The integrating factor (I.F.) is e^(∫ P(x)dx). We need to evaluate the integral of P(x). Notice that the numerator e^x(x-2) = xe^x - 2e^x is the derivative of xe^x - 2e^x. The denominator is x(x²+e^x) = x³ + xe^x. This integral is tricky, but we can observe that the derivative of the denominator x³ + xe^x is 3x² + e^x + xe^x. This doesn't directly simplify. Let's re-examine the numerator. The derivative of x² + e^x is 2x + e^x. This also doesn't directly simplify. Let's try a different approach for the integral. Consider the derivative of ln(x² + e^x). It is (2x + e^x)/(x² + e^x). This is not matching. Let's try to rewrite P(x) as (e^x(x-2))/(x(x²+e^x)) = (xe^x - 2e^x)/(x³ + xe^x). This integral is not straightforward. Let's recheck the problem statement for any simplification. The given differential equation is x(x² + e^x)dy + (e^x(x - 2)y - x³)dx = 0. This can be written as x(x² + e^x)dy = -(e^x(x - 2)y - x³)dx. So, (dy)/(dx) = (x³ - e^x(x - 2)y)/(x(x² + e^x)). This is (dy)/(dx) + (e^x(x - 2))/(x(x² + e^x))y = (x²)/(x² + e^x). Let's re-evaluate the integral of P(x). We can rewrite P(x) as (e^x(x-2))/(x(x²+e^x)) = (xe^x - 2e^x)/(x³ + xe^x). This is not easily integrable. Let's try to rewrite the original equation in a different form. Let's try to rewrite the original equation as x(x² + e^x)dy + (xe^x - 2e^x)y dx = x³ dx. Divide by x² + e^x: x dy + (xe^x - 2e^x)/(x² + e^x) y dx = (x³)/(x² + e^x) dx. This is not a standard form. Let's go back to the linear form: (dy)/(dx) + (e^x(x - 2))/(x(x² + e^x))y = (x²)/(x² + e^x). The integrating factor is e^(∫ (e^x(x - 2))/(x(x² + e^x))dx). Let's try to simplify the integrand: (e^x(x - 2))/(x(x² + e^x)) = (xe^x - 2e^x)/(x³ + xe^x). Consider the derivative of ln(x² + e^x). It is (2x + e^x)/(x² + e^x). Consider the derivative of ln(x). It is (1)/(x). Let's try to manipulate the integrand: (e^x(x-2))/(x(x²+e^x)) = (e^x(x²+e^x) - e^x(x²+2))/(x(x²+e^x)). This is not working. Let's try to rewrite the original equation as x(x² + e^x)dy + (xe^x - 2e^x)y dx = x³ dx. This is x(x² + e^x)dy + (xe^x - 2e^x)y dx = x³ dx. Divide