Question

Solve the initial value problem and find the interval of validity of the solution. $$ y^{\prime}=-2 x\left(y^{2}-3 y+2\right), \quad y(0)=3 $$

Short answer

Question: Determine the interval of validity for the particular solution of the initial value problem: $$ \frac{dy}{dx}=-2x(y^2-3y+2), \quad y(0) = 3 $$ Answer: The interval of validity for the particular solution of the given initial value problem is \((-\infty, \infty)\).

Step-by-step solution

Separate variables

First, we need to rewrite the given ODE in a form where all the terms involving y are together, and all the terms involving x are together: $$ \frac{1}{y^2 - 3y + 2}\,dy = -2x\,dx $$ Now the variables are separated.

Integrate both sides

Next, we need to integrate both sides of the equation with respect to their respective variables: $$ \int\frac{1}{y^2 - 3y + 2}\,dy = -2\int x\,dx $$ To integrate the left-hand side, perform partial fractions on the fraction. Rewrite the denominator as a product of its factors: $$ \frac{1}{(y-1)(y-2)} = \frac{A}{y-1} + \frac{B}{y-2} $$ Solve for A and B by multiplying both sides by (y-1)(y-2) and simplifying: $$ 1 = A(y-2) + B(y-1) \\ 1= (A+B)y - (2A+B) \\ A+B=0\\ 2A+B=1 $$ Solving this system of equations, we find A = 1, B = -1. So we can rewrite the left-hand side of the integral equation as: $$ \int\frac{1}{y^2 - 3y + 2}\,dy = \int\left(\frac{1}{y-1} - \frac{1}{y-2}\right)dy $$ Now, integrate both sides of the equation: $$ \int\left(\frac{1}{y-1} - \frac{1}{y-2}\right)dy = -2\int x\,dx \\ \ln|y-1|-\ln|y-2|=-x^2+C $$ Where C is the constant of integration.

Solve for y(x)

Using properties of logarithms, the left-hand side can be combined into a single logarithm: $$ \ln \left|\frac{y-1}{y-2}\right|=-x^2+C $$ Now, exponentiate both sides to eliminate the natural logarithm: $$ \left|\frac{y-1}{y-2}\right|=e^{-x^2+C}=e^{-x^2}e^{C} $$ Define a new constant K = e^C, which is positive: $$ \frac{y-1}{y-2}=Ke^{-x^2} $$ Now, solve for y: $$ y-1=(y-2)Ke^{-x^2}\\ y-1=yKe^{-x^2}-2Ke^{-x^2}\\ y(1-Ke^{-x^2})=1-2Ke^{-x^2}\\ y(x)=\frac{1-2Ke^{-x^2}}{1-Ke^{-x^2}} $$

Apply the initial condition and find the interval of validity

We need to find the particular solution using the initial condition y(0) = 3. Replace y(x) with 3 and x with 0 in the general solution: $$ 3=\frac{1-2K}{1-K} $$ Solving for K, we find K = -1. Thus, the particular solution is: $$ y(x)=\frac{1-2(-1)e^{-x^2}}{1+e^{-x^2}} $$ Since the fraction is continuous for all values of x and the denominator never becomes zero, the interval of validity is \((-\infty, \infty)\).

Key concepts

Interval of Validity

When solving an initial value problem, one must often determine the interval of validity for the solution. This refers to the range of the independent variable, typically x, for which the solution to the differential equation is defined and continuous. In this exercise, upon finding the particular solution to the differential equation with the given initial condition, the resultant fraction is continuous for all x. This is because the denominator never equals zero, allowing for the interval of validity to be all real numbers, or \( (-\infty, \infty) \). To summarize, to determine the interval of validity:

• Ensure the solution is defined for all x.
• Check that there are no discontinuities, like division by zero.
• The interval may often depend on the initial conditions provided.

Understanding this concept is crucial, as it helps specify where the solution applies in real-world scenarios.

Variable Separation

Variable separation is a common method used to solve differential equations, particularly when dealing with first-order ordinary differential equations. The goal is to rearrange the equation so that each variable appears on a separate side of the equation. In the provided exercise, the original differential equation was rearranged as:\[\frac{1}{y^2 - 3y + 2}\,dy = -2x\,dx\]This involves rearranging terms so that all y-related terms (including \(dy\)) are isolated on one side and all x-related terms (including \(dx\)) on the other. Once this is achieved, each side can be integrated independently. This technique relies on the multiplicative separation of differentials and works well for equations that can be decomposed into such a form. This step is often the first in solving separable differential equations.

Partial Fraction Decomposition

Partial fraction decomposition is a technique used to break down complex rational expressions into simpler fractions that are easier to integrate or differentiate. In this problem, this technique was applied to the integrand:\[\frac{1}{y^2 - 3y + 2}\]This expression is decomposed into partial fractions:\[\frac{1}{(y-1)(y-2)} = \frac{A}{y-1} + \frac{B}{y-2}\]Solving for constants A and B involved equating coefficients on both sides, which allows these complex terms to be separately integrated. This method is crucial because it simplifies complicated integrals, making them more manageable. Partial fraction decomposition is often used when the denominator is a product of linear factors, and it accelerates the solution process of differential equations by turning a challenging integral into a series of simpler ones.

Partial Solutions

In the realm of differential equations, a partial solution refers to any function that satisfies the differential equation, but not necessarily the initial conditions. When solving problems, particularly in step-by-step formats, partial solutions are often derived, and then adjusted to meet specific conditions. For our exercise, the derived general solution was:\[ y(x)=\frac{1-2Ke^{-x^2}}{1-Ke^{-x^2}} \]This represents the partial solution which satisfies the differential equation. To find a particular solution, the initial condition \( y(0) = 3 \) was used. This allowed for solving the constant \( K \), giving a solution tailored to the specific scenario. Understanding partial solutions enables deeper comprehension of how different solutions can be, before taking into account specific conditions or real-world applications.

Differential Equations

Differential equations involve functions and their derivatives, expressing relationships between variables. There are numerous forms and methods to solve differentials, aligning with the type and order of the equation. In this exercise, we dealt with a first-order ordinary differential equation (ODE):\[ y' = -2x(y^2 - 3y + 2) \]Differential equations can model real-world phenomena in physics, engineering, biology, and economics. The solution to these equations, especially in initial value problems, is critical to understanding the behavior of various systems over time or space. Key techniques used include:

• Separation of variables, to simplify and solve ODEs.
• Integration, both indefinite and definite, which provides solutions in functional forms.
• Partial fraction decomposition, for simplifying complex expressions within integrals.

These methods showcase the versatility and application of differential equations in solving practical and theoretical problems across disciplines.