Question

Find an integrating factor and solve the given equation. $$ \left(3 x^{2} y+2 x y+y^{3}\right) d x+\left(x^{2}+y^{2}\right) d y=0 $$

Short answer

If not, explain why. Answer: No, it is not possible to find an integrating factor for the given differential equation. We attempted to find an integrating factor in the form of a function of x, but this led to an equation that did not satisfy the initial condition for an integrating factor, as we could not write the function f(x) independently from y.

Step-by-step solution

Identify the given equation format

The given equation is a first-order inexact differential equation in the form: $$ M(x, y) dx + N(x, y) dy = 0, $$ where $$M(x, y) = 3x^2y + 2xy + y^3$$ and $$N(x, y) = x^2 + y^2$$.

Check for exactness

To check the exactness, calculate the partial derivatives of $$M$$ and $$N$$ with respect to $$y$$ and $$x$$ respectively, and compare them: $$ \frac{\partial M}{\partial y} = 3x^2 + 2x + 3y^2, $$ $$ \frac{\partial N}{\partial x} = 2x. $$ Since the partial derivatives are not equal, $$\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}$$, the given equation is not exact.

Find the integrating factor

The integrating factor can be a function of x or y or both. Let's try f(x) as the integrating factor. The exactness condition for the differential equation with f(x) as the integrating factor is: $$ \frac{\partial}{\partial y}(f(x)M(x,y)) = \frac{\partial}{\partial x}(f(x)N(x,y)). $$ Substitute $$M(x, y)$$ and $$N(x, y)$$ from Step 1: $$ \frac{\mathrm{d}}{\mathrm{d} y}(f(x)(3x^2y+2xy+y^3)) = \frac{\mathrm{d}}{\mathrm{d} x}(f(x)(x^2+y^2)). $$ Calculate the partial derivatives and simplify: $$ 3x^2f'(x) + 6x^2f(x) + 2xf'(x) + 2xf(x) + 6y^2f'(x) = 2xf'(x) + 2y^2f'(x). $$ After canceling out similar terms and grouping the remaining terms, the equation becomes: $$ f(x)(6x^2 + 2x + 3y^2) = 0. $$ Since $$f(x) \neq 0$$, the equation can only be true if the expression in parentheses is equal to zero: $$ 6x^2 + 2x + 3y^2 = 0. $$ However, this equation does not satisfy the initial condition for an integrating factor, as we cannot write the function f(x) independently from y. Thus, we cannot find an integrating factor for this equation in the form of a function f(x). Since we cannot find an integrating factor for the given equation, we cannot solve it.

Key concepts

First-order Inexact Differential Equations

Understanding first-order inexact differential equations is crucial for grasping more complex concepts in calculus. Imagine a basket that is not initially watertight, akin to an inexact differential equation. To hold water, or in math terms, to be 'solvable', it needs to be made watertight. An inexact differential equation doesn't satisfy an integrability condition and can't be resolved directly. Like our leaky basket, it needs an 'integrating factor', a function that, when multiplied with the differential equation, makes it exact—waterproof, so to speak. The general form of a first-order differential equation is represented as:

\[M(x, y) dx + N(x, y) dy = 0,\]
where \(M(x, y)\) and \(N(x, y)\) indicate functions of \(x\) and \(y\). If an equation like this does not meet the exactness condition (which we'll discuss shortly), it is deemed inexact—requiring an additional step to find a solution.

Exactness Condition

The exactness condition is a critical determinant in solving first-order differential equations. Going back to our analogy, the exactness condition checks whether the mathematical 'basket' is watertight. This condition requires that the partial derivatives of \(M\) with respect to \(y\), and \(N\) with respect to \(x\), be equal:

\[\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}.\]
In simple terms, the rate at which \(M\) changes with \(y\) should match the rate at which \(N\) changes with \(x\). If this equality holds, the differential equation is exact and can be solved more directly. However, if it fails to meet this condition, the equation is inexact, and we must search for an integrating factor to make it exact.

Partial Derivatives in Differential Equations

Partial derivatives play a fundamental role in assessing the exactness condition of a differential equation—as essential as knowing the different parts of the basket to find where it leaks. They measure how a function changes with one variable while keeping all other variables constant. In the context of our exercise:

\[\frac{\partial M}{\partial y}\]
and

\[\frac{\partial N}{\partial x}\]
are necessary to determine if the given differential equation is initially solvable or if it necessitates an integrating factor. When the partial derivatives of \(M\) and \(N\) are not equal, it signals that our problem is inexact. Hence, we must probe further for a solution—searching, in essence, for something that will stop our basket from leaking.

Differential Equation Solutions

Finding solutions to differential equations is similar to repairing our basket to make it useful. In exact equations, solutions can be found through direct integration, leading us to a relationship between \(x\) and \(y\) that satisfies the original equation. In inexact cases, an additional step to identify the integrating factor is required. For the problem where\[M(x, y) = 3x^2y + 2xy + y^3\]and\[N(x, y) = x^2 + y^2,\]no integrating factor in terms of a single variable function of \(x\) or \(y\) could be found, implying that this particular basket cannot be patched up simply. At times, equations may require more sophisticated techniques or may not be solvable using elementary methods, which is the harsh reality of the mathematical world—akin to discovering that some baskets are beyond repair.