by x²: (x + (e^x)/(x))dy + ((e^x)/(x) - (2e^x)/(x²))y dx = x dx. This is not a linear form. Let's re-examine the original equation: x(x² + e^x)dy + (e^x(x - 2)y - x³)dx = 0. This can be written as x(x² + e^x)dy + e^x(x - 2)y dx = x³ dx. Divide by x²: (x + (e^x)/(x))dy + ((e^x)/(x) - (2e^x)/(x²))y dx = x dx. This is not a linear form. Let's try to rewrite the equation as M dx + N dy = 0. M = e^x(x - 2)y - x³ and N = x(x² + e^x). ( M)/( y) = e^x(x - 2). ( N)/( x) = (x² + e^x) + x(2x + e^x) = x² + e^x + 2x² + xe^x = 3x² + e^x + xe^x. Since ( M)/( y) ≠ ( N)/( x), the equation is not exact. Let's try to find an integrating factor. Consider (1)/(N)(( M)/( y) - ( N)/( x)) = (1)/(x(x² + e^x))(e^x(x - 2) - (3x² + e^x + xe^x)) = (xe^x - 2e^x - 3x² - e^x - xe^x)/(x(x² + e^x)) = (-3x² - 3e^x)/(x(x² + e^x)) = (-3(x² + e^x))/(x(x² + e^x)) = -(3)/(x). This is a function of x only. So, the integrating factor is e^(∫ -(3)/(x)dx) = e^(-3ln x) = e^(ln x^(-3)) = x^(-3) = (1)/(x³). Multiply the original equation by (1)/(x³): (1)/(x³) [x(x² + e^x)dy + (e^x(x - 2)y - x³)dx] = 0 (x² + e^x)/(x²)dy + ((e^x(x - 2))/(x³)y - 1)dx = 0 ((e^x(x - 2))/(x³)y - 1)dx + (1 + (e^x)/(x²))dy = 0. Now, let M' = (e^x(x - 2))/(x³)y - 1 and N' = 1 + (e^x)/(x²). ( M')/( y) = (e^x(x - 2))/(x³). ( N')/( x) = (x² e^x - e^x (2x))/(x^4) = (xe^x - 2e^x)/(x³) = (e^x(x - 2))/(x³). Since ( M')/( y) = ( N')/( x), the equation is exact. Now we need to solve the exact differential equation. We know that there exists a function f(x, y) such that ( f)/( x) = M' and ( f)/( y) = N'. Integrate N' with respect to y: f(x, y) = ∫ (1 + (e^x)/(x²))dy = y + (e^x)/(x²)y + g(x). Now differentiate f(x, y) with respect to x and equate it to M': ( f)/( x) = (x² e^x - e^x (2x))/(x^4)y + g'(x) = (xe^x - 2e^x)/(x³)y + g'(x). We have M' = (e^x(x - 2))/(x³)y - 1. So, (e^x(x - 2))/(x³)y + g'(x) = (e^x(x - 2))/(x³)y - 1. This implies g'(x) = -1. Integrating g'(x) with respect to x: g(x) = ∫ -1 dx = -x + C_1. So, the solution is y + (e^x)/(x²)y - x = C. y(1 + (e^x)/(x²)) = x + C. y((x² + e^x)/(x²)) = x + C. y = (x²(x + C))/(x² + e^x).

