2nd order DE largest interval confusion

In summary, the task is to determine the largest t-interval on which theorem 3.1 guarantees the existence of a unique solution for the given second order differential equation. The theorem states that if p(t), q(t), and g(t) are continuous functions on the interval (a,b), then the IVP y'' + p(t)y' + q(t)y = g(t) has a unique solution defined on the entire interval (a,b). In this case, p(t) = 3t^2, q(t) = 2, and g(t) = sin(t), all of which are continuous functions. Therefore, the largest interval is -∞ < t < ∞. The initial conditions, y(
  • #1
fufufu
17
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2nd order DE..."largest interval" confusion

Homework Statement


Determine the largest t-interval on which therem 3.1 guarantees the existence of a unique solution:

y'' + 3t^2y' + 2y = sin(t) ...y(1) = 1 ...y'(1) = -1




Homework Equations


theroem 3.1 is the one that states if p(t) and q(t) and g(t) are continuous functions on the interval (a,b) then the IVP y'' + p(t)y' + q(t)y = g(t) has unique solution defined on the entire interval , (a,b) .


The Attempt at a Solution



if i therefore locate any discontinuities on each function in the equation, then that will determine what the intervals are.
p(t) = 3t^2 and g(t) sin(t) and q(t) = 2(*) are all continuous, so the largest interval is -infin<t<infin, which is the answer in back of book.
But what was the point of the initial values then?

(my geuss) --do i look at the tvalues and consider which is the largest interval that that t-value falls in? (in this case i only have one interval, so it makes it easy, but am i explaining it correctly?)

(*) my other question is: in this problem, is q(t) = 2? if there is no t in the term, then how can i ask if there are any discontinuities there?

thanks for any help in understanding this..
 
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  • #2


fufufu said:

Homework Statement


Determine the largest t-interval on which therem 3.1 guarantees the existence of a unique solution:

y'' + 3t^2y' + 2y = sin(t) ...y(1) = 1 ...y'(1) = -1




Homework Equations


theroem 3.1 is the one that states if p(t) and q(t) and g(t) are continuous functions on the interval (a,b) then the IVP y'' + p(t)y' + q(t)y = g(t) has unique solution defined on the entire interval , (a,b) .


The Attempt at a Solution



if i therefore locate any discontinuities on each function in the equation, then that will determine what the intervals are.
p(t) = 3t^2 and g(t) sin(t) and q(t) = 2(*) are all continuous, so the largest interval is -infin<t<infin, which is the answer in back of book.
But what was the point of the initial values then?

(my geuss) --do i look at the tvalues and consider which is the largest interval that that t-value falls in? (in this case i only have one interval, so it makes it easy, but am i explaining it correctly?)

(*) my other question is: in this problem, is q(t) = 2? if there is no t in the term, then how can i ask if there are any discontinuities there?

thanks for any help in understanding this..

Homework Statement



.

Homework Equations





The Attempt at a Solution


The point of the initial conditions is to cause the solution to be unique. If you had arbitrary constants in there, you would have lots of solutions to the equation. And yes, q(t)= 2 is a perfectly good constant function. You get 2 for every t.
 

1. What is a second order differential equation?

A second order differential equation is a mathematical equation that involves a function and its second derivative. It represents the relationship between a quantity and its rate of change.

2. How do you solve a second order differential equation?

To solve a second order differential equation, you need to find a particular solution that satisfies the given initial conditions. This can be done by using various methods such as separation of variables, integrating factors, and variation of parameters.

3. What is the largest interval of a second order differential equation?

The largest interval of a second order differential equation is the interval where the solution to the equation is valid. This interval can be determined by solving the differential equation and checking for any restrictions on the domain of the solution.

4. Why is there confusion surrounding second order differential equations?

Second order differential equations can be quite complex and involve advanced mathematical concepts. This can lead to confusion for those who are not familiar with the topic. Additionally, there are various methods and techniques for solving these equations, which can also contribute to confusion.

5. What are the real-world applications of second order differential equations?

Second order differential equations are used in many fields of science and engineering to model and understand various phenomena. They are commonly used in physics, biology, economics, and engineering to describe systems that involve acceleration, oscillation, and growth.

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