dy\dx=-(3x^2y+y^2)\2x^3+3xy

Question

ordinary diffrantional eq

 

Answer 1

Let w_=w(x)

Answer 2

& Y=X^2•W → (1)

Answer 3

Then Y'

Answer 4

→ x^2•w'+2wx

Answer 5

The given is transformed into:

Answer 6

x^2w’

Answer 7

=3x^4w+x^4w^2/2x^3+3x^3 w-2xw

Answer 8

→ x^2 w'

Answer 9

= -x(5w^2+w)/3w+2

Answer 10

→ x^2 w'= -x(5w^2+w)/3w+2

Answer 11

And (3w+2)w(5w+1) dw

Answer 12

=- dx/x → (2)

Answer 13

Now 3w+2/w(5w+1)

Answer 14

_= A/W+B/5w+1

Answer 15

Which gives,

Answer 16

A=2 &

Answer 17

B=-7

Answer 18

That is,

Answer 19

That is 3w+2/w(5w+1) dw

Answer 20

=2 dw/w-7dw/5w+1

Answer 21

=2 dw/w-7/5 d(5w+1)/5w+1 → (3)

Answer 22

From (2) and (3)

Answer 23

After performing the integration we get:

Answer 24

lnw^2 -7/5 ln(5w+1)

Answer 25

= c power ~ -lnx

Answer 26

→ 5 lnw^2+5 lnx- ln(5w+1)^7= c power ~

Answer 27

ln(w^10 X^5)+ ln(5w+1)^-7

Answer 28

= c power ~

Answer 29

→ ln [w^10 X^5/(5w+1)^7]= c power ~

Answer 30

W→ Y/ x^2:

Answer 31

W^10 X^5_= Y^10/X^20•X^5

Answer 32

= Y^10/ X^15

Answer 33

Therefore, Y^10/X^15

Answer 34

(5y/x^2+1)^7•e^ c power ~

Answer 35

→ y^10

Answer 36

= X^15 •(5y+x^2)^7/X^14 •e^c power ~

Answer 37

The solution then is given in implicit form as:

Answer 38

Y^10=C•X(5y+x^2)^7

Answer 39

Solve:dy/dx=-(3x^2 y+y^2)/(2x^3+3xy

Let v=y/x  or y=vx. Then dy/dx=v+x∙dv/dx

So then, v+x∙dv/dx=-(3x^2 y+y^2)/(2x^3+3xy)
x∙dv/dx=-(3x^2 y+y^2)/(2x^3+3xy)-(2xv+3v^2)/(2x+3v)=-(5xv+4v^2)/(2x+3v)
dv/dx=-(5xv+4v^2)/(2x^2+3xv)
dv/dx=-(5(x/v)+4(v/x)^2)/(2+3(v/x) )= -(5w+4w^2)/(2+3w),   w=v/x

Let v=wx,  then dv/dx=w+x∙dw/dx

So, w+x∙dw/dx=-(5w+4w^2)/(2+3w)
Then, x∙dw/dx=-(5w+4w^2)/(2+3w)-(2w+3w^2)/(2+3w)= -(7w+7w^2)/(2+3w)

The above DE now transforms to,

1/7 ∫(2+3w)/(w+w^2 ) dw=-∫x dx

Partial fractions

(2+3w)/(w+w^2 )=A/w+B/(1+w)
From which we find, A=2,B=1

Then,
1/7 ∫(2+3w)/(w+w^2 ) dw=1/7 {∫2/w dw+∫1/(1+w) dw}=1/7 {2.ln(w)+ln(1+w) }
1/7 ∫(2+3w)/(w+w^2 ) dw=1/7∙ln(w^2+w^3 )
And,
-∫1/x dx= -ln(x)+C

Therefore,
1/7∙ln(w^2+w^3 )=C - ln(x)=ln(A/x)  (ln(A) = C)
w^2+w^3=kx^(-7), (k = A^7)
Where w=v/x and v=y/x, giving w=y/x^2 .
So,
w^2+w^3=y^2/x^4 +y^3/x^6 =kx^(-7)
Or,
(y^2 x^3+y^3 x=k),k≠0

If we differentiate the above implicit equation and solve for y’, we will get the original DE in the problem statement.