ordinary diffrantional eq

Let w_=w(x)
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& Y=X^2•W → (1)
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Then Y'
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→ x^2•w'+2wx
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The given is transformed into:
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x^2w’
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=3x^4w+x^4w^2/2x^3+3x^3 w-2xw
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→ x^2 w'
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= -x(5w^2+w)/3w+2
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→ x^2 w'= -x(5w^2+w)/3w+2
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And (3w+2)w(5w+1) dw
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=- dx/x → (2)
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Now 3w+2/w(5w+1)
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_= A/W+B/5w+1
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Which gives,
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A=2 &
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B=-7
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That is,
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That is 3w+2/w(5w+1) dw
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=2 dw/w-7dw/5w+1
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=2 dw/w-7/5 d(5w+1)/5w+1 → (3)
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From (2) and (3)
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After performing the integration we get:
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lnw^2 -7/5 ln(5w+1)
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= c power ~ -lnx
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→ 5 lnw^2+5 lnx- ln(5w+1)^7= c power ~
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ln(w^10 X^5)+ ln(5w+1)^-7
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= c power ~
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→ ln [w^10 X^5/(5w+1)^7]= c power ~
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W→ Y/ x^2:
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W^10 X^5_= Y^10/X^20•X^5
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= Y^10/ X^15
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Therefore, Y^10/X^15
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(5y/x^2+1)^7•e^ c power ~
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→ y^10
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= X^15 •(5y+x^2)^7/X^14 •e^c power ~
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The solution then is given in implicit form as:
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Y^10=C•X(5y+x^2)^7
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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.

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