Calc Help? :\

[zammam]

Apparently these are first order linear ordinary differential equations:

x/2 + y - 1 = dy/dx

some help? my calc ab class learned this yesterday, and i was sick, so i had to stay home

 

[tkhunny]

Rewrite a little:

dy/dx - y = x/2 - 1

This should look like a form in your text. Give it a go and let's see what you get.

 

[zammam]

I can't really find this in my textbook, but here's what I've done so far:

x/2 + y - 1 = dy/dx
x/2 - 1 = dy/dx -y
x-2 / 2 = (dy - y dx) / dx
(x-2/2)dx = dy - ydx

I can integrate the left side, but the right side's still confusing :\

 

[zammam]

I'm still not getting the answer :\

i can only get up to the integration of x dx, and then y ends up being attached to dy and dx. I'm not sure if "y dx" is able to be derived.

 

[zammam]

i fail my bad D:
i forgot integration of y dx = xy

 

[Deleted member 4993]

i fail my bad D:
i forgot integration of y dx = xy <<<< No!!!:

dy/dx - y = x/2 - 1

e[sup:3nebecxm]-x[/sup:3nebecxm](dy/dx - y) = e[sup:3nebecxm]-x[/sup:3nebecxm]( x/2 -1)

integrating both sides:

\(\displaystyle e^{-x} \cdot y \ = \ \int{e^{-x}(\frac{x}{2}-1)dx\)

Now finish integrating right-hand-side.

 

[zammam]

i fail my bad D:
i forgot integration of y dx = xy <<<< No!!!:

dy/dx - y = x/2 - 1

e[sup:eb76zr08]-x[/sup:eb76zr08](dy/dx - y) = e[sup:eb76zr08]-x[/sup:eb76zr08]( x/2 -1)

integrating both sides:

\(\displaystyle e^{-x} \cdot y \ = \ \int{e^{-x}(\frac{x}{2}-1)dx\)

Now finish integrating right-hand-side.

I don't understand how you went from the second to third step. Why did you multiply* e to those values?

 

[Deleted member 4993]

i fail my bad D:
i forgot integration of y dx = xy <<<< No!!!:

dy/dx - y = x/2 - 1

e[sup:3l6zs2vf]-x[/sup:3l6zs2vf](dy/dx - y) = e[sup:3l6zs2vf]-x[/sup:3l6zs2vf]( x/2 -1)

integrating both sides:

\(\displaystyle e^{-x} \cdot y \ = \ \int{e^{-x}(\frac{x}{2}-1)dx\)

Now finish integrating right-hand-side.

I don't understand how you went from the second to third step. Why did you multiply* e to those values?

That is called "integrating factor"

That is whatmissed by not attending the class - ask your teacher.