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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

Best Answer 

 #16
avatar
+19

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
avatar+11855 
+9

is there a name of these kind of questions ? 

 Jun 5, 2014
 #2
avatar+2353 
0

Sure,

 

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

 Jun 5, 2014
 #3
avatar+11855 
+14

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

 Jun 5, 2014
 #4
avatar+27549 
+10

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!

IntFac1

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
avatar+11855 
+14

see my " all the best " works ! alan came so fast for ur helt after my " all the best " isnt it ! 

 Jun 5, 2014
 #6
avatar+890 
+10

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
avatar+2353 
+3

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
avatar+890 
+5

Almost, the c = (y0 - 2) should be multiplied by et.

 Jun 5, 2014
 #9
avatar+2353 
0

Why is that?

 

 Jun 5, 2014
 #10
avatar+890 
+10

e-ty = e-t(t2 + 2t + 2) + c, where c = y0 - 2.

Multiply throughout by et.

 Jun 5, 2014
 #11
avatar+27549 
+10

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
avatar+2353 
+5

Yes, that makes sense.

Thank you!

 Jun 6, 2014
 #13
avatar+11855 
+14

reinout u didnt say anything about my " all the best " ? i think u forgot it ! 

 Jun 6, 2014
 #14
avatar+2353 
+5

Haha, yes that was very helpful!

 

Dankjewel (That's Dutch for thank you)

 Jun 6, 2014
 #15
avatar+11855 
+11

ihr willkommen (thats german for ur welcome )

 

 Jun 6, 2014
 #16
avatar
+19
Best Answer

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
avatar+2353 
+5

Some fine german lessons here!

 

 Jun 7, 2014
 #18
avatar+11855 
+14

i have german lessons in my class too reinout ! but uh , i have to study german in here too ! 

 Jun 8, 2014

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