# Thread: 3. What method I can use on this equation?

1. ## 3. What method I can use on this equation?

It is Seperation of Variable, Homogeneous Equation, Exact Equation or Linear Equation?

Question
$(x^3 + y^3) dx + y^2(3x + ky) dy = 0$ ; $k$ is a constant

Answer
$ky^4 + 4xy^3 + x^4 = c$

2. Originally Posted by redfox2600
It is Seperation of Variable, Homogeneous Equation, Exact Equation or Linear Equation?

Question
$(x^3 + y^3) dx + y^2(3x + ky) dy = 0$ ; $k$ is a constant

Answer
$ky^4 + 4xy^3 + x^4 = c$
It's similar to others you have asked. Divide through by x^3 and re-arrange. What do you notice?

If you need more help, please post your work.

3. Is this correct? correct if I am wrong.
I used the exact method.

Question
$(x^3 + y^3) dx + y^2(3x + ky) dy = 0$

Answer
$ky^4 + 4xy^3 + x^4 = c$

Solution:
$(x^3 + y^3) dx + (3xy^2 + ky^3) dy = 0$

$M = (x^3 + y^3)$

$N = (3xy^2 + ky^3)$

$\frac {\partial M}{\partial y} = 3y^2$

$\frac {\partial N}{\partial x} = 3y^2$

$\frac {x^4}{4} + xy^3 + \frac {ky^4}{4} = c$

$ky^4 + 4xy^3 + x^4 = c$

thanks mr fantastic.

4. Yes, that is correct.