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

Dowqueuestbew1j 2021-12-17 Answered
Determine (without solving the problem) an interval in which the solution of the given initial value problem is certain to exist. \(\displaystyle{\left({t}−{3}\right)}{y}'+{\left({\ln{{t}}}\right)}{y}={2}{t},\ {y}{\left({1}\right)}={2}\)

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Expert Answer

Jordan Mitchell
Answered 2021-12-18 Author has 2556 answers
Step 1
Divide equation with \(\displaystyle{\left({t}-{3}\right)}\),
\(\displaystyle{y}'+{\frac{{{\ln{{t}}}}}{{{t}-{3}}}}\ {y}={\frac{{{2}{t}}}{{{t}-{3}}}}\)
According to Theorem 2.4.1,
\(\displaystyle{p}{\left({t}\right)}={\frac{{{\ln{{t}}}}}{{{t}-{3}}}}\) and \(\displaystyle{g{{\left({t}\right)}}}={\frac{{{2}{t}}}{{{t}-{3}}}}\)
Functions p(t), g(t) are continuos on intervals (0, 3) and \(\displaystyle{\left({3},\ \infty\right)}\) because domains are \(\displaystyle{\frac{{{\left({0},\ \infty\right)}}}{{{\left\lbrace{3}\right\rbrace}}}}\) for p and \(\displaystyle{\frac{{{\mathbb{{{R}}}}{\left\lbrace{\left\lbrace{3}\right\rbrace}\right\rbrace}}}{}}\) for g.
Since \(\displaystyle{t}_{{{0}}}={1}\), solution exists on (0, 3)
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Paul Mitchell
Answered 2021-12-19 Author has 201 answers
First we put the ODE into standard form:
\(\displaystyle{y}'+{\frac{{{\ln{{\left({t}\right)}}}}}{{{\left({t}-{3}\right)}}}}{y}={\frac{{{2}{t}}}{{{\left({t}-{3}\right)}}}}\)
Then we see that
\(\displaystyle{p}{\left({t}\right)}={\frac{{{\ln{{\left({t}\right)}}}}}{{{\left({t}-{3}\right)}}}}\),
while
\(\displaystyle{g{{\left({t}\right)}}}={\frac{{{2}{t}}}{{{\left({t}-{3}\right)}}}}\)
Now,
\(\displaystyle{p}{\left({t}\right)}={\frac{{{\ln{{\left({t}\right)}}}}}{{{\left({t}-{3}\right)}}}}\)
is continuous on the intervals
\(\displaystyle{I}_{{{p}{1}}}={\left({0},\ {3}\right)}\)
and \(\displaystyle{I}_{{{p}{2}}}={\left({3},\ +\infty\right)}\)
while
\(\displaystyle{g{{\left({t}\right)}}}={\frac{{{2}{t}}}{{{\left({t}-{3}\right)}}}}\)
is continuous on the intervals
\(\displaystyle{I}_{{{g}{1}}}={\left(-\infty,\ {3}\right)}\)
and
\(\displaystyle{I}_{{{g}{2}}}={\left({3},\ +\infty\right)}\)
Thus we see that both p(t) and g(t) are continuous on \(\displaystyle{I}_{{{p}{1}}}\ {\quad\text{and}\quad}\ {I}_{{{p}{2}}}={I}_{{{g}{2}}}\). However, the initial condition is at \(\displaystyle{t}={1},\ {y}{\left({t}\right)}={2};\ {t}={1}\) is in the interval
\(\displaystyle{I}_{{{p}{1}}}={\left({0},\ {3}\right)}\)
thus we are guranteed a solution on the interval \(\displaystyle{I}_{{{p}{1}}}={\left({0},\ {3}\right)}\)
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nick1337
Answered 2021-12-27 Author has 9467 answers

Step 1
Rewriting the above equation in the standard form, we have
\(y'+\frac{\ln t}{t-3}y=\frac{2t}{t-3}\)
So \(p(t)=\frac{\ln t}{t-3}\) and \(g(t)=\frac{2t}{t-3}\).
g is continuous for all \(t\ne3\). p is continuous for all \(t\ne0,\ 3.\)
Therefore p and g are both continuous on the interval \((-\infty,\ 0)\cup(0,\ 3)\cup(3,\ \infty)\)
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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