Can anyone help to integrate this ? thanks
\Rightarrow \frac {dy}{(y + 3)^2} = (x - 1)^2~dx
isn't this related to that thread you posted on ODE's? if you have problems with what was done in that thread, or were not sure on how to proceed in that thread then you should have made that known in THAT thread
for use a substitution,
for expand and use the power rule
(and you need to wrap math tags around your LaTeX code to get it to work, that is, you must type [tex]\frac {dy}{(y + 3)^2} = (x - 1)^2~dx[/tex] to get
It would have been tidier to post this question as part of this thread since it's a follow-up question ....
trying to arrange the equation into this format- dy/dx + P * y = Q
Integrating factor = e ^ int P dx
dy/dx + y/ 2(3-x) = -1/ 2(3-x)
so P = 1/ 2(3-x)
1/ 2(3-x) = 2(3-x)^ -1 so cannot use the normal mathod to integrate, so I use ln instead.
so 1/2 * 1/ (3-x) = 1/2 ln (3-x)
integrating factor = e^ 1/2 in (3-x) and stucked , dunno wat to do next..
correct answer is 1/ square root of (3-x)
Thanks for your help..