Questions: Consider the initial value problem [ 2 t y'=8 y, quad y(-2)=16 ] a. Find the value of the constant C and the exponent r so that y=C the solution of this initial value problem. [ y=square quad text help (formulas) ] b. Determine the largest interval of the form a<t<b on which the existence and uniqueness theorem for first order linear differential equations guarantees the existence of a unique solution. c. What is the actual interval of existence for the solution

Solution Steps

a. To solve the differential equation \(2ty' = 8y\), we first separate the variables and integrate both sides to find the general solution. Then, we use the initial condition \(y(-2) = 16\) to find the specific values of the constant \(C\) and the exponent \(r\).

b. The largest interval for the existence and uniqueness of the solution is determined by the points where the coefficients of the differential equation are continuous. We analyze the function and its derivatives to find this interval.

c. The actual interval of existence is determined by solving the differential equation and considering any restrictions imposed by the solution itself, such as division by zero or other undefined behaviors.

To solve the given initial value problem, we will follow the steps outlined below:

The given differential equation is: \[ 2t y' = 8y \] We need to find the solution \( y(t) \) that satisfies this equation along with the initial condition \( y(-2) = 16 \).

First, we separate the variables to solve the differential equation: \[ \frac{y'}{y} = \frac{8}{2t} \] \[ \frac{y'}{y} = \frac{4}{t} \]

Integrate both sides with respect to their respective variables: \[ \int \frac{1}{y} \, dy = \int \frac{4}{t} \, dt \] \[ \ln |y| = 4 \ln |t| + C \]

Exponentiate both sides to solve for \( y \): \[ |y| = e^C \cdot |t|^4 \] Let \( C_1 = e^C \), then: \[ y = C_1 t^4 \]

Use the initial condition \( y(-2) = 16 \) to find \( C_1 \): \[ 16 = C_1 (-2)^4 \] \[ 16 = C_1 \cdot 16 \] \[ C_1 = 1 \]

Thus, the solution is: \[ y = t^4 \]

The differential equation is of the form \( y' = \frac{4}{t} y \). The function \( \frac{4}{t} \) is continuous for \( t \neq 0 \). Therefore, the largest interval for existence and uniqueness is any interval that does not include \( t = 0 \). Given the initial condition \( t = -2 \), the largest interval is: \[ (-\infty, 0) \cup (0, \infty) \]

The actual interval of existence for the solution, considering the initial condition \( t = -2 \), is: \[ (-\infty, 0) \]

Final Answer