Differential Equation Analysis

The given differential equation is:

$x (x^2 + e^x) dy + (e^x (x-2) y - x^3) dx = 0$

We can write this in the form $M dx + N dy = 0$, where:

• $M(x, y) = e^x (x-2) y - x^3$
• $N(x, y) = x (x^2 + e^x) = x^3 + x e^x$

Checking for Exactness

To determine if the equation is exact, we compute the partial derivatives:

• $\frac{\partial M}{\partial y} = \frac{\partial}{\partial y} (e^x (x-2) y - x^3) = e^x (x-2)$
• $\frac{\partial N}{\partial x} = \frac{\partial}{\partial x} (x^3 + x e^x) = 3x^2 + (1 \cdot e^x + x \cdot e^x) = 3x^2 + e^x + x e^x$

Since $\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}$, the differential equation is not exact.

Finding the Integrating Factor

We look for an integrating factor $\mu(x)$ that makes the equation exact. We compute:

$\frac{1}{N}\left(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}\right)$

First, calculate the difference of partial derivatives:

• $\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x} = e^x(x-2) - (3x^2 + e^x + x e^x) = x e^x - 2e^x - 3x^2 - e^x - x e^x = -3x^2 - 3e^x = -3(x^2 + e^x)$

Now, divide by $N(x, y)$:

$\frac{1}{N}(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}) = \frac{-3(x^2 + e^x)}{x(x^2 + e^x)} = -\frac{3}{x}$

Since this expression depends only on $x$, the integrating factor is:

$\mu(x) = e^{\int -\frac{3}{x} dx} = e^{-3 \ln|x|} = x^{-3}$ (since $x > 0$)

Transforming to an Exact Equation

Multiply the original differential equation by the integrating factor $\mu(x) = \frac{1}{x^3}$:

$\frac{1}{x^3} [x (x^2 + e^x) dy] + \frac{1}{x^3} [(e^x (x-2) y - x^3) dx] = 0$

This simplifies to:

$\frac{x^2 + e^x}{x^2} dy + \left(\frac{e^x (x-2)}{x^3} y - 1\right) dx = 0$

Let the new coefficients be $M'(x,y)$ and $N'(x,y)$:

• $M'(x, y) = \frac{e^x (x-2)}{x^3} y - 1$
• $N'(x, y) = \frac{x^2 + e^x}{x^2} = 1 + \frac{e^x}{x^2}$

We verify that this new equation is exact:

• $\frac{\partial M'}{\partial y} = \frac{e^x(x-2)}{x^3}$
• $\frac{\partial N'}{\partial x} = \frac{\partial}{\partial x} (1 + x^{-2} e^x) = -2x^{-3} e^x + x^{-2} e^x = \frac{-2e^x + xe^x}{x^3} = \frac{e^x(x-2)}{x^3}$

Since $\frac{\partial M'}{\partial y} = \frac{\partial N'}{\partial x}$, the transformed equation is exact.

Finding the General Solution

The solution is of the form $F(x, y) = C$, where $\frac{\partial F}{\partial y} = N'(x, y)$ and $\frac{\partial F}{\partial x} = M'(x, y)$.

Integrate $N'(x, y)$ with respect to $y$:

$F(x, y) = \int N'(x, y) dy = \int \left(1 + \frac{e^x}{x^2}\right) dy = y \left(1 + \frac{e^x}{x^2}\right) + g(x)$

Differentiate $F(x, y)$ with respect to $x$:

$\frac{\partial F}{\partial x} = y \frac{\partial}{\partial x} \left(\frac{e^x}{x^2}\right) + g'(x) = y \left(\frac{x^2 e^x - e^x (2x)}{x^4}\right) + g'(x) = \frac{y e^x (x-2)}{x^3} + g'(x)$

Equate this to $M'(x, y)$:

$\frac{y e^x (x-2)}{x^3} + g'(x) = \frac{e^x (x-2)}{x^3} y - 1$

This implies $g'(x) = -1$. Integrating yields $g(x) = -x$.

The general solution is $F(x, y) = C$:

$y \left(1 + \frac{e^x}{x^2}\right) - x = C$

This can be written as $y \left(\frac{x^2 + e^x}{x^2}\right) - x = C$.

Applying the Initial Condition

The solution curve passes through the point $(1, 0)$, meaning $y(1) = 0$. Substitute $x=1$ and $y=0$ into the general solution:

$0 \left(\frac{1^2 + e^1}{1^2}\right) - 1 = C \implies 0 - 1 = C \implies C = -1$

The particular solution is:

$y \left(\frac{x^2 + e^x}{x^2}\right) - x = -1$

Solving for $y$ gives:

$y = \frac{x^2 (x - 1)}{x^2 + e^x}$

Calculating y(2)

To find $y(2)$, substitute $x=2$ into the particular solution:

$y(2) = \frac{2^2 (2 - 1)}{2^2 + e^2} = \frac{4(1)}{4 + e^2} = \frac{4}{4 + e^2}$

Therefore, $y(2) = \frac{4}{4+e^2}$.