The integrating factor (I.F.) is e^(∫ P(x)dx). We need to evaluate the integral of P(x). Notice that the numerator e^x(x-2) = xe^x - 2e^x is the derivative of xe^x - 2e^x. The denominator is x(x²+e^x) = x³ + xe^x. This integral is tricky, but we can observe that the derivative of the denominator x³ + xe^x is 3x² + e^x + xe^x. This doesn't directly simplify. Let's re-examine the numerator. The derivative of x² + e^x is 2x + e^x. This also doesn't directly simplify. Let's try a different approach for the integral. Consider the derivative of ln(x² + e^x). It is (2x + e^x)/(x² + e^x). This is not matching. Let's try to rewrite P(x) as (e^x(x-2))/(x(x²+e^x)) = (xe^x - 2e^x)/(x³ + xe^x). This integral is not straightforward. Let's recheck the problem statement for any simplification. The given differential equation is x(x² + e^x)dy + (e^x(x - 2)y - x³)dx = 0. This can be written as x(x² + e^x)dy = -(e^x(x - 2)y - x³)dx. So, (dy)/(dx) = (x³ - e^x(x - 2)y)/(x(x² + e^x)). This is (dy)/(dx) + (e^x(x - 2))/(x(x² + e^x))y = (x²)/(x² + e^x). Let's re-evaluate the integral of P(x). We can rewrite P(x) as (e^x(x-2))/(x(x²+e^x)) = (xe^x - 2e^x)/(x³ + xe^x). This is not easily integrable. Let's try to rewrite the original equation in a different form. Let's try to rewrite the original equation as x(x² + e^x)dy + (xe^x - 2e^x)y dx = x³ dx. Divide by x² + e^x: x dy + (xe^x - 2e^x)/(x² + e^x) y dx = (x³)/(x² + e^x) dx. This is not a standard form. Let's go back to the linear form: (dy)/(dx) + (e^x(x - 2))/(x(x² + e^x))y = (x²)/(x² + e^x). The integrating factor is e^(∫ (e^x(x - 2))/(x(x² + e^x))dx). Let's try to simplify the integrand: (e^x(x - 2))/(x(x² + e^x)) = (xe^x - 2e^x)/(x³ + xe^x). Consider the derivative of ln(x² + e^x). It is (2x + e^x)/(x² + e^x). Consider the derivative of ln(x). It is (1)/(x). Let's try to manipulate the integrand: (e^x(x-2))/(x(x²+e^x)) = (e^x(x²+e^x) - e^x(x²+2))/(x(x²+e^x)). This is not working. Let's try to rewrite the original equation as x(x² + e^x)dy + (xe^x - 2e^x)y dx = x³ dx. This is x(x² + e^x)dy + (xe^x - 2e^x)y dx = x³ dx. Divide by x²: (x + (e^x)/(x))dy + ((e^x)/(x) - (2e^x)/(x²))y dx = x dx. This is not a linear form. Let's re-examine the original equation: x(x² + e^x)dy + (e^x(x - 2)y - x³)dx = 0. This can be written as x(x² + e^x)dy + e^x(x - 2)y dx = x³ dx. Divide by x²: (x + (e^x)/(x))dy + ((e^x)/(x) - (2e^x)/(x²))y dx = x dx. This is not a linear form. Let's try to rewrite the equation as M dx + N dy = 0. M = e^x(x - 2)y - x³ and N = x(x² + e^x). ( M)/( y) = e^x(x - 2). ( N)/( x) = (x² + e^x) + x(2x + e^x) = x² + e^x + 2x² + xe^x = 3x² + e^x + xe^x. Since ( M)/( y) ≠ ( N)/( x), the equation is not exact. Let's try to find an integrating factor. Consider (1)/(N)(( M)/( y) - ( N)/( x)) = (1)/(x(x² + e^x))(e^x(x - 2) - (3x² + e^x + xe^x)) = (xe^x - 2e^x - 3x² - e^x - xe^x)/(x(x² + e^x)) = (-3x² - 3e^x)/(x(x² + e^x)) = (-3(x² + e^x))/(x(x² + e^x)) = -(3)/(x). This is a function of x only. So, the integrating factor is e^(∫ -(3)/(x)dx) = e^(-3ln x) = e^(ln x^(-3)) = x^(-3) = (1)/(x³). Multiply the original equation by (1)/(x³): (1)/(x³) [x(x² + e^x)dy + (e^x(x - 2)y - x³)dx] = 0 (x² + e^x)/(x²)dy + ((e^x(x - 2))/(x³)y - 1)dx = 0 ((e^x(x - 2))/(x³)y - 1)dx + (1 + (e^x)/(x²))dy = 0. Now, let M' = (e^x(x - 2))/(x³)y - 1 and N' = 1 + (e^x)/(x²). ( M')/( y) = (e^x(x - 2))/(x³). ( N')/( x) = (x² e^x - e^x (2x))/(x^4) = (xe^x - 2e^x)/(x³) = (e^x(x - 2))/(x³). Since ( M')/( y) = ( N')/( x), the equation is exact. Now we need to solve the exact differential equation. We know that there exists a function f(x, y) such that ( f)/( x) = M' and ( f)/( y) = N'. Integrate N' with respect to y: f(x, y) = ∫ (1 + (e^x)/(x²))dy = y + (e^x)/(x²)y + g(x). Now differentiate f(x, y) with respect to x and equate it to M': ( f)/( x) = (x² e^x - e^x (2x))/(x^4)y + g'(x) = (xe^x - 2e^x)/(x³)y + g'(x). We have M' = (e^x(x - 2))/(x³)y - 1. So, (e^x(x - 2))/(x³)y + g'(x) = (e^x(x - 2))/(x³)y - 1. This implies g'(x) = -1. Integrating g'(x) with respect to x: g(x) = ∫ -1 dx = -x + C_1. So, the solution is y + (e^x)/(x²)y - x = C. y(1 + (e^x)/(x²)) = x + C. y((x² + e^x)/(x²)) = x + C. y = (x²(x + C))/(x² + e^x).

