Dowqueuestbew1j

2021-12-17

Determine (without solving the problem) an interval in which the solution of the given initial value problem is certain to exist.

Jordan Mitchell

Step 1
Divide equation with $\left(t-3\right)$,

According to Theorem 2.4.1,
$p\left(t\right)=\frac{\mathrm{ln}t}{t-3}$ and $g\left(t\right)=\frac{2t}{t-3}$
Functions p(t), g(t) are continuos on intervals (0, 3) and because domains are for p and $\frac{\mathbb{R}\left\{\left\{3\right\}\right\}}{}$ for g.
Since ${t}_{0}=1$, solution exists on (0, 3)

Paul Mitchell

First we put the ODE into standard form:
${y}^{\prime }+\frac{\mathrm{ln}\left(t\right)}{\left(t-3\right)}y=\frac{2t}{\left(t-3\right)}$
Then we see that
$p\left(t\right)=\frac{\mathrm{ln}\left(t\right)}{\left(t-3\right)}$,
while
$g\left(t\right)=\frac{2t}{\left(t-3\right)}$
Now,
$p\left(t\right)=\frac{\mathrm{ln}\left(t\right)}{\left(t-3\right)}$
is continuous on the intervals

and
while
$g\left(t\right)=\frac{2t}{\left(t-3\right)}$
is continuous on the intervals

and

Thus we see that both p(t) and g(t) are continuous on . However, the initial condition is at is in the interval

thus we are guranteed a solution on the interval

nick1337

Step 1
Rewriting the above equation in the standard form, we have
${y}^{\prime }+\frac{\mathrm{ln}t}{t-3}y=\frac{2t}{t-3}$
So $p\left(t\right)=\frac{\mathrm{ln}t}{t-3}$ and $g\left(t\right)=\frac{2t}{t-3}$.
g is continuous for all $t\ne 3$. p is continuous for all
Therefore p and g are both continuous on the interval
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
We now turn our attention to nonlinear differential equations and modify Theorem 3.1 by a more general theorem.

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