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

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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Jordan Mitchell
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
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 $$\displaystyle{I}_{{{p}{1}}}={\left({0},\ {3}\right)}$$
nick1337

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.