### Author Topic: Q3 TUT 0701  (Read 3497 times)

#### Victor Ivrii

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##### Q3 TUT 0701
« on: October 12, 2018, 06:06:42 PM »
Find the Wronskian of two solutions of the given differential equation without solving the equation.
$$\cos(t)y''+\sin(t)y'-ty=0.$$

#### Tzu-Ching Yen

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##### Re: Q3 TUT 0701
« Reply #1 on: October 12, 2018, 06:09:14 PM »
Reformat equation into form $y'' + p(t)y' + q(t) = 0$
We can see that
$p(t) = \frac{\sin(t)}{\cos(t)}$
By the equation for wronskian
$$W(t) = \exp \bigl(\int p(t) dt\bigr) = \exp \bigl(\int \frac{\sin(t)}{\cos(t)} dt\bigr) = \exp (-\ln(\cos(t))+\ln (c_0)) = \frac{c_0}{\cos(t)}$$
where $c_0$ depends on choice of $y_2$ and $y_2$
« Last Edit: October 12, 2018, 07:29:53 PM by Victor Ivrii »

#### Michael Poon

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##### Re: Q3 TUT 0701
« Reply #2 on: October 13, 2018, 10:12:57 AM »
Thomas, your solution seems mostly correct, but I think you are missing something small. If I remember correctly, Abel's theorem has a negative factor you forgot.

$$W(t)=e^{∫-p(t)dt}=e^{∫-\tan(t)dt}=e^{\ln(\cos(t))+c}=c(\cos(t))$$, for some constant c.
« Last Edit: October 13, 2018, 02:05:30 PM by Victor Ivrii »

#### Wei Cui

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##### Re: Q3 TUT 0701
« Reply #3 on: October 13, 2018, 10:49:13 AM »
First, we divide both sides of the equation by $\cos(t)$ and we get:
$y''+\tan(t)y'-\frac{t}{\cos(t)}y = 0$

Then the equation in the form of $L(y) = y'' +p(t)y' +q(t)y=0$, then in this case $p(t) = tan(t)$
According to Abel's Theorem, then the Wronskian                         $W(y_1,y_2)(t) = Ce^{-\int \tan(t)dt}$
$=Ce^{\int \frac{1}{\cos(t)}d\cos(t)}$
$= Ce^{\ln(\cos(t))}$
$= C\cos(t)$

Therefore, the solution is $W(y_1,y_2)(t) = C\cos(t)$.
« Last Edit: October 13, 2018, 02:03:53 PM by Victor Ivrii »