Question

Solve the initial value problem and find the interval of validity of the solution. $$ y^{\prime}\left(x^{2}+2\right)+4 x\left(y^{2}+2 y+1\right)=0, \quad y(1)=-1 $$

Short answer

The particular solution to the given initial value problem is: $$ y(x) = 1 - \frac{1}{2\ln\left|\sqrt{x^{2} + 2}\right| - 2\ln\left|\sqrt{3}\right|} $$ and the interval of validity for this solution is $$(-\infty, \infty)$$.

Step-by-step solution

Rewrite the ODE

First, let's rewrite the given ODE as: $$ y^{\prime}\left(x^{2}+2\right) = -4x\left(y^{2}+2y+1\right) $$ Now, we can divide both sides by $$\left(x^{2}+2\right)$$. This gives: $$ y^{\prime} = \frac{-4x\left(y^{2}+2y+1\right)}{x^{2}+2} $$

Simplify the ODE

The right-hand side of the ODE can be simplified by a substitution. Let's note that $$y^{2} + 2y + 1 = (y + 1)^2$$. Now the ODE becomes: $$ y^{\prime} = \frac{-4x(y + 1)^2}{x^{2}+2} $$ Since this is a first-order ODE, our goal is to separate the variables. We'll look for a substitution to make this possible.

Find a suitable substitution

We can make the substitution $$z = y + 1$$ to simplify the ODE. This implies $$y = z-1$$ and $$y^{\prime} = z^{\prime}$$. Substituting these expressions into our ODE, we get: $$ z^{\prime} = \frac{-4xz^2}{x^{2}+2} $$ Now we have separated variables in the form $$z^{\prime} = f(x)g(z)$$, where $$f(x) = \frac{-4x}{x^{2}+2}$$ and $$g(z) = z^2$$.

Solve the ODE using separation of variables

To solve using separation of variables, we'll divide both sides by $$g(z)$$ and integrate with respect to $$x$$ on both sides: $$ \int\frac{z^{\prime}}{z^2} dx = \int\frac{-4x}{x^{2}+2} dx $$ Integrating, we obtain: $$ -\frac{1}{z} = -2\ln\left|\sqrt{x^{2} + 2}\right| + C $$

Solve for z and apply the initial condition

Now we'll solve for z and substitute back $$y + 1$$ for z: $$ -\frac{1}{y + 1} = -2\ln\left|\sqrt{x^{2} + 2}\right| + C $$ Applying the initial condition $$y(1)=-1$$, we find the constant $$C$$: $$ -\frac{1}{0} = -2\ln\left|\sqrt{3}\right| + C \implies C = -2\ln\left|\sqrt{3}\right| $$

Use C to obtain the particular solution and find the interval of validity

Plugging $$C$$ back into our equation, we find the particular solution: $$ -\frac{1}{y + 1} = -2\ln\left|\sqrt{x^{2} + 2}\right| - 2\ln\left|\sqrt{3}\right| $$ Multiplying both sides by $$-1$$ and adding 1 to both sides, we get: $$ y(x) = 1 - \frac{1}{2\ln\left|\sqrt{x^{2} + 2}\right| - 2\ln\left|\sqrt{3}\right|} $$ Since the logarithm is defined for positive arguments and the square root is defined for non-negative arguments, the interval of validity for this solution is $$(-\infty, \infty)$$.

Key concepts

First-Order ODE

A first-order ordinary differential equation (ODE) is an equation that involves a function, its first derivative, and a variable. In the exercise we have, the ODE is first-order because it contains the first derivative of the function: \(y'\). The general form of a first-order ODE is:

• \(y' = f(x, y)\)

In these equations, you're usually tasked with finding the function \(y(x)\) given its rate of change and some initial condition. Solving such an ODE often involves a variety of techniques, each applicable to different kinds of equations.

Separation of Variables

Separation of variables is a method for solving first-order ODEs where you aim to rearrange the equation such that all terms involving the dependent variable \(y\) are on one side, and all terms involving the independent variable \(x\) are on the other. This allows you to treat each side as an indefinite integral:

• Arrange as: \( N(y) \frac{dy}{dx} = M(x) \)
• Separate variables: \( \int N(y) \, dy = \int M(x) \, dx \)

By integrating both sides separately, we can often solve for \(y\) as a function of \(x\). In our exercise, achieving separation was aided by a substitution that simplified the original equation. Once the variables were isolated on each side, integration led to the solution, modified by any constant determined using the initial condition.

Substitution Method

When a differential equation is not readily separable, substitution can be a powerful tool. This means introducing a new variable to simplify the equation into a more straightforward form. In the exercise, we noticed that \(y^2 + 2y + 1\) could be expressed as \((y + 1)^2\), which inspired the substitution:

• Let \( z = y + 1 \)
• This transforms the original ODE to one that can be tackled more easily

Once the substitution was completed, the new variable made the equation much simpler and allowed us to use separation of variables smoothly. After solving for the substitute variable, we substituted back to express \(y\) in terms of \(x\) using the initial condition to finalize the particular solution.