Thread: Hard integral question but with hints, please solve it help!!!

please solve this question in full. very urgently need it . thanks alot in adavance

Originally Posted by unisa_students

please solve this question in full. very urgently need it . thanks alot in adavance

1. I seem to recall saying in http://www.mathhelpforum.com/math-he...t-pl-help.html:

"Note: It's good form to take the few minutes it requires to type your question if you expect help."

2. Nevertheless, I did open the file and have a look. Exactly what is it that you can't do? The hints appear more than sufficient.

3. Call me lazy but as a general rule I'm just not in the habit of solving questions in full.


If you show what work you've done and where you got stuck, I'm sure someone will give you a hint.

Actually i dont know how to type it properly, but thanks for taking interest.
another way i tried was this
y'=y-0.3sin(pi x/3)
y'-y=-0.3sin(pi x/3)
I(x)=e^anti(-1)dx , I(x)=e^-x
so (e^-x)y'-(e^-x)y=-(e^-x)0.3sin(pi x/3)
that means that Direvative((e^-x)y)=-(e^-x)0.3sin(pi x/3)
sooo (e^-x)y=anti(-(e^-x)0.3sin(pi x/3)), here im stuck i cant go any further as i have absolutely no idea how to anti-derive -(e^-x)0.3sin(pi x/3)
any hints absolutely appreciated mr fantastic or anyone else
ooh it is very hard typing maths, hope u understand what i did

i also did it this way but im not sure whether its right or not

y'=y-0.3sin(pi x/3)
by leaving out -0.3sin(pi x/3)
y'=y
(1/y)y'=1
anti(1/y)dy=1dx
In(y)=x+c
y=e^(x+c)
y=Ae^(x+c)
so answering the origional question will make
y=Ae^(x+c)+P i guess P is the antiderivative of -0.3sin(pi x/3)
so y=Ae^(x+c)+(0.9/pi)cos(pi x/3) can some one verify if this is correct
if not can someone please and PLEASE tell what P stands for in the hints that is given