Determine the largest interval in which the initial value problem (t − 3)y ′′ − ty′ + 6y = sin t has unique solution.&nbsp;y(−3) = 2, y′ (−3) = 1&nbsp;

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Determine the largest interval in which the initial value problem (t − 3)y ′′ − ty′ + 6y = sin t has unique solution.&amp;nbsp;y(−3) = 2, y′ (−3) = 1&amp;nbsp;

madeleinejames20 · Accepted Answer

To determine the largest interval in which the initial value problem (t−3)y″−ty′+6y=sint has a unique solution with the initial conditions y(−3)=2 and y′(−3)=1, we need to consider the interval in which the coefficients of the differential equation are continuous and the solutions are well-behaved.First, let&#039;s check the interval of uniqueness for the differential equation. The coefficients t−3, −t, and 6 are continuous for all real values of t. Therefore, the differential equation is defined and continuous on the entire real line.Next, let&#039;s examine the initial conditions. The initial condition y(−3)=2 ensures that the solution passes through the point (−3,2), and the initial condition y′(−3)=1 specifies the slope of the solution at t=−3.To find the largest interval of uniqueness, we need to determine if there are any other solutions that satisfy the same initial conditions. If there are no other solutions, then the interval of uniqueness is the entire real line.Given the specific initial conditions, we can solve the differential equation using standard methods, such as the method of undetermined coefficients or variation of parameters. The solution y(t) can be expressed in terms of t and the initial conditions.Since there are no constraints or limitations specified in the problem that would lead to a restricted interval, we can conclude that the largest interval in which the initial value problem has a unique solution is the entire real line: (−∞,∞).

Determine the largest interval in which the initial

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