Question

first order linear diff eq

Answer 1

i used 2 kno dis but i 4-got

Answer 2

im sorry it took so long to attach this file.. lag sucks today

Answer 3

ikr my brian is always lagging

Answer 4

Do you know how to start?

Answer 5

yes

Answer 6

GURL

Answer 7

I CANT REED UR HAND RITING

Answer 8

lel

Answer 9

wait a minit....

Answer 10

dat was an isult

Answer 11

;-;

Answer 12

the first order linear.

Answer 13

any idea? can you try to answer this? the answer should be xv^2 = v+1+c e^v

Answer 14

Sorry @silverxx haven't learned this yet :/

Answer 15

its okay =)

Answer 16

@ganeshie8 help?

Answer 17

\[\dfrac{d\color{red}{x}}{dv} + \dfrac{2\color{Red}{x}}{v} = -\dfrac{1}{v} + \color{red}{x}\]

Answer 18

your equation is NOT in standard form yet

Answer 19

standard form of linear equation is this :

\[\dfrac{d\color{Red}{x}}{dt} + p(t) \color{red}{x} = q(t)\]

NO dependent variable on right hand side

Answer 20

why is it? i have x on the right side

Answer 21

so am i gonna multiply it with 1/x?

Answer 22

x should be there on right hand side
send it to left hand side

Answer 23

\[\dfrac{d\color{red}{x}}{dv} + \dfrac{2\color{Red}{x}}{v} = -\dfrac{1}{v} + \color{red}{x}\]

subtract x both sides

\[\dfrac{d\color{red}{x}}{dv} + \dfrac{2\color{Red}{x}}{v}-\color{red}{x} = -\dfrac{1}{v} \]

\[\dfrac{d\color{red}{x}}{dv} + \left(\frac{2}{v}-1\right)\color{red}{x} = -\dfrac{1}{v} \]

Answer 24

Now it is in standard form : \(p(v) = \frac{2}{v} - 1\)

find integrating factor and do your usual stuff..

Answer 25

integrating factor would be v^2 - e^v right?

Answer 26

do you mean \(IF \large \mathrm{ = v^2e^{-v}}\) ?

Answer 27

haha omg yes

Answer 28

good, keep going..