Question

Tell if the DE is exact and solve: (3x^2+1)+(3y^2+2y)y'=0

Answer 1

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 \( C \).