Y^3 dx+2(x^3-xy^2) dy=0
in Calculus Answers by Level 12 User (101k points)

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40 Answers

→ dy/dx=(y/x)^3 /2(y/x)^2 -2 → (1)
by Level 12 User (101k points)
Transform: y/x
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F(x)_=F: y'
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→ F+xF' → (2)
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(1),(2)→ x dF/dx=F^3/2F^2-2 -F
by Level 12 User (101k points)
→ x dF/dx=F(2-F^2)/2(F^2 -1) ,
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That is,
by Level 12 User (101k points)
{2F^2 -2/ F(2-F^2) dF
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={dx/x → (3)
by Level 12 User (101k points)
The integrand on the left can be written as:
by Level 12 User (101k points)
2F^2 -2/F(2-F^2)=A/F+B/√2+F+c/√2-F
by Level 12 User (101k points)
Where the A,B,C are given from the system,
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-A-B+C=2
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B+C=0
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& 2A=-2
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Solving, we get
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A=-1, B=-1/2 , C=1/2
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{2F^2 -2/F(2-F^2) dx=-{ dF/F-1/2•d(F+√2)/F+√2-1/2• { d(F-√2)/F-√2
by Level 12 User (101k points)
=-lnF-1/2 ln(F+√2)-1/2 ln(F-√2)
by Level 12 User (101k points)
=- lnF-1/2 ln(F^2-2)
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= -lnF•√F^2 -2
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And back in (3) we have
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- ln F√F^2 -2
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=lnx+c base 0
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→ ln F•√F^2-2
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=ln(1/xe^co)
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→F^2•(F^2 -2)
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= 1/x^2 e^2co
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Inverse transform:
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(Y/x)4 -2(y/x)^2
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=y/x^2 e^-2co
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= 1/x^2 ^-2co
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→ y^4-2x^2 y^2-cx^2=0
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We get an incomplete quartic equation.
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This can be reduced to a quadratic
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t^2-2x^2•t-cx^2=0
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With D=4x^2•(x^2+c)
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t=x^2 ± x|√x^2 + c|
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t = y^2
by Level 12 User (101k points)
± y=ab•sq•rt{x^2±x•ab•sq•rt^(x^2+c)}
by Level 12 User (101k points)

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