helpp (1 Viewer)

[vanush]

If (0,2) lies on the curve y = Ae^(kx) and y'' - 3(y') + 4y = 0, find the values of A and k.

 

[Trev]

(0,2) satisfies y=Ae^kx; so 2=Ae⁰; A=2.
y'=2ke^kx
y"=2k²e^kx
Since y"-3y'+4y=0
2k²e^kx-6ke^kx+8e^kx=0
2e^kx[k²-3k+4]=0
Since e^kx can't equal zero;
When k²-3k+4=0; k=4, or, -1.

So y=2e^4x or y=2e^-x

 

[vanush]

ffs, i was integrating to get y'. no wonder.

thanks

 

[vanush]

damnit, ive spent nearly all day on this one.

anybody?

 

[sikeveo]

suuuure u havnot missed somting as why is there a u in it?

 

[vanush]

as in units squared.

 

[SoulSearcher]

Let b be the length of OA, and h be the length of AB
Since the area of a rectangle is base multiplied by height, Area = OA * AB
Since OA is the length of the rectangle, the area of the rectangle is equal to b * AB, but as the length of OA is x units from the point (0,0), area is x * AB
To find AB, we have to find the point at which the rectangle touches the curve y = e^-x, which is when the x-co-ordinate is b.
So when x = b, y = e^-b, but as x = b, y = e^-x, which gives us the height of the rectanlge.
So the area of the rectangle is x * e^-x = xe^-x units²

 

[vanush]

Cool. I was thinking x and the power on the exponential function (-x) were the same variable.