Let $y = y(x)$ be the solution curve of the differential equation $x(x^2 + e^x)dy + (e^x(x - 2)y - x^3)dx = 0,\ x > 0$ , passing through the point $(1, 0)$ . Then $y(2)$ is equal to:

Answer: $\frac{4}{4 + e^2}$

Step-by-step solution

Step 1: Rearrange the differential equation into standard form

First, we rearrange the given differential equation to isolate the derivative term and group terms involving $y$ . This helps us identify if it's a linear differential equation, which can be solved using an integrating factor.

Step 2: Convert to linear first-order form

Divide the entire equation by $x(x^2 + e^x)$ to bring it into the standard form of a linear first-order differential equation: $\frac{dy}{dx} + P(x)y = Q(x)$ . Here, $P(x) = \frac{e^x(x - 2)}{x(x^2 + e^x)}$ and $Q(x) = \frac{x^2}{x^2 + e^x}$ .

Step 3: Calculate the integrating factor

The integrating factor (I.F.) is $e^{\int P(x)dx}$ . We need to evaluate the integral of $P(x)$ . Notice that the numerator $e^x(x-2) = xe^x - 2e^x$ is the derivative of $xe^x - 2e^x$ . The denominator is $x(x^2+e^x) = x^3 + xe^x$ . This integral is tricky, but we can observe that the derivative of the denominator $x^3 + xe^x$ is $3x^2 + e^x + xe^x$ . This doesn't directly simplify. Let's re-examine the numerator. The derivative of $x^2 + e^x$ is $2x + e^x$ . This also doesn't directly simplify. Let's try a different approach for the integral. Consider the derivative of $\ln(x^2 + e^x)$ . It is $\frac{2x + e^x}{x^2 + e^x}$ . This is not matching. Let's try to rewrite $P(x)$ as $\frac{e^x(x-2)}{x(x^2+e^x)} = \frac{xe^x - 2e^x}{x^3 + xe^x}$ . This integral is not straightforward. Let's recheck the problem statement for any simplification. The given differential equation is $x(x^2 + e^x)dy + (e^x(x - 2)y - x^3)dx = 0$ . This can be written as $x(x^2 + e^x)dy = -(e^x(x - 2)y - x^3)dx$ . So, $\frac{dy}{dx} = \frac{x^3 - e^x(x - 2)y}{x(x^2 + e^x)}$ . This is $\frac{dy}{dx} + \frac{e^x(x - 2)}{x(x^2 + e^x)}y = \frac{x^2}{x^2 + e^x}$ . Let's re-evaluate the integral of $P(x)$ . We can rewrite $P(x)$ as $\frac{e^x(x-2)}{x(x^2+e^x)} = \frac{xe^x - 2e^x}{x^3 + xe^x}$ . This is not easily integrable. Let's try to rewrite the original equation in a different form. Let's try to rewrite the original equation as $x(x^2 + e^x)dy + (xe^x - 2e^x)y dx = x^3 dx$ . Divide by $x^2 + e^x$ : $x dy + \frac{xe^x - 2e^x}{x^2 + e^x} y dx = \frac{x^3}{x^2 + e^x} dx$ . This is not a standard form. Let's go back to the linear form: $\frac{dy}{dx} + \frac{e^x(x - 2)}{x(x^2 + e^x)}y = \frac{x^2}{x^2 + e^x}$ . The integrating factor is $e^{\int \frac{e^x(x - 2)}{x(x^2 + e^x)}dx}$ . Let's try to simplify the integrand: $\frac{e^x(x - 2)}{x(x^2 + e^x)} = \frac{xe^x - 2e^x}{x^3 + xe^x}$ . Consider the derivative of $\ln(x^2 + e^x)$ . It is $\frac{2x + e^x}{x^2 + e^x}$ . Consider the derivative of $\ln(x)$ . It is $\frac{1}{x}$ . Let's try to manipulate the integrand: $\frac{e^x(x-2)}{x(x^2+e^x)} = \frac{e^x(x^2+e^x) - e^x(x^2+2)}{x(x^2+e^x)}$ . This is not working. Let's try to rewrite the original equation as $x(x^2 + e^x)dy + (xe^x - 2e^x)y dx = x^3 dx$ . This is $x(x^2 + e^x)dy + (xe^x - 2e^x)y dx = x^3 dx$ . Divide by $x^2$ : $(x + \frac{e^x}{x})dy + (\frac{e^x}{x} - \frac{2e^x}{x^2})y dx = x dx$ . This is not a linear form. Let's re-examine the original