what is the differential equation of (3x^2-2y^2)+(1-4xy)dy/dx=0
That is the differential equation! Now, if you want the solution, it is an exact equation, since you have
M(x,y) dx + N(x,y) dy = 0
and
∂M/∂y = ∂N/∂x = -4y
That means there is f(x,y) = c such that
f(x,y) = x^3-2xy^2 + p(y) = y - 2xy^2 + q(x)
so, p(y) = y and q(x) = x^3 and
f(x,y) = x^3-2xy^2+y = c
Thank you :-)
To find the differential equation of the given equation, we need to rearrange it such that it is expressed in terms of \(\frac{{dy}}{{dx}}\).
The given equation is:
\(3x^2 - 2y^2 + (1 - 4xy)\frac{{dy}}{{dx}} = 0\)
Let's begin solving for \(\frac{{dy}}{{dx}}\):
\(3x^2 - 2y^2 = (4xy - 1)\frac{{dy}}{{dx}}\)
Next, isolate \(\frac{{dy}}{{dx}}\):
\(\frac{{dy}}{{dx}} = \frac{{3x^2 - 2y^2}}{{4xy - 1}}\)
Therefore, the differential equation of the given equation is:
\(\frac{{dy}}{{dx}} = \frac{{3x^2 - 2y^2}}{{4xy - 1}}\)
To find the differential equation of the given expression, we need to isolate dy/dx.
Step 1: Rearrange the equation.
(3x^2 - 2y^2) + (1 - 4xy)dy/dx = 0
Step 2: Move the term (3x^2 - 2y^2) to the right-hand side.
(1 - 4xy)dy/dx = 2y^2 - 3x^2
Step 3: Divide both sides by (1 - 4xy).
dy/dx = (2y^2 - 3x^2) / (1 - 4xy)
Therefore, the differential equation of the given expression is dy/dx = (2y^2 - 3x^2) / (1 - 4xy).