Solve the DE by Exact method:-(y^2. cosx- 3x^2y - 2x)dx + (2ysinx-x^3. + In y) dy,
y (o) = e

To solve the differential equation by the exact method, we first need to verify that the given differential equation is exact. An exact differential equation can be written in the form:

\[
M(x, y) \, dx + N(x, y) \, dy = 0
\]

Given equation:

\[
(y^2 \cos x - 3x^2 y - 2x) \, dx + \left(2y \sin x - x^3 + \ln y\right) \, dy = 0
\]

Here, we identify:

\[
M(x, y) = y^2 \cos x - 3x^2 y - 2x
\]

\[
N(x, y) = 2y \sin x - x^3 + \ln y
\]

### Step 1: Check for Exactness

To check if the equation is exact, compute the partial derivatives:

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

\[
\frac{\partial N}{\partial x} = \frac{\partial}{\partial x} \left(2y \sin x - x^3 + \ln y\right) = 2y \cos x - 3x^2
\]

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

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

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

\[
\frac{\partial \psi}{\partial x} = M(x, y)
\]

\[
\frac{\partial \psi}{\partial y} = N(x, y)
\]

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

\[
\psi(x, y) = \int (y^2 \cos x - 3x^2 y - 2x) \, dx
\]

\[
\psi(x, y) = y^2 \sin x - x^3 y - x^2 + h(y)
\]

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

Now, differentiate \(\psi(x, y)\) with respect to \(y\) and equate to \(N(x, y)\):

\[
\frac{\partial \psi}{\partial y} = 2y \sin x - x^3 + h'(y) = N(x, y) = 2y \sin x - x^3 + \ln y
\]

Equating terms, we find:

\[
h'(y) = \ln y
\]

Integrating with respect to \(y\):

\[
h(y) = y \ln y - y + C
\]

### Step 3: Write the General Solution

Substituting \(h(y)\) back into \(\psi(x, y)\), the potential function becomes:

\[
\psi(x, y) = y^2 \sin x - x^3 y - x^2 + y \ln y - y + C
\]

The general solution is given by:

\[
\psi(x, y) = C
\]

So, the solution to the differential equation is:

\[
y^2 \sin x - x^3 y - x^2 + y \ln y - y = C
\]

### Step 4: Apply the Initial Condition

Given \(y(0) = e\), substitute \(x = 0\) and \(y = e\) into the equation:

\[
(e)^2 \sin(0) - (0)^3 e - (0)^2 + e \ln e - e = C
\]

\[
0 + 0 + 0 + e \cdot 1 - e = C
\]

\[
C = 0
\]

So, the solution simplifies to:

\[
y^2 \sin x - x^3 y - x^2 + y \ln y - y = 0
\]

This is the required solution.

Would you like more details or have any questions? 

Here are five related questions:

1. What is an exact differential equation, and how do you recognize one?
2. What are the steps to solve a non-exact differential equation?
3. How does the method of integrating factors apply to exact equations?
4. Can every differential equation be made exact, and if so, how?
5. How do you interpret the potential function \(\psi(x, y)\) in the context of physics?

**Tip:** Always verify the exactness of a differential equation by checking the partial derivatives of \(M(x, y)\) and \(N(x, y)\) before attempting to solve it.

Learn how to solve the exact differential equation (y^2 cos(x) - 3x^2 y - 2x) dx + (2y sin(x) - x^3 + ln(y)) dy step-by-step. Verify exactness, find potential function, apply initial conditions. Suitable for advanced undergraduate level students.

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