equation: $x(x^2 + e^x)dy + (e^x(x - 2)y - x^3)dx = 0$ . This can be written as $x(x^2 + e^x)dy + e^x(x - 2)y dx = x^3 dx$ . Divide by $x^2$ : $(x + \frac{e^x}{x})dy + (\frac{e^x}{x} - \frac{2e^x}{x^2})y dx = x dx$ . This is not a linear form. Let's try to rewrite the equation as $M dx + N dy = 0$ . $M = e^x(x - 2)y - x^3$ and $N = x(x^2 + e^x)$ . $\frac{\partial M}{\partial y} = e^x(x - 2)$ . $\frac{\partial N}{\partial x} = (x^2 + e^x) + x(2x + e^x) = x^2 + e^x + 2x^2 + xe^x = 3x^2 + e^x + xe^x$ . Since $\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}$ , the equation is not exact. Let's try to find an integrating factor. Consider $\frac{1}{N}(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}) = \frac{1}{x(x^2 + e^x)}(e^x(x - 2) - (3x^2 + e^x + xe^x)) = \frac{xe^x - 2e^x - 3x^2 - e^x - xe^x}{x(x^2 + e^x)} = \frac{-3x^2 - 3e^x}{x(x^2 + e^x)} = \frac{-3(x^2 + e^x)}{x(x^2 + e^x)} = -\frac{3}{x}$ . This is a function of $x$ only. So, the integrating factor is $e^{\int -\frac{3}{x}dx} = e^{-3\ln x} = e^{\ln x^{-3}} = x^{-3} = \frac{1}{x^3}$ . Multiply the original equation by $\frac{1}{x^3}$ : $\frac{1}{x^3} [x(x^2 + e^x)dy + (e^x(x - 2)y - x^3)dx] = 0$ $\frac{x^2 + e^x}{x^2}dy + (\frac{e^x(x - 2)}{x^3}y - 1)dx = 0$ $(\frac{e^x(x - 2)}{x^3}y - 1)dx + (1 + \frac{e^x}{x^2})dy = 0$ . Now, let $M' = \frac{e^x(x - 2)}{x^3}y - 1$ and $N' = 1 + \frac{e^x}{x^2}$ . $\frac{\partial M'}{\partial y} = \frac{e^x(x - 2)}{x^3}$ . $\frac{\partial N'}{\partial x} = \frac{x^2 e^x - e^x (2x)}{x^4} = \frac{xe^x - 2e^x}{x^3} = \frac{e^x(x - 2)}{x^3}$ . Since $\frac{\partial M'}{\partial y} = \frac{\partial N'}{\partial x}$ , the equation is exact. Now we need to solve the exact differential equation. We know that there exists a function $f(x, y)$ such that $\frac{\partial f}{\partial x} = M'$ and $\frac{\partial f}{\partial y} = N'$ . Integrate $N'$ with respect to $y$ : $f(x, y) = \int (1 + \frac{e^x}{x^2})dy = y + \frac{e^x}{x^2}y + g(x)$ . Now differentiate $f(x, y)$ with respect to $x$ and equate it to $M'$ : $\frac{\partial f}{\partial x} = \frac{x^2 e^x - e^x (2x)}{x^4}y + g'(x) = \frac{xe^x - 2e^x}{x^3}y + g'(x)$ . We have $M' = \frac{e^x(x - 2)}{x^3}y - 1$ . So, $\frac{e^x(x - 2)}{x^3}y + g'(x) = \frac{e^x(x - 2)}{x^3}y - 1$ . This implies $g'(x) = -1$ . Integrating $g'(x)$ with respect to $x$ : $g(x) = \int -1 dx = -x + C_1$ . So, the solution is $y + \frac{e^x}{x^2}y - x = C$ . $y(1 + \frac{e^x}{x^2}) = x + C$ . $y(\frac{x^2 + e^x}{x^2}) = x + C$ . $y = \frac{x^2(x + C)}{x^2 + e^x}$ .

Step 4: Apply the initial condition

The solution curve passes through the point $(1, 0)$ . Substitute $x = 1$ and $y = 0$ into the general solution to find the value of the constant $C$ .

Step 5: Write the particular solution

Substitute the value of $C = -1$ back into the general solution to obtain the particular solution for the given initial condition.

Step 6: Calculate y(2)

Finally, substitute $x = 2$ into the particular solution to find the value of $y(2)$ .

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Let y=y(x)y = y(x)y=y(x) be the solution curve of the differential equation x(x2+ex)dy+(ex(x−2)y−x3)dx=0, x>0x(x^2 + e^x)dy + (e^x(x - 2)y - x^3)dx = 0,\ x > 0x(x2+ex)dy+(ex(x−2)y−x3)dx=0, x>0, passing through the point (1,0)(1, 0)(1,0). Then y(2)y(2)y(2) is equal to: