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When solving the differential equation

$$ (3x^2+6xy^2) dx + (6x^2y+4y^2)dy = 0 $$

I verified that it is an Exact Equation. By saying that

$$ F_x = 3x^2+6xy^2\\ F_y = 6x^2y+4y^2 $$

and integrating both sides I found $$F(x,y) = 3x^2y^2 + \frac{4}{3}y^3+x^3 = C$$ and since $y(0)=2$ we have $$3x^2y^2+ \frac{4}{3}y^3+x^3 = \frac{32}{3}$$

The problem is that the textbook's solution is $$x^3+3x^2y^2+y^3=16$$

Where did I go wrong?

Thank you.

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    $\begingroup$ Your answer is correct. The book must be wrong. $\endgroup$
    – KittyL
    Commented Aug 24, 2016 at 10:38

1 Answer 1

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Your book solution is not correct $${ x }^{ 3 }+3{ x }^{ 2 }{ y }^{ 2 }+{ y }^{ 3 }=16\\ 3{ x }^{ 2 }+6{ x }{ y }^{ 2 }+6{ x }^{ 2 }y{ y }^{ \prime }+3{ y }^{ 2 }{ y }^{ \prime }=0\\ 3{ x }^{ 2 }+6{ x }{ y }^{ 2 }+\frac { dy }{ dx } \left( 6{ x }^{ 2 }y+3{ y }^{ 2 } \right) =0\\ \left( 3{ x }^{ 2 }+6{ x }{ y }^{ 2 } \right) dx+\left( 6{ x }^{ 2 }y+\color{red}3{ y }^{ 2 } \right) dy=0\\ \\ \\ $$ which is a differ from original equation form($(3x^2+6xy^2) dx + (6x^2y+\color{red}4y^2)dy = 0$)

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  • $\begingroup$ Thank you! Best regards my friend. $\endgroup$
    – bru1987
    Commented Aug 24, 2016 at 11:58

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