https://r2.erweima.ai/mathbot/image/1725512195921p9bp0kor.jpg

The image shows a differential equation with an initial condition:

\[
(6xy - 2y^2 + 1)dx + (3x^2 - 4xy)dy = 0, \quad y(1) = 2
\]

This appears to be a first-order differential equation. To solve it, we'll first check if it's exact or if it can be made exact.

### Step 1: Check if the equation is exact
The general form of an exact equation is:
\[
M(x, y)dx + N(x, y)dy = 0
\]
where \( M(x, y) = 6xy - 2y^2 + 1 \) and \( N(x, y) = 3x^2 - 4xy \).

For the equation to be exact, the following condition must hold:
\[
\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}
\]

Let's calculate these partial derivatives:

1. \( \frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(6xy - 2y^2 + 1) = 6x - 4y \)
2. \( \frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(3x^2 - 4xy) = 6x - 4y \)

Since \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \), the equation is exact.

### Step 2: Solve the exact equation
To solve an exact equation, we find a potential function \( \psi(x, y) \) such that:
\[
\frac{\partial \psi}{\partial x} = M(x, y), \quad \frac{\partial \psi}{\partial y} = N(x, y)
\]

Integrating \( M(x, y) = 6xy - 2y^2 + 1 \) with respect to \( x \):
\[
\psi(x, y) = \int (6xy - 2y^2 + 1) dx = 3x^2y - 2y^2x + x + h(y)
\]
where \( h(y) \) is an arbitrary function of \( y \).

Now, differentiate \( \psi(x, y) \) with respect to \( y \) and equate it to \( N(x, y) = 3x^2 - 4xy \):
\[
\frac{\partial \psi}{\partial y} = 3x^2 - 4xy + h'(y)
\]
Equating this with \( N(x, y) = 3x^2 - 4xy \), we get \( h'(y) = 0 \), so \( h(y) \) is a constant. Let’s call it \( C \).

Thus, the potential function is:
\[
\psi(x, y) = 3x^2y - 2y^2x + x + C
\]

### Step 3: Solve for the constant \( C \)
Using the initial condition \( y(1) = 2 \), substitute \( x = 1 \) and \( y = 2 \) into \( \psi(x, y) = 0 \):
\[
3(1)^2(2) - 2(2)^2(1) + 1 + C = 0
\]
\[
6 - 8 + 1 + C = 0 \quad \Rightarrow \quad C = 1
\]

### Final Solution:
The solution to the differential equation is:
\[
3x^2y - 2y^2x + x = -1
\]

Would you like more details or have any questions about the solution?

### Related Questions:
1. How do you verify if a differential equation is exact?
2. What is the significance of the initial condition in solving differential equations?
3. How do you solve non-exact differential equations?
4. Can all first-order differential equations be solved using the exact method?
5. How do you interpret the geometric meaning of an exact differential equation?

### Tip:
When checking if an equation is exact, always compare the mixed partial derivatives of \( M(x, y) \) and \( N(x, y) \) to ensure they are equal.

(6xy - 2y^2 + 1)dx + (3x^2 - 4xy)dy = 0, y(1) = 2

Learn how to solve the exact differential equation (6xy - 2y^2 + 1)dx + (3x^2 - 4xy)dy = 0 with an initial condition y(1) = 2. This problem covers key concepts in differential equations, including checking if an equation is exact and finding a potential function. Suitable for undergraduate students studying Calculus II or Differential Equations.

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