To solve the differential equation
\[ (3x^2 + 6xy^2) , dx + ((6x^2)y + 4y^2) , dy = 0, \]
we can rewrite it in the form
\[ M(x, y) , dx + N(x, y) , dy = 0, \]
where \[ M(x, y) = 3x^2 + 6xy^2 \] and \[ N(x, y) = (6x^2y + 4y^2). \]
To check if the equation is exact, we need to verify if
\[ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}. \]
First, we compute the partial derivatives:
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Compute \(\frac{\partial M}{\partial y}\): \[ M(x, y) = 3x^2 + 6xy^2 \implies \frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(3x^2) + \frac{\partial}{\partial y}(6xy^2) = 0 + 12xy = 12xy. \]
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Compute \(\frac{\partial N}{\partial x}\): \[ N(x, y) = 6x^2y + 4y^2 \implies \frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(6x^2y) + \frac{\partial}{\partial x}(4y^2) = 12xy + 0 = 12xy. \]
Since \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} = 12xy\), the equation is exact.
Next, we find a function \(\Psi(x, y)\) such that
\[ \frac{\partial \Psi}{\partial x} = M(x, y) \quad \text{and} \quad \frac{\partial \Psi}{\partial y} = N(x, y). \]
Step 1: Integrate \(M\) with respect to \(x\):
\[ \Psi(x, y) = \int (3x^2 + 6xy^2) , dx = x^3 + 3xy^2 + h(y), \] where \(h(y)\) is an arbitrary function of \(y\).
Step 2: Differentiate \(\Psi\) with respect to \(y\):
\[ \frac{\partial \Psi}{\partial y} = \frac{\partial}{\partial y}(x^3 + 3xy^2 + h(y)) = 6xy + h'(y). \]
Step 3: Set this equal to \(N(x,y)\):
\[ 6xy + h'(y) = 6x^2y + 4y^2. \] From this, we obtain: \[ h'(y) = 4y^2. \]
Step 4: Integrate \(h'(y)\) with respect to \(y\):
\[ h(y) = \int 4y^2 , dy = \frac{4y^3}{3} + C, \] where \(C\) is a constant.
Step 5: Substitute \(h(y)\) back into \(\Psi\):
\[ \Psi(x, y) = x^3 + 3xy^2 + \frac{4y^3}{3} + C. \]
Conclusion
Setting \(\Psi(x, y) = k\) (where \(k\) is a constant), we have the implicit solution to the differential equation: \[ x^3 + 3xy^2 + \frac{4y^3}{3} = C. \]
This represents the family of solutions to the given differential equation.
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