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# differential equations

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I'm trying to figure out how to solve differential equations of the type y' = a(t)y(t) + b(t)

(such as the simple one y'(t) = t-y(t))

I find the book quite vague on this.

Is there anyone that can give me a proper explanation of this?

Reinout

Jun 5, 2014

#16
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The German word you would use in this context is  "Bitte" or "Bitte sehr"

"ihr willkommen" is equivelent in the context of "welcome" to my home.

Jun 6, 2014

#1
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is there a name of these kind of questions ?

Jun 5, 2014
#2
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Sure,

they are called non-autonomous linear first order differential equations.

Jun 5, 2014
#3
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okay , the name sounds a bit scary but ill try to find out ! all the best for ur search !

Jun 5, 2014
#4
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Here's how it's done for your simple example.  It isn't always possible to get nice results for situations where a(t) and b(t) are very complicated.  For those, use a numerical method!

I need to correct one item above.  In general the integrating factor is $$e^{\int{-a(t)}dt}$$  not $$e^{-a(t)t}$$.

Just noticed Bertie has said this below.

Jun 5, 2014
#5
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see my " all the best " works ! alan came so fast for ur helt after my " all the best " isnt it !

Jun 5, 2014
#6
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The usual approach is to use what is known as an Integrating Factor.

For your example, start by moving a(t)y(t) to the lhs and then multiply throughout by

$$\displaystyle \exp\{\int-a(t)dt\}$$

The lhs will then be a perfect derivative,

$$\displaystyle \frac{d}{dt}\left[\exp\{\int-a(t)dt\}.y(t)\right]$$

and, (hopefully), you can then integrate both sides, having already (hopefully), integrated a(t).

Jun 5, 2014
#7
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Thank you Alan

So I gave it a try (not really involving a(t) yet)

and this is what I have

$$\begin{array}{rll} \frac{dy}{dt} = y-t^2\\ \frac{dy}{dt} - y = -t^2\\ e^{-t}\frac{dy}{dt}-e^{-t}y = -e^{-t}t^2\\ \frac{d}{dt}(e^{-t}y) = -e^{-t}t^2\\ \end{array}$$

$$\begin{array}{rll} e^{-t}y = e^{-t}(t^2+2t+2)+c\\ \mbox{Since }y(0)=y_0\\ e^0y_0 = e^0(0^2-2*0+2)+c\\ c = y_0-2\\ y(t) = t^2+2t+2+y_0-2\\ y(t) = t^2+2t+y_0 \end{array}$$

Does that make sense?

edit: (I also did it for a(t) = t, but I just found my mistake)

Jun 5, 2014
#8
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Almost, the c = (y0 - 2) should be multiplied by et.

Jun 5, 2014
#9
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Why is that?

Jun 5, 2014
#10
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e-ty = e-t(t2 + 2t + 2) + c, where c = y0 - 2.

Multiply throughout by et.

Jun 5, 2014
#11
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Bertie is right.

Also, in the line below "Since y(0) = y0" you have the term -2*0.  This should be +2*0.  However, since the result is zero anyway, this doesn't have any impact on the result!

Jun 5, 2014
#12
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Yes, that makes sense.

Thank you!

Jun 6, 2014
#13
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reinout u didnt say anything about my " all the best " ? i think u forgot it !

Jun 6, 2014
#14
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Haha, yes that was very helpful!

Dankjewel (That's Dutch for thank you)

Jun 6, 2014
#15
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ihr willkommen (thats german for ur welcome )

Jun 6, 2014
#16
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The German word you would use in this context is  "Bitte" or "Bitte sehr"

"ihr willkommen" is equivelent in the context of "welcome" to my home.

Guest Jun 6, 2014
#17
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Some fine german lessons here!

Jun 7, 2014
#18
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i have german lessons in my class too reinout ! but uh , i have to study german in here too !

Jun 8, 2014