$$(2xy + 3y^2)dx + (x^2 + 6xy - 2y)dy = 0$$ $$y(1) = -1/2$$
How do you solve this? I have just started learning Differential equations and I have some trouble.
Is this equivalent with this?
$$2y + (6x - 2) = 0$$ $$y(1) = -1/2$$
$$(2xy + 3y^2)dx + (x^2 + 6xy - 2y)dy = 0$$ $$y(1) = -1/2$$
How do you solve this? I have just started learning Differential equations and I have some trouble.
Is this equivalent with this?
$$2y + (6x - 2) = 0$$ $$y(1) = -1/2$$
You can rewrite your equation as $$ df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy = 0, $$ where $df$ is the total differential of a function $f\left(x,y\right)$ that you should determine. Then $df = 0$ implies $f\left(x,y\right)$ is a constant. Determine this constant with the condition on $y\left(1\right)$.