Author Topic: Problem 10 in ex 2.1  (Read 1238 times)

Anushree Kariwal

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Problem 10 in ex 2.1
« on: September 27, 2018, 09:47:46 PM »
Has anyone solved this question? I can't seem to find what I am doing wrong.

What question? You need to post it, and your attempts to solve it. V.I.
« Last Edit: September 27, 2018, 11:20:14 PM by Victor Ivrii »

Nick Callow

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Re: Problem 10 in ex 2.1
« Reply #1 on: September 28, 2018, 12:03:38 AM »
I believe the problem your talking about is finding the general solution of: $$ty'(t)-y(t) = t^2e^{-t}, t > 0$$
In that case, this differential is first order but not separable, therefore we will use the method of integrating factors.
Be more specific, what kind of equation is this one.
The first problem that arises is this is not in the standard form we use for integrating factors where $$y'(t) + a(t)y(t) = b(t)$$
So we need to fix that. We can do that by diving every term by $t$ to isolate the derivative term. This produces the equation $$y'(t) - \frac{1}{t}y(t) = te^{-t}$$
Now, we can find the integrating factor which we will denote by $\mu(t)$. $$\mu(t) = e^{\int\frac{-1}{t}dt} = e^{\ln(\frac{1}{t})} = \frac{1}{t}$$
Now we multiply each side of the differential by $\mu(t)$ and isolate for $y(t)$ as follows:
\begin{gather*}
\frac{1}{t}y'(t) - \frac{1}{t^2}y(t) = e^{-t}\implies\\
\frac{d}{dt} (\frac{1}{t}y(t)) = e^{-t}\implies\\
\int \frac{d}{dt} (\frac{1}{t}y(t))dt = \int e^{-t}dt\implies\\
 \qquad\qquad\qquad\frac{1}{t}y(t) + c_1 = -e^{-t} + c_2\implies\qquad\qquad\qquad\color{red}{\checkmark}\\
 \frac{1}{t}y(t) = -e^{-t} + k\implies\\
 y(t) = -te^{-t} + kt.
\end{gather*}
No need to introduce $c_1$ and $c_2$ at all: marked line is excessive

Therefore, the general solution of the above differential is given by: $$ y(t) = -te^{-t} + kt $$

Note that this is only valid when $t > 0$. You can see from above if $t = 0$ this makes no sense as we had to divide by $t$ several times.
« Last Edit: September 28, 2018, 02:41:37 AM by Victor Ivrii »

Victor Ivrii

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Re: Problem 10 in ex 2.1
« Reply #2 on: September 28, 2018, 02:40:24 AM »
Obtained solution is a solution for all $t$, including $t\le 0$. And it is the only solution which is differentiable at $t=0$. Indeed, ignoring $t=0$ first we get
$$
y(t)=-te^{-t}+\left\{\begin{aligned} &c_1t &&t>0,\\ &c_2t &&t<0;\end{aligned}\right.
$$
this is a continuous function at $t=0$ no matter if $c_1=c_2$ but it is differentiable only if $c_1=c_2$.

So the last line in your statement is not exactly correct. To avoid such details the authors put $t>0$.

Pengyun Li

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Re: Problem 10 in ex 2.1
« Reply #3 on: October 01, 2018, 11:06:04 AM »
Hi! I have attached the detailed answer in the image. Hope it helps :)

Victor Ivrii

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Re: Problem 10 in ex 2.1
« Reply #4 on: October 01, 2018, 01:45:38 PM »
Hi! I have attached the detailed answer in the image. Hope it helps :)
ROTFL