# Toronto Math Forum

## MAT244--2018F => MAT244--Tests => Quiz-3 => Topic started by: Victor Ivrii on October 12, 2018, 06:04:55 PM

Title: Q3 TUT0401
Post by: Victor Ivrii on October 12, 2018, 06:04:55 PM
If the Wronskian $W$ of $f$ and $g$ is $t^2e^t$ , and if $f(t)=t$, find $g(t)$.
Title: Re: Q3 TUT0401
Post by: Yunqi(Yuki) Huang on October 12, 2018, 06:20:48 PM
in the attachment
Title: Re: Q3 TUT0401
Post by: Monika Dydynski on October 12, 2018, 07:09:36 PM
If the Wronskian $W$ of $f$ and $g$ is $t^{2}e^{t}$, and if $f(t)=t$, find $g(t)$.

Suppose that $W(f,g)=t^{2}e^{t}$ and $f(t)=t \Rightarrow f'(t)=1$

Then from $W(f,g)=fg'-gf'$, we get a first order DE

$$tg'-g\cdot 1=t^{2}e^{t}\tag{1}$$

Dividing both sides of $(1)$ by $t$ and multiplying by integrating factor, $\mu(t)=e^{\int{p(t)}dt}=\frac{1}{t}$, we  have

$$(\frac{1}{t} g)'=e^{t}$$

$$\int{(\frac{1}{t} g)'}=\int{e^{t}}dt$$

$$\frac{1}{t} g=e^{t}+c$$

$$g(t)=te^{t}+ct.$$

Title: Re: Q3 TUT0401
Post by: Yunqi(Yuki) Huang on October 12, 2018, 07:16:55 PM
$$f(t)=t$$
$$So f'(t)=1$$
$$W=tg'(t)-g(t)=t^2e^t$$
$$g'(t)-\frac{1}{t}g(t)=te^t$$
$$p(t)=-\frac{1}{t}$$
$$u(t)=e^{\int1p(t)dt}$$
$$wherep(t)=-\frac{1}{t}$$
$$u(t)=\frac{1}{t}$$
$$\frac{1}{t}g(t)=\int e^tdt$$
$$g(t)=te^t+ct$$where let c=1

Thus g(t)=te^t+t