Consider the initial value problem $$ 2 t y^{\prime}=8 y, \quad y(-2)=32 \text {. } $$ a. Find the value of the constant $$ C $$ and the exponent $$ r $$ so that $$ y=C t^{r} $$ is the solution of this initial value problem. $$ y= $$ b. Determine the largest interval of the form $$ a

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a. Assume a solution of the form
$$
y(t)=C t^{r}.
$$
Then,
$$
y'(t)= C r t^{r-1}.
$$
Substitute these into the differential equation
$$
2t\,y' = 8y:
$$
$$
2t\,(C r t^{r-1}) = 8(C t^{r}).
$$
This simplifies to
$$
2C r t^{r} = 8C t^{r}.
$$
For $$t\neq 0$$ and $$C\neq 0$$, we can cancel $$C t^{r}$$ to obtain
$$
2r=8,
$$
so
$$
r=4.
$$
Now, use the initial condition $$y(-2)=32$$:
$$
32 = C(-2)^{4}.
$$
Since
$$
(-2)^{4}=16,
$$
we have
$$
32= 16C,
$$
which gives
$$
C=2.
$$
Thus, the solution is
$$
y(t)=2t^4.
$$

b. To apply the existence and uniqueness theorem for first order linear differential equations, we first rewrite the ODE in standard form. Divide the given equation by $$2t$$ (noting that $$t\neq 0$$) to obtain
$$
y' = \frac{8}{2t}y = \frac{4}{t}y.
$$
In standard form, this is written as
$$
y' - \frac{4}{t}y=0.
$$
The coefficient function $$p(t)=-\frac{4}{t}$$ is not continuous at $$t=0$$. Since the initial condition $$t_0=-2$$ is given, the theorem guarantees existence and uniqueness on the largest interval containing $$-2$$ that does not include the singular point $$t=0$$. Thus, the largest interval is
$$
(-\infty,0).
$$

c. Although the explicit solution $$y(t)=2t^4$$ is defined for all $$t$$, the original differential equation is not defined at $$t=0$$ (because of the factor $$2t$$). Therefore, in the context of the given initial value problem, the actual interval of existence for the solution is the same as the interval guaranteed by the theorem:
$$
(-\infty,0).
$$
a. The solution is $$ y = 2t^4 $$. b. The largest interval for existence and uniqueness is $$ (-\infty, 0) $$. c. The actual interval of existence for the solution is $$ (-\infty, 0) $$.

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