dy/dx=x+y^2 ,y(-2)=0

Question

dy/dx=x+y^2 ,given that y(-2)=0

Answer 1

General form:

Answer 2

Pdx+Qdy=0

Answer 3

P_=x+y^2

Answer 4

Q_= -1

Answer 5

We have Q,x=0

Answer 6

And Py=2y

Answer 7

The integrating factor

Answer 8

(If) is If=exp ^{-2ydx ,

Answer 9

If=e^-2xy :

Answer 10

PIf dx+QIf dy_=du(x,y)=0

Answer 11

→ u(x,y)=c

Answer 12

æu/æy=-e^-2xy

Answer 13

→ u(x,y)=1/2x e^-2xy+w(x)=c →(1)

Answer 14

æu/æx=1/2(-e^-2xy/x^2+1/x•(-2y)e^-2xy)+dw/dx

Answer 15

→Ux=dw/dx-(1/2x^2+y/x) e^-2xy_=P•If

Answer 16

w(x)={(x+y^2+1/2x^2+y/x) e^-2xy dx

Answer 17

We evaluate the integral

Answer 18

(I) {x•If dx

Answer 19

=-1/2y{xd(If)

Answer 20

=-x/2y If +1/2y(-1/2y) If

Answer 21

=-(x/2y+1/4y^2)If,

Answer 22

(ii) y^2 {If dx

Answer 23

=y^2(-1/2y) If

Answer 24

=-y/2 If,

Answer 25

(iii) {If/2x^2 dx

Answer 26

=1/2{If/x^2 dx

Answer 27

=-1/2{If d(1/x)

Answer 28

=- If/2x+1/2{1/x If dx

Answer 29

We observe that

Answer 30

{If/2x^2 dx+y•{ If/x dx_=-If/2x+(y+1/2)•{ If/x dx

Answer 31

By substitution in (1)

Answer 32

(x/2y+1/4y^2+y/2)•e^-2xy

Answer 33

=c+(y+1/2)•{e^-2xy/x dx

Answer 34

This is the general solution

Answer 35

And the integral can be evaluated by numerical methods

Answer 36

Now let 2y _= k ,

Answer 37

Then because we know (EULER)

Answer 38

That e^f=∞ summation n=0 F^n/ n ! ,

Answer 39

(F_=-kx in our case

Answer 40

And dividing by x :

Answer 41

{1/x e^-kx dx

Answer 42

={(1/x-k/1!+k^2x/2!—+...) dx

Answer 43

lnx-{∞ summation n=1 (-1)^n+1 (2xy)^n/n!}

Answer 44

Therefore (x/2y+1/4y^2+y/2) e^-2xy

Answer 45

=c+(y+1/2)•[lux-{∞ summation n=1 (-1)^n+1/n!•(2xy)^n}]

Answer 46

Y(-2)=0 :

Answer 47

c+(0+1/2)•[ln(-2)-0]=0

Answer 48

→ c= -1/2ln(-2)

Answer 49

ln(-2)=ln(2cisπ)

Answer 50

lu|2|+ I (π±2mπ) ,

Answer 51

m= 0,1,2, ...

Answer 52

For m=0 : principal value that is :

Answer 53

C=-1/2 ln|2| - iπ/2