Differential equations

[Guest]

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Convert to difference equations

[Guest]

y΄->dy/dx

[HallsofIvy]

If [tex]\frac{dy}{dx}= -ay[/tex] then [tex]\frac{dy}{y}= -adx[/tex]. Integrating both sides, [tex]ln(y)= -ax+ C[/tex]. Taking the exponential of both sides, [tex]y= e^{-ax+ C}= e^Ce^{-ax}= C'e^{-ax}[/tex] where [tex]C'= e^C[/tex] is an arbitrary constant.
Since we want [tex]y(0)= 1[/tex], [tex]y(0)= 1= C'e^0= C'[/tex]. C'= 1 so the solution to the differential equation satisfying the initial condition is [tex]y(x)= e^{-ax}[/tex].

Check: if [tex]y(x)= e^{-ax}[/tex] then [tex]y'= -ae^{-ax}= -ay[/tex] and [tex]y(0)= e^{-a(0)}= 1[/tex].

Do the second problem in the same way.

(I would expect this to be covered in the first few weeks of a differential equations course. Have you just started the course?)

[Guest]

Was the "convert to difference equations" added? I didn't see it before.

To convert to a difference equation replace "dy/dx" with the corresponding "difference quotient", [tex]\frac{\Delta y}{\Delta x}= \frac{y(x+ h)- y(x)}{h}[/tex]. Typically, where the differential would take the limit as h goes to 0, the difference quotient takes h= 1.

The differential equation y'= -ay becomes the difference equation y(x+ 1)- y(x)= ay(x) so that y(x+1)= (a+ 1)y(x). Since y(0)= 1, y(1)= a+1, [tex]y(2)= (a+1)(a+ 1)= (a+1)^2[/tex], and, for n any positive integer, [tex]y(n)= (a+1)^n[/tex]. With the given information we can only determine y(n) for n an integer. Normally difference equations give the value of y(x) for from 0 to 1 or on some other interval of length 1.

[Guest]

The second derivative converts to a "difference quotient" in a similar way. The "first difference" is u(x)= y(x+1)- y(x) so the "second difference" is u(x+1)- u(x)= [y(x+2)- y(x+1)]- [y(x+1)- y(x)]= y(x+2)- 2y(x+1)+ y(x). The differential equation y''= -y converts to the difference equation y(x+2)- 2y(x+1)+ y(x)= -y(x) or y(x+2)- 2y(x+1)+ 2y(x)= 0.