To determine if the given differential equation
\[ (3x^2 + 1) + (3y^2 + 2y)y' = 0 \]
is exact, we need to rewrite it in the standard form:
\[ M(x, y) + N(x, y) \frac{dy}{dx} = 0 \]
where \( M(x, y) = 3x^2 + 1 \) and \( N(x, y) = 3y^2 + 2y \).
Step 1: Check if the equation is exact.
For the equation to be exact, it must satisfy the condition:
\[ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \]
Calculating the partial derivatives:
- \( M(x, y) = 3x^2 + 1 \) → \( \frac{\partial M}{\partial y} = 0 \)
- \( N(x, y) = 3y^2 + 2y \) → \( \frac{\partial N}{\partial x} = 0 \)
Since both partial derivatives are equal (\( 0 = 0 \)), the equation is exact.
Step 2: Solve the exact equation.
To solve an exact differential equation, we can find a potential function \( \Phi(x, y) \) such that:
\[ \frac{\partial \Phi}{\partial x} = M(x, y) \quad \text{and} \quad \frac{\partial \Phi}{\partial y} = N(x, y) \]
Step 2.1: Finding \( \Phi(x, y) \).
Integrate \( M(x, y) \) with respect to \( x \):
\[ \Phi(x, y) = \int (3x^2 + 1) , dx = x^3 + x + h(y) \]
where \( h(y) \) is an arbitrary function of \( y \).
Step 2.2: Differentiate \( \Phi(x, y) \) with respect to \( y \):
\[ \frac{\partial \Phi}{\partial y} = h'(y) \]
Now set this equal to \( N(x, y) \):
\[ h'(y) = 3y^2 + 2y \]
Integrate \( h'(y) \) with respect to \( y \):
\[ h(y) = \int (3y^2 + 2y) , dy = y^3 + y^2 + C \]
where \( C \) is a constant.
Step 2.3: Combine terms to find \( \Phi(x, y) \):
Now we have:
\[ \Phi(x, y) = x^3 + x + y^3 + y^2 + C \]
Step 3: Set the potential function to a constant for the solution.
The solution to the differential equation is given by:
\[ \Phi(x, y) = k \quad \text{(where \( k \) is a constant)} \]
Thus,
\[ x^3 + x + y^3 + y^2 = C \]
where \( C \) is an arbitrary constant.
Conclusion
The solution to the given exact differential equation is:
\[ x^3 + x + y^3 + y^2 = C \]
This implicitly defines \( y \) in terms of \( x \) and \( C \).