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Consider the initial value problem:

2tyâ²=8y, y(1)=â2. 2tyâ²=8y, y(1)=â2.

Required:
a. Find the value of the constant C and the exponent r so that y=Ct^r is the solution of this initial value problem.
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 (from part a)?

Answer :

Answer:

Following are the solution to the given point:

Step-by-step explanation:

In point a:

[tex]\to 2ty'=8y\to \frac{1}{y} dy = 4 \frac{1}{t} dt\\\\\to \int \frac{1}{y} dy = 4 \int \frac{1}{t} dt \\\\\to \log y = 4 \log t + \log C\\\\\to \log y = \log ct^4\\\\\to y=ct^4 \\\\\to y(-2)=16 \\\\\to 16= c(-2)^4\\\\\to 16= 16c\\\\\to c= \frac{16}{16}\\\\\to c=1\\\\\to r=4[/tex]

In point b:

[tex]\to 2ty' = 8y \\\\\to y'- \frac{4}{t} y=0 \\\\here \ p(t) = \frac{1}{t}[/tex] it's a discontinuous point at t=0, and the largest internal containing -2 since [tex]y=(-2) = 16 \ is\ (-\infty, 0)[/tex]

In point c:

[tex](-\infty,0 ) \cup (0, \infty)[/tex]

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