MHB Finding the explicit solution and the Interval of Validity

[shamieh]

Find the explicit solution to $tyy' − 1 = 0$, $y(1) = 4$ and give the interval
of validity.

$ty \frac{dy}{dt} - 1 = 0$

$y \frac{dy}{dt} - 1 = 0$ ==> $ydy - {1/t} dt = 0$

$ydy = 1/t dt$

$\frac{y^2}{2} = ln(t) + c$$y = \sqrt{2ln(t) + c}$

applying $y(1) = 4$

so the explicit solution is:
$C = 4$
But I'm not sure how to get the interval of validity?

 

[shamieh]

so for the interval of validity I got

$2ln(t) + 4 >0$

$1/e^2 < t < \infty$

so $\sqrt{C} <=> C$ right? since $C$ is just some arbitrary constant??

 

[I like Serena]

Hi Shamieh,

Let's pick $c=16$, otherwise it won't fit.
That makes the explicit solution $y=\sqrt{2\ln t + 16}$.
And let's pick $2\ln t + 16 \ge 0$ (including equality).

Btw, any $y=\pm\sqrt{2\ln(-t)+C_2}$ is also valid in combination with the solution you've found.

 

[shamieh]

Thanks Serena,

That being the case...would that imply that my Interval of Validity is: $\frac{1}{e^8} \le t < \infty$ ?

 

[MarkFL]

...That being the case...would that imply that my Interval of Validity is: $\frac{1}{e^8} \le t < \infty$ ?

Yes...we require:

$$\ln|t|+8\ge0$$

And so we see that $t\ne0$ and:

$$e^{-8}\le|t|$$

And so we must pick from:

$$\left(-\infty,-e^{-8}\right]\,\cup\,\left[e^{-8},\infty\right)$$

the sub-interval containing $t=1$ which is $\left[e^{-8},\infty\right)$.

 

[I like Serena]

$$\ln|t|+8\ge0$$

For negative t we can have a (any) different constant of integration...