# Exact differential equation

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• November 13th 2009, 05:00 AM
mormor83
Exact differential equation
hello all

I am having problem understanding this:
i solved (a) :
(4x^3 y^2 + 3x^2 y^4)dx+(2x^4 y + 4x^3 y^3)dy=0
and got this:
x^4 y^2 +x^3 Y^4 = c
the question is (b) to solve for IC :
y(0)=0
and give all the solution. what is there beside x=0 ; y=0 ??
i think maybe x=-y^2 but I'm not sure
thanks for the help
mormor83
• November 13th 2009, 06:38 AM
Calculus26
From the IC:

c= 0

Therfore you would think

x^4 y^2 + x^3 y^4 = 0 is the solution

However

dy/dx = (4x^3 y^2 + 3x^2 y^4)/2x^4 y + 4x^3 y^3)

reduces to:

dy/dx = (4x^3 y + 3x^2 y^3)/2x^4 + 4x^3 y^2)

which has the equiilibrium solution y = 0 as well

Another way of looking at this is

x^4 y^2 + x^3 y^4 = 0

y^2( x^4 +x^3y^2) =0

which yields y = 0 or x = -y^2 as you suggested.

Note

dy/dx = (4x^3 y + 3x^2 y^3)/2x^4 + 4x^3 y^2)

does not satisfy the conditions of the uniqueness theorem.
• November 13th 2009, 06:40 AM
Peritus
This is the resulting equation after applying the IC:

$
x^3 y^2 \left( {x + y^2 } \right) = 0
$

so all the possible solutions are on the following curve:

$
y = \left\{ {\left. { \pm \sqrt { - x} } \right|x \geqslant 0} \right\}
$

which is a canonical parabola rotated 90 degrees to the left.
• November 13th 2009, 10:44 AM
mormor83
thank you very much for that. ( just as i thought but i was not sure)
by the way what do you use to write in "math font"?