Question

Find the general solution to each differential equation. Try some by calculator. $$y^{\prime}=\frac{x^{2}}{y^{3}}$$

Short answer

The general solution to the differential equation is: y = 1 \textpm C' * e^{-(x^3)/3}, where C' is an arbitrary constant.

Step-by-step solution

Identify the Type of Differential Equation

The given differential equation is of the form: dy/dx = f(x, y), which suggests that it may be separable. A separable differential equation is one that can be written in the form g(y)dy = h(x)dx. To solve this, we need to see if we can separate the variables y and x.

Separate the Variables

Rearrange the equation to separate the variables y and x on different sides of the equation. The given differential equation is: y' = x^2 - x^2*y. This can be rewritten and separated as: (1/(1 - y))dy = x^2dx.

Integrate Both Sides

Now, integrate both sides with respect to their own variables to find the general solution. The integral of 1/(1 - y) with respect to y is -ln|1 - y|, and the integral of x^2 with respect to x is (x^3)/3. So, -ln|1 - y| = (x^3)/3 + C, where C is the constant of integration.

Solve for y

Solve the integrated equation for y to express the general solution. Take the exponential of both sides to eliminate the natural logarithm: e^{-ln|1 - y|} = e^{(x^3)/3 + C}. Simplifying, we get |1 - y| = e^{-C} * e^{-(x^3)/3}. The absolute value indicates two possibilities: 1 - y = e^{-C} * e^{-(x^3)/3} or 1 - y = -e^{-C} * e^{-(x^3)/3}. Let C' = e^{-C} since C is an arbitrary constant. So the general solution will be: y = 1 - C' * e^{-(x^3)/3} and y = 1 + C' * e^{-(x^3)/3}.

Key concepts

Separable Differential Equations

Understanding separable differential equations is essential for tackling a variety of problems in calculus and differential equations coursework. A differential equation is considered separable if it can be arranged into a form where all terms involving the dependent variable (typically denoted as y or f(x)) and its differentials are on one side of the equation, and all terms involving the independent variable (usually x) and its differentials are on the other side.

Let's see this in action with the provided example: \( y' = x^{2} - x^{2} y \). When rearranged, we obtain \( \frac{1}{1 - y}dy = x^{2}dx \) which clearly displays the variables y and x on separate sides. This separation allows us to integrate each side independently, which is a powerful technique for solving these kinds of equations.

Why is Separation Powerful?

Separation of variables leverages the fundamental theorem of calculus, which connects the concepts of differentiation and integration. By isolating each variable with its differential, we set up two integrals that can be solved to find the general solution to the original differential equation, offering a straightforward path to solution.

Integrating Factors

In many cases of solving differential equations, the terms may not separate easily, or the equation might not be written in a recognizable separable form. This is where integrating factors come into play – they are a method used to transform a non-separable equation into one that can be integrated more easily.

An integrating factor is typically a function, often denoted by \( \mu(x) \), that, when multiplied with the original differential equation, renders the equation integrable. The magic behind integrating factors lies in their ability to convert the left-hand side of a differential equation into an exact differential. This transformation allows for direct integration of both sides of the equation and ultimately leads to finding a solution.

Though our example did not require the use of an integrating factor, understanding how they function is fundamental for tackling more complex differential equations that are not innately separable.

General Solution of Differential Equations

The ultimate goal of solving a differential equation is often to find its general solution, which encompasses all possible solutions to the equation. For separable equations, after integrating and finding the antiderivatives on each side, one typically arrives at a relationship involving the dependent variable y, the independent variable x, and an arbitrary constant C.

In our exercise, once we performed integration, we obtained \( -\ln|1 - y| = \frac{x^3}{3} + C \). By applying properties of logarithms and exponentials, we can isolate y to find the general solution, revealing all the infinite ways y can behave as a function of x, given different initial conditions reflected in the arbitrary constant C.

Importance of the Arbitrary Constant

The arbitrary constant in the general solution represents the infinite unique solutions that satisfy the original differential equation. By setting initial conditions or specific criteria, one can 'pin down' this constant to find a particular solution, connecting abstract mathematical concepts with practical, real-world applications. It's the combination of these general and particular solutions that make differential equations a powerful tool in modeling and understanding phenomena across science and engineering fields.