Question

solve the differential equation \[\frac{\text dy}{\text dx}-y=xy^{1/2}\]

Answer 1

\[\frac{\text dy}{\text dx}-y=xy^{1/2}\]
\[\text{let }v=y^{1-1/2}=y^{1/2}\]\[\text dv=\frac 1{2y^{1/2}}\text dy\]\[\text dy={2v}\text dv\]
\[2v\frac{\text dv}{\text dx}-v^2=xv\]\[\frac{\text dv}{\text dx}-\frac 12 v=\frac x2\]

\[R(x)=e^{\int -\frac 12\text dx}=e^{-\frac x2}\]
\[\frac{\text d}{\text dx}\left(ve^{-\frac x2}\right)=\frac{xe^{-\frac x2}}2\]\[ve^{-\frac x2}=\int\frac{xe^{-\frac x2}}2\text dx\]

Answer 2

sqrt(y(x))+2+x-exp((1/2)*x)*_C1 = 0

Answer 3

how did you ?

Answer 4

@timo86m did you use any of my working ? how did you get that result/

Answer 5

\[ve^{-\frac12}=\frac 12\int xe^{-\frac x2}\text dx\]

Answer 6

It's correct:)

Answer 7

i guess i have to integrate by parts now

Answer 8

Yeah:) use by parts

Answer 9

\(\int uv'=\left.uv\right|-\int vu'\)

Answer 10

Yeah:D you're doing great

Answer 11

\[u=x\qquad u'=1\]\[v'=e^{-x/2}\qquad v=\frac{-e^{-x/2}}{2}\]

Answer 12

\[ve^{-\frac x2}=\left.-\frac{xe^{-x/2}}{2}\right|+\int \frac{e^{-x/2}}2\text dx\]\[ve^{-\frac x2}=\left.\frac{-xe^{-x/2}}{2}\right|-{e^{-x/2}}+c\]

Answer 13

correction ..
\[\large{}v' =e^{-\frac{x}{2}} \] \[\large{v=-\frac{e^{-\frac{x}{2}}}{\frac{1}{2}}}=-2e^{-\frac{x}{2}}\]

Answer 14

Ah!

Answer 15

Thank YOU

Answer 16

:D

Answer 17

\[ve^{-\frac x2}=\left.-{2}{xe^{-x/2}}\right|+2\int {e^{-x/2}}\text dx\]
\[ve^{-\frac x2}=\left.-2xe^{-x/2}\right|-4{e^{-x/2}}+c\]

Answer 18

yes and v=sqrt(y)

Answer 19

what to i do with the limit \(\left. \right|\)

Answer 20

\[\sqrt y={-2x}-4+ce^{x/2}\]

Answer 21

This is the answer in the back of the book
_______________________\[\sqrt y=-2-x+ce^{x/2}\]_______________________

Answer 22

can u give me a min i want to try

Answer 23

v=sqrt(y)
dv=1/(2sqrt(y))

\[v'-\frac{v}{2}=\frac{x}{2}\]
\[\large{I=e^{\int\limits{-\frac{1}{2}}dx}}=e^{-\frac{x}{2}}\]multiplying the DE by I
\[e^{-\frac{x}{2}}v'-\frac{e^{-\frac{x}{2}}}{2}v=\frac{xe^{-\frac{x}{2}}}{2}\]\[ve^{-\frac{x}{2}}=\frac{1}{2} \int\limits{xe^{-\frac{x}{2}}dx}\]now integrating

u=x dv=e^(-x/2)dx
du=dx v=-2e^(-x/2)

\[ve^{-\frac{x}{2}}=\frac{1}{2}(-2xe^{-\frac{x}{2}}+2\int\limits\limits\limits{e^{-\frac{x}{2}}dx})\]\[ve^{-\frac{x}{2}}=\frac{1}{2}(-2xe^{-\frac{x}{2}}-4e^{-\frac{x}{2}}+C)\]now divide everything by e^(-x/2)
\[v=\frac{1}{2}(-2x-4+ce^{\frac{x}{2}})\]\[v=-x-2+\frac{c}{2}e^{\frac{x}{2}}\]

Answer 24

c/2 is any constant so it can be just C

Answer 25

no need for any limits in integration ,,, and u forgot the 1/2 u put at the beginning

Answer 26

@UnkleRhaukus

Answer 27

so the answer is \[\sqrt{y}=-x-2+ce^{\frac{x}{2}}\]

Answer 28

yeah forgot about the \(\frac 12\) when i integrated by parts
thanks for all your work

Answer 29

anytime

Answer 30

in the integration by parts bit we are ment to drop the limit because we are adding the arbitrary constant of integration/ is this right?

\(\int\limits_a^b uv'=\left.uv\right|_a^b-\int\limits_a^b vu'\)

\(\int uv'=uv-\int\limits_a^b vu'+c\)

Answer 31

?

Answer 32

yes its an indefinite integral