.
Differentiating this equation with respect to y and comparing it with the second equation:
∂F/∂y = 2x^2y^3 + g'(y) = 2x^2y^3 + 2xy + 2y^5
Therefore, g'(y) = 2xy + 2y^5. Integrating this equation with respect to y:
g(y) = xy^2 + (1/3)y^6 + C
where C is a constant of integration.
Substituting g(y) back into F(x,y):
F(x,y) = (1/5)x^5y^4 + (1/2)x^2y^2 + xy^2 + (1/3)y^6 + C
The general solution of the differential equation is given by:
(1/5)x^5y^4 + (1/2)x^2y^2 + xy^2 + (1/3)y^6 = C
where C is an arbitrary constant.
Conclusion
We have successfully solved the non-exact differential equation by finding an integrating factor and transforming it into an exact equation. The final solution is given in the implicit form as shown above. This method provides a systematic way to solve a large class of differential equations, highlighting the importance of identifying their properties and applying appropriate techniques.