Solve (2x+siny)dx+(xcosy+4y³)dy=0 using exact differential equation 

To solve the differential equation:

\[
(2x + \sin y)dx + (x \cos y + 4y^3)dy = 0
\]

we first check if it is an exact differential equation. A differential equation of the form \( M(x, y)dx + N(x, y)dy = 0 \) is **exact** if the following condition holds:

\[
\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}
\]

### Step 1: Identify \( M(x, y) \) and \( N(x, y) \)
From the given equation:

\[
M(x, y) = 2x + \sin y
\]
\[
N(x, y) = x \cos y + 4y^3
\]

### Step 2: Compute the partial derivatives
- \(\frac{\partial M}{\partial y}\):
\[
\frac{\partial}{\partial y}(2x + \sin y) = \cos y
\]

- \(\frac{\partial N}{\partial x}\):
\[
\frac{\partial}{\partial x}(x \cos y + 4y^3) = \cos y
\]

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

### Step 3: Solve for the potential function \( \psi(x, y) \)

We now find the potential function \( \psi(x, y) \) such that:

\[
\frac{\partial \psi}{\partial x} = M(x, y) = 2x + \sin y
\]
\[
\frac{\partial \psi}{\partial y} = N(x, y) = x \cos y + 4y^3
\]

#### Step 3.1: Integrate \( M(x, y) \) with respect to \( x \)
To find \( \psi(x, y) \), first integrate \( M(x, y) = 2x + \sin y \) with respect to \( x \):

\[
\psi(x, y) = \int (2x + \sin y) \, dx = x^2 + x \sin y + h(y)
\]

Here, \( h(y) \) is an arbitrary function of \( y \) (since the derivative of a function of \( y \) with respect to \( x \) is 0).

#### Step 3.2: Differentiate \( \psi(x, y) \) with respect to \( y \)
Now, differentiate \( \psi(x, y) = x^2 + x \sin y + h(y) \) with respect to \( y \):

\[
\frac{\partial \psi}{\partial y} = x \cos y + h'(y)
\]

Set this equal to \( N(x, y) = x \cos y + 4y^3 \):

\[
x \cos y + h'(y) = x \cos y + 4y^3
\]

This gives:

\[
h'(y) = 4y^3
\]

#### Step 3.3: Integrate \( h'(y) \)
Now, integrate \( h'(y) = 4y^3 \) with respect to \( y \):

\[
h(y) = y^4 + C
\]

Thus, the potential function \( \psi(x, y) \) is:

\[
\psi(x, y) = x^2 + x \sin y + y^4 + C
\]

### Step 4: General solution
Since the differential equation is exact, the general solution is obtained by setting the potential function \( \psi(x, y) \) equal to a constant:

\[
x^2 + x \sin y + y^4 = C
\]

This is the implicit solution to the differential equation.

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Would you like more details on a specific part of this solution? 

Here are some related questions to further explore the topic:
1. How can you verify if a differential equation is exact?
2. What are the conditions for a differential equation to be exact?
3. What happens when a differential equation is not exact?
4. How can integrating factors be used to solve non-exact equations?
5. Can we convert a non-exact equation into an exact one?

**Tip:** When solving exact differential equations, always verify the exactness by checking if \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\) before proceeding.

Solve (2x + siny)dx + (xcosy + 4y³)dy = 0 using exact differential equation

Discover how to solve the exact differential equation (2x + siny)dx + (xcosy + 4y³)dy = 0. This step-by-step guide covers the exactness condition, partial derivatives, and finding the potential function for the solution. Suitable for undergraduate students studying differential equations.

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