Determine the longest interval in which the given initial value problem is certain to have a unique twice-differentiable solution. Do not attempt to find the solution. (Enter your answer using interval notation.) (t – 1)y” – 3ty’ + 3y = sin t, y(-8) = 8, y'(-8) = 1 1 GO Tutorial
Determine the longest interval in which the given initial value problem is certain to have a unique twice-differentiable solution. Do not attempt to find the slotuion. (Enter your answer using interval notation.)
(t-1)y”-3ty’+3y=sin t
y(-8)=8
y'(-8)=
The Correct Answer and Explanation is:
We are given a second-order linear nonhomogeneous differential equation: (t−1)y′′−3ty′+3y=sint,(t – 1)y” – 3t y’ + 3y = \sin t,
with initial conditions: y(−8)=8,y′(−8)=1.y(-8) = 8, \quad y'(-8) = 1.
Step 1: Standard Form
To apply the existence and uniqueness theorem for second-order linear differential equations, rewrite the equation in standard form by dividing the entire equation by t−1t – 1 (assuming t≠1t \neq 1): y′′−3tt−1y′+3t−1y=sintt−1.y” – \frac{3t}{t – 1} y’ + \frac{3}{t – 1} y = \frac{\sin t}{t – 1}.
Step 2: Identify Points of Discontinuity
The functions:
- p(t)=−3tt−1p(t) = -\frac{3t}{t – 1},
- q(t)=3t−1q(t) = \frac{3}{t – 1},
- g(t)=sintt−1g(t) = \frac{\sin t}{t – 1},
must all be continuous for the existence and uniqueness theorem to guarantee a unique, twice-differentiable solution.
All of these functions are discontinuous at t=1t = 1 (division by zero). So, the equation is not defined at t=1t = 1.
Step 3: Determine the Interval of Validity
The initial value is given at t=−8t = -8, which lies to the left of the singular point t=1t = 1.
Therefore, the largest interval centered at t=−8t = -8 and not containing the singular point t=1t = 1 is: (−∞,1)(-\infty, 1)
On this interval:
- The coefficients p(t)p(t), q(t)q(t), and the nonhomogeneous term g(t)g(t) are continuous,
- The initial point t=−8∈(−∞,1)t = -8 \in (-\infty, 1),
So by the existence and uniqueness theorem, there is a unique, twice-differentiable solution on the interval (−∞,1)(-\infty, 1).
✅ Final Answer:
(−∞,1)\boxed{(-\infty, 1)}
To determine where a unique, twice-differentiable solution to a second-order differential equation exists, we rely on the existence and uniqueness theorem for linear differential equations. This theorem states that if the equation is written in standard form: y′′+p(t)y′+q(t)y=g(t),y” + p(t)y’ + q(t)y = g(t),
and if the functions p(t)p(t), q(t)q(t), and g(t)g(t) are continuous on an interval containing the initial value point, then a unique twice-differentiable solution exists on that interval.
Our equation, (t−1)y′′−3ty′+3y=sint(t – 1)y” – 3t y’ + 3y = \sin t, becomes: y′′−3tt−1y′+3t−1y=sintt−1,y” – \frac{3t}{t – 1} y’ + \frac{3}{t – 1} y = \frac{\sin t}{t – 1},
after dividing by t−1t – 1. Here, all coefficients become undefined at t=1t = 1, creating a singularity.
The initial condition is given at t=−8t = -8, well to the left of this singularity. Therefore, the largest open interval containing t=−8t = -8 that excludes the singularity is (−∞,1)(-\infty, 1). Within this interval, all functions in the standard form are continuous, and hence the solution is guaranteed to exist and be unique.
Thus, the longest interval in which the solution is certain to exist and be unique is (−∞,1)(-\infty, 1).
