Question

plzzzz need help ODEs

\(\frac{dy}{dx}=\frac{xy^2-cos x sin x}{y(1-x^22)}\)
\(y(0)=2\)

Answer 1

check if his is correct
\(y dy -x^2dy = xy^2dx -cos x sinx dx\)

Answer 2

\(\frac{1}{2}y^2 -x^2y=\frac {1}{2}x^2 y^2-\int\cos x \sin x \dx \)

Answer 3

??
i dont know if im on the right direction

Answer 4

nope,
y is a function of x
u cannot integrate \(\large \int xy^2 dx\)

Answer 5

\(\large \frac{dy}{dx}=\frac{xy^2-cos x sin x}{y(1-x^2)}\)

Answer 6

like this right ?

Answer 7

yep

Answer 8

\(\large y' =\frac{xy^2-cos x sin x}{y(1-x^2)}\)


multiply both sides by \(y\)

\(\large yy' =\frac{xy^2-cos x sin x}{1-x^2}\)

Answer 9

isnt y'=dx/dy ?

Answer 10

sorry im not really good :'(

Answer 11

substitute \(\large v = y^2\)

that gives,
\(\large v' = 2yy'\)

Answer 12

the equation becomes :


\(\large \frac{v'}{2} =\frac{xv-cos x sin x}{1-x^2}\)


\(\large v' =\frac{2xv-\sin (2x)}{1-x^2}\)

Answer 13

fine so far ?

Answer 14

yes

Answer 15

good, knw how to solve a "linear equation" using IF method ?

Answer 16

no

Answer 17

show mw if pspl ill understand quick

Answer 18

this is a bernouli equation

Answer 19

after substitution it changes to a linear equation

Answer 20

you should know how solve "linear equations" and the Intgeration Factor method, before starting bernouli

Answer 21

well , im sure i learned them but my basecs are rusty ( been long time )
so if u show me i might remember

Answer 22

\(\large y' =\frac{xy^2-cos x sin x}{y(1-x^2)}\)


multiply both sides by \(y\)

\(\large yy' =\frac{xy^2-cos x sin x}{1-x^2}\)

substitute \(\large v = y^2\)
\(\large v' = 2yy'\)

the equation becomes :

\(\large \frac{v'}{2} =\frac{xv-cos x sin x}{1-x^2} \)

\(\large v' =\frac{2xv-\sin (2x)}{1-x^2} \)

\(\large v' + \frac{2x}{1-x^2}v = \frac{\sin(2x)}{x^2-1} \)

Answer 23

thats what we have so far^

Answer 24

i with u so far

Answer 25

\(\large y' =\frac{xy^2-cos x sin x}{y(1-x^2)}\)


multiply both sides by \(y\)

\(\large yy' =\frac{xy^2-cos x sin x}{1-x^2}\)

substitute \(\large v = y^2\)
\(\large v' = 2yy'\)

the equation becomes :

\(\large \frac{v'}{2} =\frac{xv-cos x sin x}{1-x^2} \)

\(\large v' =\frac{2xv-\sin (2x)}{1-x^2} \)

\(\large v' + \frac{2x}{x^2-1}v = \frac{\sin(2x)}{x^2-1} \)

Answer 26

im*

Answer 27

just corrected a typo in the last line

Answer 28

Multiply the IF \((x^2-1)\) through out the equaiton

Answer 29

the equation becomes :

\(\large (x^2-1)v' + 2xv = \sin(2x) \)

Answer 30

Using product rule, left hand side can be compressed to :

\(\large ((x^2-1)v)' \)

Answer 31

\(\large (x^2-1)v' + 2xv = \sin(2x) \)

becomes :

\(\large ((x^2-1)v)' = \sin(2x) \)

Answer 32

Now is the good time to integrate both sides

Answer 33

how ?

Answer 34

integrate both sides

Answer 35

\(\large \int ((x^2-1)v)'~ dx = \int \sin(2x)~dx \)

Answer 36

By FTC, \(\large \int f' dx = f\),
so left hand sides gives u back \((x^2-1)v\)

Answer 37

\(\large \int ((x^2-1)v)'~ dx = \int \sin(2x)~dx \)


\(\large (x^2-1)v = \int \sin(2x)~dx \)

Answer 38

you can integrate the right hand side

Answer 39

the only thing that i can do by myself :P
thank you for your time I really appreciate it :D

Answer 40

np :)
dont forget substituting the v's back to y^2

Answer 41

okeyy i wont

Answer 42

\(y=\sqrt {\frac{cos 2x}{2 x^2-2 }}+c\)
\(y(0)=\sqrt {\frac{cos 0}{ -2 }}+c=2\rightarrow c=2-\sqrt{\frac{1}{2}}\)

Answer 43

\(\large \large (x^2-1)v = \int \sin(2x)~dx\)


\(\large \large (x^2-1)y^2 = -\frac{\cos (2x)}{2} + C\)

Answer 44

plugin (0, 2) :

\(\large C = \frac{-7}{2}\)

Answer 45

\(\large \large (x^2-1)y^2 = -\frac{\cos (2x)}{2} - \frac{7}{2}\)

Answer 46

im really stink xD
thanks again

Answer 47

you can solve y if u want