# Toronto Math Forum

## MAT244--2018F => MAT244--Tests => Quiz-3 => Topic started by: Victor Ivrii on October 12, 2018, 06:07:20 PM

Title: Q3 TUT 0801
Post by: Victor Ivrii on October 12, 2018, 06:07:20 PM
Find the Wronskian of two solutions of the given differential equation without solving the equation.
$$t^2y''-t(t+2)y'+(t+2)y=0.$$
Title: Re: Q3 TUT 0801
Post by: Qianhao Lu on October 12, 2018, 06:23:52 PM
Title: Re: Q3 TUT 0801
Post by: Yunqi(Yuki) Huang on October 12, 2018, 06:28:18 PM
the new following attachment is right. sorry for my previous mistake to the answer
Title: Re: Q3 TUT 0801
Post by: Nick Callow on October 12, 2018, 06:35:23 PM
To find the Wronskian of the equation without solving we can apply Abel's Theorem. However, we must first isolate the second derivative term in $t^2y''(t) - t(t+2)y'(t) + (t+2)y(t) = 0$. We can do this by dividing all terms by $t^2$. Doing so yields the equation $$y'(t) - \frac{t+2}{t}y'(t) + \frac{t+2}{t^2} = 0$$ Now we will compute the Wronskian $$W = ce^{-\int p(t)dt }$$ where $p(t) = -\frac{t+2}{t}$. Aside: $- \int -\frac{t+2}{t}dt = t + 2ln(t)$.

Therefore, we get that $$W = ce^{t + 2ln(t)} = ct^2e^t$$
Title: Remarks
Post by: Victor Ivrii on October 12, 2018, 07:39:36 PM
Qianhao, NO SNAPSHOTS. Next time -- will delete. SCAN http://forum.math.toronto.edu/index.php?topic=1078.0
(http://forum.math.toronto.edu/index.php?topic=1078.0)
Yunqi, should not post identical solution to the previous!

Nick, escape ln: \ln