solve the following equations for y'
a. xy' + y =1+ y'
b.2x + 2xy' + 2y + 3y^2y' = 0
c.x/y' + y = 5/y'
please please help
2007-08-05 04:21:47 · 4 answers · asked by Paul C 1 in Science & Mathematics ➔ Mathematics
sorry c is x/(y' + y) = 5/y'
2007-08-05 04:43:55 · update #1
Hi,
Here are general rules to use to solve for y'.
To solve for y', get all terms with y' on one side of the equation and all terms without y' on the other side.
If only 1 term has y', divide by anything multiplied times it to get y' alone.
If more than one term has a y', factor out y' as a common factor. Divide by the expression in the parentheses to get y' alone.
a. xy' + y =1+ y'
Move all terms with y' to one side, without y' to the other side.
xy' + y' = 1- y
Factor out the y'.
y'(x + 1) = 1 - y
Divide both sides by x + 1.
1.-.y
------- = y' <== answer
x.+.1
b.2x + 2xy' + 2y + 3y^2y' = 0
Move all terms with y' to one side, without y' to the other side.
2xy' + 3y²y' = -2x - 2y
Factor out the y'.
y'(2x + 3y²) = -2x - 2y
Divide both sides by 2x + 3y².
-2x - 2y
------------ = y' <==answer
2x + 3y²
c. x/(y' + y) = 5/y'
Cross multiply this equation.
5(y' + y) = xy'
Distribute the 5.
5y' + 5y = xy'
Move all terms with y' to one side, without y' to the other side.
5y' - xy' = -5y
Factor out the y'.
y'(5 - x) = -5y
Divide both sides by (5 - x).
.-5y
------ = y' <== answer
5 - x
I hope those help!! :-)
2007-08-08 06:16:20 · answer #1 · answered by Pi R Squared 7 · 0⤊ 0⤋
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2016-11-11 06:58:08 · answer #2 · answered by ? 4 · 0⤊ 0⤋
a. xy' +y = 1 + y'
y =1+y' -xy'
y-1 =y'(1-x)
y'=(y-1)/(1-x)
b. 2x+2xy'+2y+3y^2y'=0
2x +2y = -2xy'-3y^2y'
(2x+2y)/-(2x+3y^2)=y'
-(x+y)/(x+1.5y^2)=y'
c. x/y' +y =5/y'
y =5/y'-x/y'
y'=(5-x)/y
2007-08-05 04:35:27 · answer #3 · answered by bignose68 4 · 0⤊ 1⤋
a) y´(x-1)=1-y so y´=(1-y)/(x-1)
b)y´(2x+3y^2)=(-2x-2y) so y´= (-2x-2y)/(2x+3y^2)
c)(x-5)/y´= -y so y´= (5-x)/y
2007-08-05 04:33:50 · answer #4 · answered by santmann2002 7 · 0⤊ 1⤋