Question

A differential equation of the form $$ y^{\prime}=p_{1}(t)+p_{2}(t) y+p_{3}(t) y^{2} $$ is known as a Riccati equation. \({ }^{5}\) Equations of this form arise when we model onedimensional motion with air resistance; see Section 2.9. In general, this equation is not separable. In certain cases, however (such as in Exercises 24-26), the equation does assume a separable form. Solve the given initial value problem and determine the \(t\)-interval of existence. $$ y^{\prime}=t\left(5+4 y+y^{2}\right), \quad y(0)=-3 $$

Short answer

Question: Solve the initial value problem involving the Riccati equation: \(ty' = 5t + 4ty + ty^2\) for \(y(0) = -3\). Determine the t-interval of existence for the solution. Answer: The solution of the initial value problem is given by: \(\left|\frac{1+y}{5+y}\right|^{1/4} = -\sqrt[4]{2}e^t\). The t-interval of existence is \((-\infty, +\infty)\).

Step-by-step solution

Check if the equation is separable

Let's first rewrite the given equation in its standard form by dividing both sides by t, so that we can check if it's separable: $$ y' = 5 + 4y + y^2 $$ In this case, the differential equation becomes separable, as we can write it as $$ y' = (1+y)(5+ y) $$

Solve the separable equation

Now, since we have a separable equation, we can separate the variables and integrate both sides: $$ \frac{dy}{(1+y)(5+y)} = dt $$ Next we integrate both sides with respect to their respective variables: $$ \int \frac{dy}{(1+y)(5+y)} = \int dt $$

Performing partial fractions decomposition

Now we perform partial fraction decomposition to integrate the left-hand side. For that, we will represent the integrand in terms of its partial fractions: $$ \frac{1}{(1+y)(5+y)} = \frac{A}{1+y} + \frac{B}{5+y} $$ Solving for A and B using a system of linear equations, we find that: $$ A = -\frac{1}{4}, \quad B = \frac{1}{4} $$ Now, we can rewrite the fractions as: $$ \int\left(-\frac{1}{4}\frac{1}{1+y}+\frac{1}{4}\frac{1}{5+y}\right)dy = \int dt $$

Integrating both sides

Now we integrate both sides: $$ -\frac{1}{4}\int\frac{1}{1+y}dy+\frac{1}{4}\int\frac{1}{5+y}dy = \int dt $$ We have: $$ -\frac{1}{4}\ln|1+y|+\frac{1}{4}\ln|5+y|=t+C $$

Finding the general solution

Taking the exponent of both sides and simplifying, we have: $$ \left|\frac{1+y}{5+y}\right|^{1/4} = e^{t+C} $$ Replacing \(e^C\) by another constant \(K\), we have: $$ \left|\frac{1+y}{5+y}\right|^{1/4} = Ke^t $$

Applying the initial condition

Now apply the initial condition \(y(0) = -3\): $$ \left|\frac{1-3}{5-3}\right|^{1/4} = K\cdot e^{0} $$ This gives us \(K=-\sqrt[4]{2}\).

Finding the particular solution

Now, we plug the value of K back into the general solution: $$ \left|\frac{1+y}{5+y}\right|^{1/4} = -\sqrt[4]{2}e^t $$ We find the t-interval of existence for the solution is \((-\infty, +\infty)\) as there aren't any specific restrictions on t in the above expression.

Key concepts

Initial Value Problem

An Initial Value Problem (IVP) in the context of differential equations is a problem where not only a differential equation is provided, but also the value of the unknown function at a particular point, which is known as the initial condition.

Specifically, in the case of the Riccati differential equation provided, the IVP requires us to find a solution to the equation that satisfies the condition \(y(0) = -3\). This additional piece of information is crucial because it pinpoints exactly one particular solution out of the potentially infinite family of solutions that satisfy the differential equation.

When solving an IVP, the objective is first to find the general solution to the differential equation, and then use the initial condition to find the particular solution that exactly fits the initial value. The initial condition essentially determines the constant of integration in the solution process, turning a broad solution into a precise one. The existence and uniqueness theorem for first-order IVPs also tells us that, under suitable conditions, there is one and only one solution to an IVP, which will exist in some interval around the initial point.

Separable Differential Equations

In differential equations, a separable equation is one that can be rewritten in a way that allows all terms involving the independent variable (usually \(t\)) to be separated on one side of the equation and all terms involving the dependent variable (usually \(y\)) on the other.

For example, the differential equation \(y' = t(5 + 4y + y^2)\) from the exercise can be manipulated into a separable form as follows:\[\frac{{dy}}{{dt}} = (1+y)(5+y)\]By rearranging the terms, the variables can be separated:\[\frac{{dy}}{{(1+y)(5+y)}} = dt\]Separation of variables is a powerful technique because it transforms the problem of solving a differential equation into a problem of integrating two separate functions of single variables. After separation, we integrate both sides with respect to their respective variables to find the general solution. The separable form is especially valuable when the differential equation corresponds to a physically interpretable process, often simplifying complex dynamic systems into easier-to-solve integrals.

Partial Fraction Decomposition

The method of Partial Fraction Decomposition is used in integral calculus to simplify complex rational expressions before integrating. It's particularly useful when dealing with integrals of fractions that have polynomial expressions in the denominator.

For the given exercise, the integrand \(\frac{1}{{(1+y)(5+y)}}\) is a rational expression that is not straightforward to integrate. Through partial fraction decomposition, we break it down into simpler terms that can be easily integrated:\[\frac{1}{{(1+y)(5+y)}} = \frac{A}{{1+y}} + \frac{B}{{5+y}}\]To find the constants \(A\) and \(B\), we'll set up and solve a system of equations by equating coefficients or plugging in suitable values for \(y\) that simplify the equation.

Once \(A\) and \(B\) are determined, the integral becomes a sum of simpler integrals that can often be solved by direct integration or by recognizing them as standard integral forms. Through this process, we are able to integrate complex rational functions that might otherwise be difficult to tackle.