Question
Asked Jun 17, 2019
15 views

Using the existence and uniqueness theorem for second order linear ordinary differential equations, find the largest interval in which the solution to the initial value is certain to exist.

t(t2-4)y"-ty'+3t2y=0, y(1)=1, y'(1)=3

check_circle

Expert Answer

Step 1

The given initial value problem is:

(P-4)y-ty'+3ty=0, y(1)=1, y'(1) =3
help_outline

Image Transcriptionclose

(P-4)y-ty'+3ty=0, y(1)=1, y'(1) =3

fullscreen
Step 2

Existence and Uniqueness Theorem:

If p(x),q(x) and g(x) are continuous on the interval [a, b], then the second order differential
equation
y'p(x)xy=g(x), y(x)y, y(x,)=
has a unique solution defined for all x in [a, b].
help_outline

Image Transcriptionclose

If p(x),q(x) and g(x) are continuous on the interval [a, b], then the second order differential equation y'p(x)xy=g(x), y(x)y, y(x,)= has a unique solution defined for all x in [a, b].

fullscreen
Step 3

Redefine the differential e...

t
-y'+
y0
(P-4)(4)
1
3t
y=0
( -4)*
help_outline

Image Transcriptionclose

t -y'+ y0 (P-4)(4) 1 3t y=0 ( -4)*

fullscreen

Want to see the full answer?

See Solution

Check out a sample Q&A here.

Want to see this answer and more?

Solutions are written by subject experts who are available 24/7. Questions are typically answered within 1 hour.*

See Solution
*Response times may vary by subject and question.
Tagged in

Math

Advanced Math

Related Advanced Math Q&A

Find answers to questions asked by student like you

Show more Q&A add