(y³-1)e^xdx+3y²(e^x+1)dy=0, y(0)=0 solve the exact differential equation 

To solve the given exact differential equation, we start by verifying whether it is indeed exact, then proceed with finding a potential function \( F(x, y) \) such that \( dF = 0 \).

The given differential equation is:

\[
(y^3 - 1)e^x \, dx + 3y^2(e^x + 1) \, dy = 0
\]

### Step 1: Verify Exactness

An equation of the form \( M(x, y) \, dx + N(x, y) \, dy = 0 \) is exact if:

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

Here,

\[
M(x, y) = (y^3 - 1)e^x, \quad N(x, y) = 3y^2(e^x + 1)
\]

Compute the partial derivatives:

\[
\frac{\partial M}{\partial y} = \frac{\partial}{\partial y} \left( (y^3 - 1)e^x \right) = 3y^2 e^x
\]

\[
\frac{\partial N}{\partial x} = \frac{\partial}{\partial x} \left( 3y^2(e^x + 1) \right) = 3y^2 e^x
\]

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

### Step 2: Find the Potential Function \( F(x, y) \)

Since the equation is exact, there exists a function \( F(x, y) \) such that:

\[
\frac{\partial F}{\partial x} = M(x, y) = (y^3 - 1)e^x, \quad \frac{\partial F}{\partial y} = N(x, y) = 3y^2(e^x + 1)
\]

#### Finding \( F(x, y) \)

Integrate \( M(x, y) \) with respect to \( x \):

\[
F(x, y) = \int (y^3 - 1)e^x \, dx = (y^3 - 1)e^x + g(y)
\]

where \( g(y) \) is a function of \( y \) alone.

Next, differentiate \( F(x, y) \) with respect to \( y \) and set it equal to \( N(x, y) \):

\[
\frac{\partial F}{\partial y} = \frac{\partial}{\partial y} \left( (y^3 - 1)e^x + g(y) \right) = 3y^2 e^x + g'(y)
\]

Set this equal to \( N(x, y) \):

\[
3y^2 e^x + g'(y) = 3y^2(e^x + 1)
\]

Simplify the equation:

\[
3y^2 e^x + g'(y) = 3y^2 e^x + 3y^2
\]

\[
g'(y) = 3y^2
\]

Integrate \( g'(y) \) with respect to \( y \):

\[
g(y) = \int 3y^2 \, dy = y^3 + C
\]

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

\[
F(x, y) = (y^3 - 1)e^x + y^3 + C
\]

### Step 3: Solve for the Implicit Solution

The general solution is given by \( F(x, y) = C \). Set \( F(x, y) \) equal to a constant \( C_1 \):

\[
(y^3 - 1)e^x + y^3 = C_1
\]

### Step 4: Apply Initial Condition

Use the initial condition \( y(0) = 0 \) to find \( C_1 \):

\[
(y^3 - 1)e^0 + y^3 = C_1
\]

\[
(0^3 - 1) \cdot 1 + 0^3 = -1 = C_1
\]

So, \( C_1 = -1 \).

### Step 5: Write the Final Solution

Substitute \( C_1 = -1 \) into the general solution:

\[
(y^3 - 1)e^x + y^3 = -1
\]

Simplify the equation:

\[
(y^3 - 1)e^x + y^3 + 1 = 0
\]

\[
(y^3 - 1)(e^x + 1) = 0
\]

This implies two cases:

1. \( y^3 - 1 = 0 \) which gives \( y = 1 \).
2. \( e^x + 1 = 0 \) which has no real solution since \( e^x > 0 \) for all \( x \).

Therefore, the solution is:

\[
\boxed{y = 1}
\]

This is the solution satisfying the initial condition \( y(0) = 0 \).

Would you like more details or have any questions?

### Related Questions:

1. What are the conditions for a differential equation to be exact?
2. How do you verify the exactness of a differential equation?
3. What is a potential function in the context of exact differential equations?
4. How does an initial condition influence the solution of a differential equation?
5. Can all exact differential equations be solved using the method of finding a potential function?

### Tip:

When solving differential equations, always check if the equation is exact or can be made exact by multiplying by an integrating factor. This can simplify the solving process significantly.

Learn how to solve the exact differential equation (y³-1)e^x dx + 3y²(e^x+1) dy = 0 step-by-step. Understand the conditions for exactness, finding potential functions, and applying initial conditions. Suitable for advanced undergraduate students studying differential equations.

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