Question

(a) For what value of the constant \(C\) and exponent \(r\) is \(y=C t^{r}\) the solution of the initial value problem $$ 2 t y^{\prime}-6 y=0, \quad y(-2)=8 ? $$ (b) Determine the largest interval of the form \((a, b)\) on which Theorem \(2.1\) guarantees the existence of a unique solution. (c) What is the actual interval of existence for the solution found in part (a)?

Short answer

Answer: The values of \(C\) and \(r\) for the solution are \(C = 512\) and \(r = 3t\), so the solution is \(y = 512t^{3t}\). The largest interval of the existence of a unique solution is \((-\infty, 0) \cup (0, \infty)\).

Step-by-step solution

Find the derivative of y

Find \(\frac{dy}{dt}\) of the function \(y=C t^r\) using the power rule: \(\frac{d}{dt}(t^r) = rt^{r-1}\). So, we have $$ \frac{dy}{dt} = Cr^{r-1}. $$

Substitute the function and its derivative into the given differential equation

We substitute \(y=C t^{r}\) and its derivative \(y' = Cr^{r-1}\) into the differential equation \(2ty' - 6y = 0\): $$ 2t(Cr^{r-1}) - 6(Ct^r) = 0 . $$

Simplify and find the values of r

Simplify the expression: $$ 2tCr^{r-1} - 6Ct^r = 0 . $$ We can factor out \(2Ct^{r-1}\) from both terms: $$ 2Ct^{r-1}(r - 3t) = 0 . $$ Now we have two possibilities: 1. \(2Ct^{r-1} = 0\), which leads to \(C=0\) or \(t=0\), but having \(C=0\) or \(t=0\) creates constant or trivial solutions, so we discard this case. 2. \(r - 3t = 0\), which gives us the value of \(r\): $$ r = 3t. $$ We must now find the value of \(C\) using the initial condition.

Find the value of C using the initial condition

Plug in the initial condition \(y(-2) = 8\) into \(y=Ct^r\) and use \(r = 3t\): $$ 8 = C(-2)^{3(-2)}. $$ We get: $$ 8 = C\frac{1}{64} . $$ Solve for \(C\): $$ C = 512 . $$ So, \(y = 512 t^r\) with \(r = 3t\).

Analyze the given differential equation

Now we analyze the given differential equation to find the largest interval of the existence of the solution: $$ 2 t y^{\prime}-6 y=0 . $$ This is a first-order linear homogeneous differential equation. According to the existence and uniqueness theorem, there exists a unique solution on the interval where the coefficients of the differential equation are continuous.

Determine the largest interval of unique solution

The coefficients are continuous for all values of \(t\). However, as per the denominator formula, we must exclude any values of \(t\) for which \(y(t)\) is infinite or not defined. Since \(y = 512 t^r\) with \(r = 3t\), the function is defined for all \(t \ne 0\). Hence, the largest interval of the existence of unique solution is $$ (-\infty, 0) \cup (0, \infty) . $$

Determine the actual interval of existence

As \(y = 512 t^r\) with \(r = 3t\) is continuous and differentiable everywhere on its domain, its actual interval of existence is $$ (-\infty, 0) \cup (0, \infty) . $$ In summary: (a) \(C = 512\) and \(r = 3t\), so the solution is \(y = 512t^{3t}\). (b) The largest interval of the existence of a unique solution is \((-\infty, 0) \cup (0, \infty)\). (c) The actual interval of existence for the solution found in part (a) is \((-\infty, 0) \cup (0, \infty)\).

Key concepts

Differential Equations

Differential equations are mathematical statements that relate a function with its derivatives. They are a fundamental tool in mathematics and the sciences because they can describe a wide variety of phenomena, from physics to economics. In the context of the given exercise, the differential equation presented was a first-order linear homogeneous differential equation, where the term 'first-order' indicates that only the first derivative of the function appears in the equation.

The equation \[2 t y' - 6 y = 0\] represents a relationship between a function, its derivative, and time. Solving a differential equation typically means finding a function or a set of functions that satisfy the equation. In the given problem, the solution sought is of the form \(y=C t^r\), where \(C\) and \(r\) are constants to be determined.

Existence and Uniqueness Theorem

The existence and uniqueness theorem is a key concept when it comes to solving differential equations. It assures us that under certain conditions, there is precisely one solution to a differential equation that passes through a given point, which is referred to as an initial condition. Specifically, the theorem states that if the functions in the equation are continuous and satisfy certain other technical conditions in a region around an initial point, then there exists a unique solution that passes through that point.

Using Theorem 2.1 in the exercise, one had to determine the interval within which a unique solution to the initial value problem \(2 t y' - 6 y = 0\) is guaranteed. Due to the continuity of the coefficients, the unique solution exists on the interval \((-\text{\infty}, 0) \bigcup (0, +\text{\infty})\), which excludes the singularity point where \(t = 0\). This theorem is vital for ensuring that the solution not only exists but also is the only solution within that interval.

Power Rule Derivative

The power rule is a basic derivative rule that is used to find the derivative of a function of the form \(t^r\), where \(r\) is a real number. The power rule states that the derivative of \(t^r\) with respect to \(t\) is \(rt^{r-1}\). Applying this rule, the derivative of the function \(y=C t^r\) from the exercise is \(y' = Cr^{r-1}\), which is a key step in solving the differential equation provided.

By utilizing the power rule, students can quickly find derivatives of power functions, which is an essential component in solving differential equations like the one in the given problem. Power rule derivatives are a fundamental part of calculus, and fluency with this rule can greatly simplify the process of working with differential equations.