Question

PLEASE HELP dy/dx+ 2xy + xy^4 = 0

Answer 1

burnoodle

Answer 2

\[y' + 2xy = -xy^4\]
\[(y' + 2xy = -xy^4)y^{-4}\]
\[y^{-4} y' + 2xy^{-3 } = -x\]

Answer 3

Not sure im getting it

Answer 4

y^-4?
whres that coming from

Answer 5

substitute a new funtion of y in place of y;

\[\begin{array}\\
z=y^{-3}\\\\

z^{-1/3}=y& :z^{4/3}=y^{-4}\\\\
-\frac{1}{3}z^{-4/3}z' =y'
\end{array}
\]

\[y^{-4} y' + 2xy^{-3 } = -x\]
\[z^{4/3} (-\frac{1}{3}z^{-4/3}z') + 2x\ z = -x\]
\[-\frac{1}{3}z' + 2x\ z = -x\]

Answer 6

it comes from dividing off the y^4 to get all the ys to one side

Answer 7

ok

Answer 8

multiply thru by -3 to clear our z'

\[-3(-\frac{1}{3}z' + 2x\ z = -x)\]
\[z' + (-6x)\ z = 3x\]

and we are to a form that we should be familiar with now

Answer 9

solve for z and then revert back to y

z = (.......) = y^-3

Answer 10

wow..you good..i need to look at it a few mins to understand

Answer 11

thanx, if you have questions, feel free to ask

Answer 12

how do i find an intersection of 2 vectors

Answer 13

the question actualy says find the S intesect Tand hence find a basis and the dimention, where S={(a+b,a,-a,-b)} and T={(x,2y,-y,-x)}

Answer 14

IF they intersect, then there component parts will be in the same place at the same time

Answer 15

S=:
1a+1b | 1| | 1|
1a+0b = a | 1|+ b | 0|
-1a+0b |-1| | 0|
0a-1b | 0| |-1|

all linear combinations of the span of independant vectors form a vector space.

\[col\ S=span\left\{
\begin{pmatrix} 1\\1\\-1\\0 \end{pmatrix},
\begin{pmatrix}1\\0\\0\\-1 \end{pmatrix}\right\}\]

T=:
1x+0y | 1| | 0|
0x+2y = x | 0|+ y | 2|
0x-1y | 0| |-1|
-1x+0y |-1| | 0|

same reasoning as for S


\[col\ T=span\left\{
\begin{pmatrix}1\\0\\0\\-1 \end{pmatrix},
\begin{pmatrix}0\\2\\-1\\0 \end{pmatrix}\right\}\]

Answer 16

I notice that the span of each vector space contains the same vector; [1,0,0,-1] so this is at least a line of intersection; for when a=0 and y=0 and b=x; we have then same object defined

Answer 17

2 vectors will span at most a plane; the planes themselves stretch thru R^4, so these are a subspace of R^4 ..

Answer 18

the other 2 vectors, if they form equal ratios from their component parts; they are parallel; for any vector that is a scalar multiple of another retains the same ratio of parts

2 3 5 8
14 21 35 56
-----------------
1/7 1/7 1/7 1/7


0,2,-1,0
1,1,-1,0
--------
0,2, 1,und ; since their ratios are not consistent they are not scalar multiples and therefore NOT parallel