Question

Solution of the differential equation \(\left(2 x-10 y^{3}\right) \frac{d y}{d x}+y=0\) is (1) \(\left(x-2 y^{3}\right) \cdot y^{2}=c\) (2) \(\left(x-\frac{2 y^{3}}{5}\right) y^{2}=c\) (3) \(\left(x+2 y^{3}\right) \cdot y^{2}=c\) (4) \(\left(x+\frac{2 y^{3}}{5}\right) y^{2}=c\)

Short answer

The correct solution is \ \(left\( x - \frac {2y^3}{5}) y^{2} = Const\

Step-by-step solution

Rewrite the given differential equation

The given differential equation is \[(2x - 10y^3) \frac{dy}{dx} + y = 0 \].\ Rewrite it to isolate \( \frac{dy}{dx} \):\[(2x - 10y^3) \frac{dy}{dx} = -y \].\ \[ \frac{dy}{dx} = \frac{-y}{2x - 10y^3} \].

Separate the variables

Separate the variables so that all terms involving \(y\) are on one side and all terms involving \(x\) are on the other side:\ \[ \frac{dy}{y} = \frac{-dx}{2x - 10y^3} \].

Integrate both sides

Integrate both sides of the equation:\ \[ \int \frac{1}{y} dy = \int \frac{-1}{2x - 10y^3} dx \].

Simplify the integral

Finding the integral can be tricky, but we notice a specific substitution can simplify this. Let's substitute \[ u = x - 2y^3 \].\ Then, \[ du = dx - 6y^2 dy \]. Given the separation of variables and integrating accordingly will yield the solution as:\ \[ \frac{d}{dy} (x - 2y^3) = \frac{dx}{2y^2}, \].\ Using this substitution simplifies and integrates to:\ \[ ln|y| = \frac{1}{5}ln|x - \frac{2y^3}{5}| + C \]

Exponentiate to solve for y

Exponentiating both sides removes the logarithms and puts the equation in a solvable form\ \[ y = e^{C} (left term form) \]; thus, we see the cleaned result becomes\ \left( x - \frac{ 2y^3}{5} \right) y^{2} = Const (integrating factor of solution.)

Key concepts

Simplifying Complex Differential Equations through Separation of Variables

The 'separation of variables' technique is a powerful method used to tackle first-order differential equations. The goal is to rearrange the equation so that each variable appears on a different side of the equation.

This approach starts by isolating the derivative term, such as \(\frac{dy}{dx}\), making it easier to split the variables.
For instance, in the given equation:
\((2x - 10y^3) \frac{dy}{dx} = -y\)
We can rewrite it as:
\(\frac{dy}{y} = \frac{-dx}{2x - 10y^3}\)
Notice that all terms involving y are now on one side, and all terms involving x are on the other side.
After separating the variables, the next step is to integrate both sides. This leads us into the field of integration, another core concept in solving these equations efficiently.

Mastering Integration in Mathematics for Differential Equations

Integration is a mathematical process of finding the antiderivative or the area under a curve. In the context of differential equations, integration allows us to solve for the function that satisfies the equation.

When we have our separated variables, like:
\(\frac{dy}{y} = \frac{-dx}{2x - 10y^3}\)
We integrate both sides to find the unknown function.
\(\[\begin{equation} \int \frac{1}{y} dy = \int \frac{-1}{2x - 10y^3} dx \end{equation}\]\)

Finding these integrals often requires recognizing special patterns or using clever methods.
For instance, the integral of \(\frac{1}{y} dy\) simplifies nicely to \( ln|y|\).
However, the right side might be more complicated, requiring substitutions to simplify.

Substitution Method: Simplifying Integrals and Differential Equations

The substitution method is a technique used to simplify complex integrals by introducing a new variable. This makes the integral more manageable.

In our given differential equation:
\(\frac{dy}{dx} = \frac{-y}{2x - 10y^3}\),
We notice that substituting \(u = x - 2y^3\) can simplify the equation.
We have:
\(du = dx - 6y^2 dy\)
Though this substitution simplifies our integration significantly:
\( \frac{du}{dy} = \frac{dx}{2y^2} \),
We continue by integrating both sides and solving for the variable.
This is a powerful method that helps us deal with otherwise intractable integrals and differential equations, revealing solutions that are easier to interpret.