Question

\(x y^{2} \frac{d y}{d x}=1-x^{2}+y^{2}-x^{2} y^{2}\)

Short answer

Based on the step-by-step solution above, create a short answer for the given first-order differential equation. Short Answer: To solve the given first-order differential equation \(x y^{2} \frac{dy}{dx} = 1 - x^2 + y^2 - x^2 y^2\), we first rewrite the equation and use the substitution \(v = y^2\). This results in a first-order linear differential equation in terms of \(v(x)\). We can then use an integrating factor to solve for \(v(x)\). Finally, substitute back \(v = y^2\) and solve for the solution for the original equation.

Step-by-step solution

Analyze the given equation

First, let's examine the given equation: \(x y^{2} \frac{dy}{dx} = 1 - x^2 + y^2 - x^2 y^2\) Notice that this is a first-order differential equation in the form of \(\frac{dy}{dx} = f(x, y)\), and it does not look like it can be directly separated into variables nor fit into a more apparent form like linear-first order or exact-first order. However, we can try to rewrite the equation to see if we can find a substitution that will help us solve the differential equation.

Rewrite the equation

We can factor out terms in the given equation, which can potentially help us identify a substitution: \(x y^{2} \frac{dy}{dx} = (1 - x^2) + (1 - x^2)y^2\) Rewrite it as: \(\frac{dy}{dx} = \frac{1 - x^2 + (1 - x^2)y^2}{xy^2}\) Now we can use substitution. Let \(v = y^2\), then we have: \(\frac{dv}{dx} = 2y\frac{dy}{dx}\) And the given equation becomes: \(\frac{1}{x} \frac{dv}{dx} = \frac{2(1 - x^2) + (1 - x^2)v}{2}\) Next, we can simplify this equation:

Simplify the equation

We now need to simplify the equation to make the solving process easier: \(\frac{1}{x} \frac{dv}{dx} = \frac{2 - 2x^2 + v - x^2v}{2}\) Multiply both sides by \(x\): \(\frac{dv}{dx} = x [2 - 2x^2 + v - x^2v]\) Now, it is a first-order linear differential equation in terms of \(v(x)\) and can be solved using an integrating factor.

Solve the first-order linear differential equation

We will now solve the following first-order linear differential equation: \(\frac{dv}{dx} + (x^3)v = 2x - 2x^3\) The integrating factor can be found by finding the function \(\mu(x)\), where: \(\mu(x) = e^{\int{x^3}dx}\) Compute the integral: \(\mu(x) = e^{\frac{1}{4}x^4}\) Now, we can multiply both sides of the equation by \(\mu(x)\): \(e^{\frac{1}{4}x^4}\frac{dv}{dx} + x^3 e^{\frac{1}{4}x^4}v = 2x e^{\frac{1}{4}x^4} - 2x^3 e^{\frac{1}{4}x^4}\) The left-hand side of the equation is now the derivative of the product: \(\frac{d}{dx}[ve^{\frac{1}{4}x^4}] = 2xe^{\frac{1}{4}x^4} - 2x^3e^{\frac{1}{4}x^4}\) Integrate both sides with respect to \(x\): \(ve^{\frac{1}{4}x^4} = \int(2xe^{\frac{1}{4}x^4} - 2x^3e^{\frac{1}{4}x^4})dx\) At this point, we can use integration by parts to find the integral of each term. After integrating, we solve for \(v(x)\), and eventually substitute back \(v = y^2\) to find the solution for the original equation.

Key concepts

Integrating Factor Method

The integrating factor method is a powerful technique to solve first-order linear differential equations, which are equations of the form \(\frac{dy}{dx} + p(x)y = q(x)\). In these cases, the integrating factor, denoted by \(\mu(x)\), is an exponential function of the integral of \(p(x)\). Specifically, \(\mu(x) = e^{\int p(x)dx}\).

To apply the integrating factor method, you multiply every term of the differential equation by \(\mu(x)\), which allows you to express the left side of the equation as the derivative of the product of \(\mu(x)\) and \(y\). This simplifies the equation, making it easier to solve since you can then integrate both sides with respect to \(x\). Finally, you isolate \(y\) to get the solution to the original differential equation. The utility of the integrating factor lies in its ability to turn a difficult-to-solve differential equation into a form where direct integration is possible.

Integration by Parts

Integration by parts is a technique based on the product rule for differentiation and is used when you are dealing with the integral of the product of two functions, say \(u\) and \(dv\). The formula for integration by parts is \(\int u dv = uv - \int v du\).

The method involves choosing \(u\) and \(dv\) intelligently such that \(du\) and \(v\) (after integrating \(dv\)) lead to an integral that is simpler than the original. This method is particularly useful when you encounter an integral that is the product of a polynomial and an exponential function or trigonometric functions within the integrating factor method. For the equation in the exercise, after applying the integrating factor, you are left with integrals that can be solved by employing integration by parts, further facilitating the finding of a solution to the differential equation.

Substitution Method in Differential Equations

The substitution method in differential equations involves introducing a new variable, say \(v\), to replace an existing expression involving the original variables, often with the goal of simplifying the equation. This method is helpful when you spot patterns or relationships between the variables that can be exploited.

For example, if \(y\) is raised to a power or a function of \(y\) is multiplied by \(\frac{dy}{dx}\), you might choose \(v = y^2\) or a similar substitution. After substituting, the key is to express every part of the original differential equation in terms of the new variable and \(\frac{dv}{dx}\), simplifying to a form that is more manageable. When the new equation is solved for \(v\), you must remember to substitute back to the original variable to obtain the final solution to the problem. This method is especially beneficial when the original equation does not easily fit into other standard forms for differential equations.