To solve the differential equation
\[ (3x^2 + 6xy^2) dx + (6x^2 + 4y^2) dy = 0, \]
we can rearrange it into a more standard form:
\[ (3x^2 + 6xy^2) dx + (6x^2 + 4y^2) dy = 0 \implies \frac{dy}{dx} = -\frac{3x^2 + 6xy^2}{6x^2 + 4y^2}. \]
Now, let's check if this equation is exact. An equation of the form \( M(x, y) dx + N(x, y) dy = 0 \) is exact if
\[ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}. \]
Here we have:
- \( M = 3x^2 + 6xy^2 \)
- \( N = 6x^2 + 4y^2 \)
Now we compute the partial derivatives:
\[ \frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(3x^2 + 6xy^2) = 0 + 12xy = 12xy, \]
\[ \frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(6x^2 + 4y^2) = 12x + 0 = 12x. \]
Since \( \frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x} \), the equation is not exact.
Next, let’s see if we can simplify it further or potentially use an integrating factor. We can try separating variables or checking for integrating factors based on the form.
Attempt to Simplify
Looking at the equation, we can factorize:
\[ 3x^2 + 6xy^2 = 3x^2(1 + 2\frac{y^2}{x}), \] \[ 6x^2 + 4y^2 = 2(3x^2 + 2y^2). \]
Now trying to rewrite the equation:
\[ 3x^2(1 + 2\frac{y^2}{x}) dx + 2(3x^2 + 2y^2) dy = 0. \]
This doesn't lead to an obvious simplification.
Use of Substitution
Consider making a substitution such as \( v = \frac{y}{x} \) (i.e., \( y = vx \)). This gives \( dy = v dx + x dv \).
Substituting into the equation:
Now, replace \( y \):
- \( y^2 = v^2 x^2 \)
- \( dy \) becomes \( vdx + xdv \)
Let's express \( M \) and \( N \) using the substitution in terms of \( x \) and \( v \):
\[ 3x^2 + 6xy^2 = 3x^2 + 6x(v^2 x^2) = 3x^2 + 6v^2 x^3, \]
\[ 6x^2 + 4y^2 = 6x^2 + 4(v^2 x^2) = 6x^2 + 4v^2 x^2. \]
Combining:
This method can lead to a complex simplification, but it suggests non-linearity in \( y \).
Integration
Alternatively, express it symmetrically and try to integrate directly or indirectly by using properties or direct integration if the method is yielding a constant or simplified differential.
As a simpler approach, notice relationships could be approached through \( v \) substitution leading back to any algebraic manipulation straight.
After approaching complexities using a substitution \( y= kx^n \) might yield clarity on a specific genus or analytic forms as sections lead through arc-tangent or log transforms upon integration.
For a polished solution method, numerical methods or computational releases help ground operations further or later confirm an exact nested method through advanced solvability base grounds.
Based on symmetry and continuity of \( M \) and \( N \) with regard to separation, one might yield cosine/sine or explicit forms as \( I(x)^2 + J(y)^2 = C \), leading through derived quadrantics showing forms of expected mathematical outputs.
Would you like to explore a numerical or methodical approach through roots or custom solutions based on direct definitions across the subsistent?
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