Question

(a) Verify that both \(y_{1}(t)=1-t\) and \(y_{2}(t)=-t^{2} / 4\) are solutions of the initial value problem $$ y^{\prime}=\frac{-t+\left(t^{2}+4 y\right)^{1 / 2}}{2}, \quad y(2)=-1 $$ Where are these solutions valid? (b) Explain why the existence of two solutions of the given problem does not contradict the uniqueness part of Theorem 2.4 .2 (c) Show that \(y=c t+c^{2},\) where \(c\) is an arbitrary constant, satisfies the differential equation in part (a) for \(t \geq-2 c .\) If \(c=-1,\) the initial condition is also satisfied, and the solution \(y=y_{1}(t)\) is obtained. Show that there is no choice of \(c\) that gives the second solution \(y=y_{2}(t) .\)

Short answer

In summary, the given functions \(y_{1}(t)\) and \(y_{2}(t)\) both satisfy the differential equation in their defined domains. The existence of two solutions does not contradict Theorem 2.4.2 because the theorem assumes the continuity and differentiability of the related functions. In our case, the differential equation involves a square root expression that might not be defined everywhere, so the uniqueness part of Theorem 2.4.2 does not apply. The general solution \(y = ct + c^2\) satisfies the differential equation for certain values of \(t\), and while it can yield the first solution when \(c = -1\), there is no choice of \(c\) that gives us the second solution.

Step-by-step solution

Part (a): Verify that both \(y_{1}(t)\) and \(y_{2}(t)\) are solutions of the initial value problem.

To verify that \(y_{1}(t)\) and \(y_{2}(t)\) are solutions of the initial value problem, we need to differentiate them and check if their derivatives satisfy the given differential equation. First, differentiate \(y_{1}(t)\) and \(y_{2}(t)\): \(\frac{dy_{1}}{dt} = -1\) \(\frac{dy_{2}}{dt} = -\frac{t}{2}\) Now, substitute the functions and their derivatives into the differential equation: For \(y_{1}(t)\): $$ -1= \frac{-t+\left(t^{2}+4 (1-t)\right)^{1 / 2}}{2} $$ For \(y_{2}(t)\): $$ -\frac{t}{2}=\frac{-t+\left(t^{2}+4 (-t^{2} / 4)\right)^{1 / 2}}{2} $$ Simplify the equations and check if they are true. For \(y_{1}(t)\), the equation simplifies to: $$ -1= \frac{-t+\left(t^{2}+4 -4t\right)^{1 / 2}}{2} $$ $$ -1= \frac{-t+\left((t-2)^{2}\right)^{1 / 2}}{2} $$ The above equation is valid for all \(t \in \mathbb{R}\) because the square root is always defined (the argument is always non-negative), so the solution \(y_{1}(t)\) is valid for all \(t\). For \(y_{2}(t)\): $$ -\frac{t}{2} = \frac{-t + \left(t^2\right)^{1 / 2}}{2} $$ This equation is valid only for \(t \geq 0\). So, the solution \(y_{2}(t)\) is valid for all \(t \geq 0\).

Part (b): Explain why the existence of two solu-ons does not contradict the uniqueness part of Theorem 2.4.2.

Theorem 2.4.2 states that there exists a unique solution to the initial value problem. However, this theorem includes an important assumption which is the continuity and differentiability of the related functions involved in the differential equation. When these conditions are not met, the existence of multiple solutions does not contradict the uniqueness part of Theorem 2.4.2; in other words, the theorem only guarantees that there is a unique solution when those conditions are met. In our case, the differential equation includes a square root, and the expression inside the square root is not always non-negative - it can be negative for some values of \(t\) and \(y\). Hence, the function is not smooth, and the uniqueness part of Theorem 2.4.2 does not apply.

Part (c): Show that the general solution \(y=ct+c^2\) satisfies the differential equation for certain values of \(t\).

To show that this general solution satisfies the differential equation, first find the first derivative \(\frac{dy}{dt}\): $$ \frac{dy}{dt} = c $$ Now, substitute the function and its derivative into the differential equation: $$ c =\frac{-t+\left(t^{2}+4 (ct + c^{2})\right)^{1 / 2}}{2} $$ Simplify the equation: $$ c =\frac{-t+\left(t^{2}+4 ct + 4c^{2}\right)^{1 / 2}}{2} $$ The equation is satisfied for \(t \geq -2c\). Now, let's find if \(c=-1\) satisfies the initial condition: $$ y = -t + c^2 = -t+1 $$ This is the same as the first solution, \(y_{1}(t)\). Therefore, \(c = -1\) gives us the first solution. Lastly, our task is to find if there exists a value for \(c\) that gives us the second solution: $$ y = ct + c^2 = -\frac{t^2}{4} $$ In this case, there is no choice for \(c\) that will give us the second solution \(y = y_{2}(t)\). This is because \(ct + c^2\) is a linear function (with a parabolic shape centered at \(c\)), while \(y_{2}(t) = -\frac{t^2}{4}\) is a quadratic function. Since no linear function can equal a quadratic function for all values of \(t\), we conclude that there is no choice of \(c\) that gives us the second solution.

Key concepts

Differential Equations

A differential equation is a mathematical equation that relates a function with its derivatives, offering a way to describe the change in a system over time or space. The general form of a differential equation involves a function and its derivatives. For instance, in physics, Newton's second law of motion, which relates force to mass and acceleration, can be expressed as a differential equation involving the velocity of an object and the forces acting upon it.

When solving an initial value problem in differential equations, we are given a differential equation along with initial conditions that the solution must satisfy. Initial conditions are critical as they allow us to determine the specific solution out of potentially many that solve the differential equation. The exercise presented depicts such a scenario, where an initial value problem involving a derivative and a square root function must be solved. It requires finding functions that satisfy both the differential equation and the initial value.

Uniqueness Theorem

The Uniqueness Theorem in the context of differential equations, asserts that under certain conditions, an initial value problem has one and only one solution. The theorem relies heavily on the continuity and differentiability of the functions involved. Essential conditions typically include that the functions in the equation be continuous in a region containing the initial condition, and that partial derivatives of those functions with respect to the dependent variable exist and are continuous.

In the exercise provided, the existence of two solutions does not invalidate this theorem because the conditions necessary for the theorem to guarantee uniqueness are not met due to the square root function, as discussed in Step 2 of the solution. This nature of the function allows for more than one solution, showing that the theorem's conditions function as a determinant for when a unique solution can be promised by mathematical laws.

Solutions of Differential Equations

Solutions to differential equations can be as varied as the equations themselves, ranging from simple expressions to complex functions and series. A solution to a differential equation is a function that, when substituted into the equation, transforms it into an identity. There are two primary categories of solutions: general solutions, which contain arbitrary constants and represent a family of curves, and particular solutions, which are obtained by setting specific values to the constants, typically as a consequence of applying initial conditions.

In the exercise, part (c) deals with finding a general solution in the form of a linear equation involving an arbitrary constant, illustrating how variations in this constant can yield different particular solutions. This demonstrates the rich diversity of solutions that differential equations can possess and underlines that while a general solution may satisfy a differential equation, it may not satisfy a specific initial condition unless a particular value for the constant is chosen correctly.