Another Primitive Functions, (1 Viewer)

Juliaan · Feb 17, 2009

Given y'' = 8, y' = 0 and y=3, when x=1, find the equation of y in terms of x..

I just dont understnad what has to be done, thanks in advance.

lolokay · Feb 17, 2009

lol this was in the other thread

y" = 8, for all x values. then you integrate and evaluate constants

Aerath · Feb 18, 2009

$y'' = 8 (when y' = 0, y = 3, x = 1)$

Integrate both sides:
$y' = 8x + c$
$from above, when y' = 0, x = 1$
$0 = 8*1 + c$
$c = -8$

Therefore, y' = 8x-8

Integrate again:
$y = 4x^{2}- 8x + c$
$From the first line, when y = 3, x = 1$
$3 = 4 - 8 + c$
$c = 7$

$Therefore y = 4x^{2}- 8x + 7$

Finx · Feb 18, 2009

Little mistake there Aerath;

3 = 4 - 8 + c
3 = -4 + c
c = 7

eq: 4x² - 8x + 7

The rest is all good =]

Aerath · Feb 18, 2009

Yeah, that's what happens at 5am in the morning. =P

Another Primitive Functions, (1 Viewer)

Juliaan

Member

lolokay

Active Member

Aerath

Retired

Finx

Member

Aerath

Retired

Users Who Are Viewing This Thread (Users: 0, Guests: 1)