Question

Solve the given differential equations. $$ d y+\left(4 y-8 y^{-3}\right) x d x=0 $$

Short answer

The general solution for the given differential equation \(dy + (4y - 8y^{-3})x\,dx = 0\) is given by: $$ y(x) = e^{-4x}(-8 e^{4x} g(y^{-3}) + C) $$ where \(g(y^{-3})\) is a function that depends on the specific initial conditions provided for the differential equation.

Step-by-step solution

Rewrite the equation in standard form

To rewrite the given equation in the standard form, we need to separate the terms involving \(y\) and its derivatives to one side. Divide both sides by \(x dx\): $$ \frac{dy}{dx} + \left(4y - 8 y^{-3}\right) = 0 $$ Now, identify the functions \(P(x)\) and \(Q(x)\) in the equation: $$ P(x) = 4 , \hspace{10mm} Q(x) = -8 y^{-4} $$

Calculate the integrating factor

The integrating factor, denoted by \(\mu(x)\), is given by the formula: $$ \mu(x) = e^{\int P(x) dx} $$ Compute the integral: $$ \int P(x) dx = \int 4 dx = 4x + C $$ We only need the function, not the constant, so the integrating factor \(\mu(x)\) is: $$ \mu(x) = e^{4x} $$

Multiply equation by the integrating factor

We need to multiply the entire differential equation by the integrating factor in order to make the left-hand side an exact derivative. Do so: $$ e^{4x} \frac{dy}{dx} + 4 e^{4x} y = -8 e^{4x} y^{-3} $$ Now, the left side of the equation should be the derivative of the product \(\mu(x) y(x)\), which is \(e^{4x} y\). Check that it is the case: $$ \frac{d}{dx}(e^{4x} y) = e^{4x} \frac{dy}{dx} + 4 e^{4x} y $$ Indeed, the left side is an exact derivative.

Integrate both sides

Now that we have the proper derivative form on the left side, we can integrate both sides of the equation with respect to \(x\): $$ \int \frac{d}{dx} (e^{4x} y) dx = \int -8 e^{4x} y^{-3} dx $$ Perform the integration: $$ e^{4x} y = \int -8 e^{4x} y^{-3} dx + C $$ To integrate the right side, notice that we cannot integrate it directly as it involves both \(x\) and \(y\). Instead, we will treat it as a new function that we will solve for later. Call this new function \(F(x, y)\): $$ e^{4x} y = F(x, y) + C $$

Solve for y and find F(x, y)

Solve the equation for \(y(x)\): $$ y(x) = e^{-4x}(F(x, y) + C) $$ Now we need to find the function \(F(x, y)\). Notice that F(x, y) should satisfy: $$ \frac{dF}{dx}=-8 y^{-3} e^{4x} $$ To satisfy the dependence between \(x\) and \(y\), we can assume: $$ F(x,y) = -8 e^{4x} y^{-3} g(y) = -8e^{4x} g(y^{-3}) $$ for some function \(g(y^{-3})\). This can be verified by differentiating with respect to x and noticing the given term: $$ \frac{dF}{dx} = -32 e^{4x} y^{-3} g(y^{-3}) $$ We can now rewrite the general solution of the equation as: $$ y(x) = e^{-4x}(-8 e^{4x} g(y^{-3}) + C) $$ The function \(g(y^{-3})\) will depend on the specific initial conditions provided for the differential equation.

Key concepts

Integrating Factor

In solving differential equations, particularly linear first-order ones, the integrating factor is a critical tool. It helps transform a non-exact equation into an exact one, making it easier to solve. The integrating factor, \( \mu(x) \), is calculated with the formula \( e^{\int P(x) \, dx} \). This factor, when multiplied throughout the equation, allows us to treat the left side as a derivative of a product.
This transformation is significant because it simplifies the equation into a form we can integrate easily. For example, in our original problem, the integrating factor was found to be \( e^{4x} \).

• This factor converts the left-hand side into the derivative of \( e^{4x} y \).
• Ensures that the differential equation is now something we can evaluate for a solution.

Thus, always check a differential equation's structure to determine if an integrating factor will simplify the solution process.

Exact Differential Equations

Exact differential equations are types of equations where a specific condition allows them to be solved by integration. This condition states that there exists a function, say \( F(x, y) \), for which the total differential \( dF = 0 \) emerges when combined with the equation's terms.
In the context of our problem, after determining the integrating factor and rewriting the equation, it becomes exact. This means:

• You can identify a function involving derivatives of both \( y \) and \( x \) that equals zero.
• The solution can then be obtained by integrating this exact differential.

Once you achieve this form, solving becomes a straightforward process of unwinding the derivative through integration. Recognizing the transformation to an exact equation is vital as it sets up the right steps to reach the final solution efficiently.

Separation of Variables

Separation of Variables is a fundamental concept used to solve certain types of differential equations. It involves separating the variables \( x \) and \( y \) onto different sides of the equation to facilitate integration. However, this method may not always be directly applicable, particularly if the equation doesn’t initially allow such separation.
In our exercise, we didn’t use this method directly, but the principles behind it are worth understanding:

• Separate \( dy \) and \( dx \) terms so both variables are with their respective differential.
• Integrate both sides to solve the equation.
• This method is often more applicable to simpler forms of differential equations compared to the one we solved.

It is important to recognize when this tactic can be applied and when an integrating factor or other approaches are more suitable.

Initial Conditions

Initial conditions provide specific values for the variables in a differential equation, typically ensuring a unique solution. They serve as boundary conditions that specify what the solution function must satisfy at a particular point. When solving differential equations, these conditions can be used to determine unknown constants or particular solutions.
With reference to our exercise, initial conditions were implicitly mentioned in the context of determining the function \( g(y^{-3}) \). While not explicitly given in the problem, usually:

• Initial conditions help narrow down the solution to one specific function among infinitely many possibilities.
• They provide critical evaluations that make the solution fit real-world scenarios or specified constraints.

In practice, once a general solution is obtained, plug in the initial conditions to find specific values for any constants included in the solution.