Question

Solve the given differential equations. $$ x \frac{d y}{d x}+y=-2 x^{6} y^{4} $$

Short answer

The solution to the given differential equation \(x \frac{d y}{d x}+y=-2 x^{6} y^{4}\) is \(y(x) = \left(\frac{1}{\frac{1}{2}x^6 - 3C + 3\ln|x|}\right)^{\frac{1}{3}}\).

Step-by-step solution

Rearrange and Separate Variables

We are given the differential equation: \( x \frac{dy}{dx} + y = -2x^6y^4 \). To separate variables, divide both sides by \(x y^4\) and then multiply both sides by \(dx\): \( \frac{1}{y^4} \frac{dy}{dx} +\frac{1}{x} = -2x^5 dx\).

Integrate both sides

Now, we will integrate both sides of the equation with respect to x: \( \int \frac{1}{y^4} \frac{dy}{dx} dx + \int \frac{1}{x} dx = \int -2x^5 dx \). Let's integrate each term: \( \int y^{-4} dy + \int \frac{1}{x} dx = \int -2x^5 dx \).

Compute the integrals

Compute the integrals: \( -\frac{1}{3}y^{-3} + \ln|x| = -\frac{1}{6}x^6 + C \), where C is the constant of integration.

Solve for y(x)

Now we want to solve for \( y(x) \). We can do this by isolating the \( y \) term: \( -\frac{1}{3}y^{-3} = -\frac{1}{6}x^6 + C - \ln|x| \), \( y^{-3} = 3\left(\frac{1}{6}x^6 - C + \ln|x|\right) \), \( y(x) = \left(\frac{1}{\frac{1}{2}x^6 - 3C + 3\ln|x|}\right)^{\frac{1}{3}} \). The solution to the given differential equation is: \( y(x) = \left(\frac{1}{\frac{1}{2}x^6 - 3C + 3\ln|x|}\right)^{\frac{1}{3}} \).

Key concepts

Separation of Variables

Separation of Variables is a fundamental technique used in solving differential equations. It involves rearranging an equation so that each variable appears on only one side. This makes it easier to integrate. In this exercise, we started with the differential equation \( x \frac{dy}{dx} + y = -2x^6y^4 \).
To separate the variables, divide by \( xy^4 \) and multiply through by \( dx \), leading to the equation \( \frac{1}{y^4} dy + \frac{1}{x} dx = -2x^5 dx \).

This allows us to clearly distinguish the parts of the equation containing \( y \) and \( x \), preparing for the next step of integration.

Integration

Integration is the process of finding the integral of a function, essentially reversing differentiation. Once the variables are separated, you integrate each side independently. For the equation from our exercise,
we integrate:

• \( \int y^{-4} dy \)
• \( \int \frac{1}{x} dx \)
• \( \int -2x^5 dx \)

Integrating each term results in:

• \( -\frac{1}{3}y^{-3} \) for \( \int y^{-4} dy \)
• \( \ln|x| \) for \( \int \frac{1}{x} dx \)
• \( -\frac{1}{6}x^6 \) for \( \int -2x^5 dx \)

These expressions are then combined to form one equation, with a constant of integration added to account for any potential vertical shift.

Solving Differential Equations

Once you've integrated and combined all terms, you solve the resulting equation for one of the variables. Here is where it's crucial to manipulate the equation correctly. From our integrated expressions, we have the equation:
\( -\frac{1}{3}y^{-3} + \ln|x| = -\frac{1}{6}x^6 + C \).
We solve for \( y(x) \) by first isolating \( y^{-3} \), then finding \( y \) itself. Step by step manipulation leads us to:

• \( y^{-3} = 3\left(\frac{1}{6}x^6 - C + \ln|x|\right) \)
• Finally yielding \( y(x) = \left(\frac{1}{\frac{1}{2}x^6 - 3C + 3\ln|x|}\right)^{\frac{1}{3}} \)

This method results in a function \( y(x) \), the solution to our original differential equation.

Constant of Integration

The constant of integration, often represented as \( C \), plays a crucial role when solving indefinite integrals. It represents an unknown constant that could have been present before differentiation, capturing the family of all potential solutions.
In our case, after integration, we expressed \( C \) within our combined equation:
\( -\frac{1}{3}y^{-3} + \ln|x| = -\frac{1}{6}x^6 + C \).
This constant can be adjusted based on any initial conditions provided in related problems, ensuring our solution fits specific scenarios or constraints imposed initially. Without it, the equation loses generality, failing to provide all possible solutions.