Determine (without solving the problem) an interval in which the

Dowqueuestbew1j

Dowqueuestbew1j

Answered question

2021-12-17

Determine (without solving the problem) an interval in which the solution of the given initial value problem is certain to exist. (t3)y+(lnt)y=2t, y(1)=2

Answer & Explanation

Jordan Mitchell

Jordan Mitchell

Beginner2021-12-18Added 31 answers

Step 1
Divide equation with (t3),
y+lntt3 y=2tt3
According to Theorem 2.4.1,
p(t)=lntt3 and g(t)=2tt3
Functions p(t), g(t) are continuos on intervals (0, 3) and (3, ) because domains are (0, ){3} for p and R{{3}} for g.
Since t0=1, solution exists on (0, 3)
Paul Mitchell

Paul Mitchell

Beginner2021-12-19Added 40 answers

First we put the ODE into standard form:
y+ln(t)(t3)y=2t(t3)
Then we see that
p(t)=ln(t)(t3),
while
g(t)=2t(t3)
Now,
p(t)=ln(t)(t3)
is continuous on the intervals
Ip1=(0, 3)
and Ip2=(3, +)
while
g(t)=2t(t3)
is continuous on the intervals
Ig1=(, 3)
and
Ig2=(3, +)
Thus we see that both p(t) and g(t) are continuous on Ip1 and Ip2=Ig2. However, the initial condition is at t=1, y(t)=2; t=1 is in the interval
Ip1=(0, 3)
thus we are guranteed a solution on the interval Ip1=(0, 3)
nick1337

nick1337

Expert2021-12-27Added 777 answers

Step 1
Rewriting the above equation in the standard form, we have
y+lntt3y=2tt3
So p(t)=lntt3 and g(t)=2tt3.
g is continuous for all t3. p is continuous for all t0, 3.
Therefore p and g are both continuous on the interval (, 0)(0, 3)(3, )
The interval (0, 3) contains the initial point t=1. Therefore Theorem 3.1 guarantees that the problem has  a unique solution on the interval 0<t<3
We now turn our attention to nonlinear differential equations and modify Theorem 3.1 by a more general theorem.

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