# Toronto Math Forum

## MAT244--2018F => MAT244--Tests => Quiz-3 => Topic started by: Victor Ivrii on October 12, 2018, 06:03:47 PM

Title: Q3 TUT 0201
Post by: Victor Ivrii on October 12, 2018, 06:03:47 PM
Find the Wronskian of the given pair of functions: $x$ and $xe^x$.
Title: Re: Q3 TUT 0201
Post by: Pengyun Li on October 12, 2018, 06:07:28 PM
$W(x, xe^x) = \left|\begin{matrix}x & xe^x \\ x' & (xe^x)'\end{matrix}\right|= \left|\begin{matrix}x & xe^x \\ 1 & x^2e^x+e^x\end{matrix}\right| = x(x^2e^x+e^x) - xe^x = x^3e^x$
Title: Re: Q3 TUT 0201
Post by: Victor Ivrii on October 12, 2018, 07:47:31 PM
Who taught you differentiate like this?!
Title: Re: Q3 TUT 0201
Post by: Monika Dydynski on October 12, 2018, 07:56:39 PM
(Pengyun's solution with corrected derivative of $xe^{x}$)

Find the Wronskian of the given pair of functions: $x$ and $xe^{x}$

$$W(x, xe^x) = \left|\begin{matrix}x & xe^{x} \\ x' & (xe^{x})'\end{matrix}\right|= \left|\begin{matrix}x & xe^{x} \\ 1 & xe^{x}+e^{x}\end{matrix}\right| = x^{2}e^{x}+xe^{x}-xe^{x}=x^{2}e^{x}.$$