To solve the unsolvable differential equation

[the_dialogue]

Homework Statement

Solve:
(2x^3 - y)dx + (2x^2y + x)dy = 0

2. The attempt at a solution
I've tried looking at this equation as a separable, homogeneous, linear, Bernoulli, and exact equation. None of them work.

Any direction is greatly appreciated. Thank you.

 

[skies222]

hmm, that is a tricky problem. So you're trying to differentiate it right? Find the deriative? It looks like an implicit differentiation problem, but I am not sure how the "dx" and "dy" fit in there...wouldn't it be an integration problem if dy and dx are in the equation?

I'm probably not much help since I've never seen a problem like this before. Sorry dude!

 

[d_leet]

hmm, that is a tricky problem. So you're trying to differentiate it right? Find the deriative? It looks like an implicit differentiation problem, but I am not sure how the "dx" and "dy" fit in there...wouldn't it be an integration problem if dy and dx are in the equation?

I'm probably not much help since I've never seen a problem like this before. Sorry dude!

He's not trying to differentiate it, he's trying to solve it, it's a differential equation. It looks like it might be exact if you could find an integrating factor.

 

[HallsofIvy]

It looks like it might be exact if you could find an integrating factor.

I love that! ANY first order differential equation is exact if you multiply by the integrating factor! The trick is to find the integrating factor!

Try putting it into polar coordinates: x= r cos(\theta), y= r sin(\theta). I believe it reduces to a separable equation.

 

[silver-rose]

If you sub those polar coordinates, then there's only one type of differential (i mean there's only d(theta) right?

With only one differential, how do you make a separable equation; sorry, I'm quite confused!

 

[TMFKAN64]

You also have dr in polar coordinates.

My question now is how do you think of polar coordinates in the first place? Is there something about the equation that leads you to believe it might work, or is it just another standard substitution that you try to see if it leads anywhere?

 

[Azurath]

(2x^(3)-y)dx+(2x^(2)y+x)dy=0

Arrange the variables alphabetically within the expression (2x^(2)y+x)dy. This is the standard way of writing an expression.
dx(2x^(3)-y)+dy(2x^(2)y+x)=0

Factor out the GCF of d from each term in the polynomial.
d(x(2x^(3)-y))+d(y(2x^(2)y+x))=0

Factor out the GCF of d from dx(2x^(3)-y)+dy(2x^(2)y+x).
d(x(2x^(3)-y)+y(2x^(2)y+x))=0

Multiply x by each term inside the parentheses.
d((2x^(4)-xy)+y(2x^(2)y+x))=0

Multiply y by each term inside the parentheses.
d((2x^(4)-xy)+(2x^(2)y^(2)+xy))=0

Remove the parentheses that are not needed from the expression.
d(2x^(4)-xy+2x^(2)y^(2)+xy)=0

Since -xy and xy are like terms, subtract xy from -xy to get 0.
d(2x^(4)+2x^(2)y^(2))=0

Factor out the GCF of 2x^(2) from each term in the polynomial.
d(2x^(2)(x^(2))+2x^(2)(y^(2)))=0

Factor out the GCF of 2x^(2) from 2x^(4)+2x^(2)y^(2).
d(2x^(2)(x^(2)+y^(2)))=0

Multiply d by 2x^(2) to get 2dx^(2).
(2dx^(2))(x^(2)+y^(2))=0

Remove the parentheses.
2dx^(2)(x^(2)+y^(2))=0

If any individual factor on the left-hand side of the equation is equal to 0, the entire expression will be equal to 0.
2dx^(2)=0_(x^(2)+y^(2))=0

Set the first factor equal to 0 and solve.
2dx^(2)=0

Divide each term in the equation by 2d.
(2dx^(2))/(2d)=(0)/(2d)

Simplify the left-hand side of the equation by canceling the common terms.
x^(2)=(0)/(2d)

0 divided by any number or variable is 0.
x^(2)=0

Take the square root of both sides of the equation to eliminate the exponent on the left-hand side.
x=\~(0)

Pull all perfect square roots out from under the radical. In this case, remove the 0 because it is a perfect square.
x=\0

\0 is equal to 0.
x=0

Set the next factor equal to 0 and solve.
(x^(2)+y^(2))=0

Remove the parentheses around the expression x^(2)+y^(2).
x^(2)+y^(2)=0

Since y^(2) does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting y^(2) from both sides.
x^(2)=-y^(2)

Take the square root of both sides of the equation to eliminate the exponent on the left-hand side.
x=(-y^(2))^((1)/(2))

The final solution is all the values that make 2dx^(2)(x^(2)+y^(2))=0 true.
x=0,(-y^(2))^((1)/(2))