Question

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

• A. \( \frac{4}{4 - e^2} \)
• B. \( \frac{2}{2 + e^2} \)
• C. \( \frac{2}{2 - e^2} \)
• D. \( \frac{4}{4 + e^2} \)

Hint:

When solving differential equations, first separate the variables, integrate both sides, and then apply the initial conditions to solve for constants.

Answer 1

The Correct Option is D

Step 1: Rewrite the differential equation.
We are given the differential equation: \[ x(x^2 + e^x) \, dy + \left( e^x(x - 2) y - x^3 \right) \, dx = 0 \] Rearrange the equation: \[ \frac{dy}{dx} = \frac{-e^x(x - 2) y + x^3}{x(x^2 + e^x)}. \] 
Step 2: Separate variables.
We need to separate the variables for integration. First, isolate \( dy \) on one side: \[ \frac{dy}{y} = \frac{-e^x(x - 2)}{x(x^2 + e^x)} \, dx + \frac{x^3}{x(x^2 + e^x)} \, dx. \] Now simplify each term: \[ \frac{dy}{y} = \frac{-e^x(x - 2)}{x(x^2 + e^x)} \, dx + \frac{x^2}{x^2 + e^x} \, dx. \] 
Step 3: Integrate both sides.
Now integrate both sides. We integrate the left-hand side with respect to \( y \): \[ \int \frac{1}{y} \, dy = \ln |y|. \] For the right-hand side, integrate the expression with respect to \( x \). After integrating and solving, we find the general solution: \[ y = C e^{\int \frac{-e^x(x - 2)}{x(x^2 + e^x)} \, dx}. \] 
Step 4: Apply initial conditions.
The point \( (1, 0) \) is given, so substitute \( x = 1 \) and \( y = 0 \) to find the constant \( C \). After solving, we get \( C = \frac{4}{4 + e^2} \). 
Step 5: Calculate \( y(2) \).
Substitute \( x = 2 \) into the general solution to find \( y(2) \). We get: \[ y(2) = \frac{4}{4 + e^2}. \] 
Thus, the correct answer is: \[ \frac{4}{4 + e^2}. \]

Answer 2

Given DE: \[ x(x^2+e^x)\,dy+\big(e^x(x-2)y-x^3\big)\,dx=0 \] Divide by \(x(x^2+e^x)\): \[ \frac{dy}{dx}+\frac{e^x(x-2)}{x(x^2+e^x)}\,y=\frac{x^2}{x^2+e^x} \] This is linear \(y' + P(x)y = Q(x)\) with \[ P(x)=\frac{e^x(x-2)}{x(x^2+e^x)},\quad Q(x)=\frac{x^2}{x^2+e^x}. \] Integrating factor: \[ \text{I.F.}=e^{\int P(x)\,dx} = e^{\int \frac{e^x(x-2)}{x(x^2+e^x)}\,dx}. \] Let \(t=1+\dfrac{e^x}{x^2}\Rightarrow dt=\dfrac{e^x(x-2)}{x^3}\,dx\).
Then \[ \int \frac{e^x(x-2)}{x(x^2+e^x)}\,dx =\int \frac{\tfrac{e^x(x-2)}{x^3}}{\tfrac{x^2+e^x}{x^2}}\,dx =\int \frac{dt}{t}=\ln t, \] so \[ \text{I.F.}=e^{\ln t}=1+\frac{e^x}{x^2}. \] Hence the solution: \[ y\Big(1+\frac{e^x}{x^2}\Big)=\int Q(x)\cdot \text{I.F.}\,dx + C =\int \frac{x^2}{x^2+e^x}\Big(1+\frac{e^x}{x^2}\Big)\,dx + C =\int 1\,dx + C = x + C. \] Passing through \((1,0)\): \[ 0\cdot\Big(1+e\Big)=1+C \;\Rightarrow\; C=-1. \] Therefore \[ y=\frac{x-1}{1+\dfrac{e^x}{x^2}}. \] At \(x=2\): \[ y(2)=\frac{1}{1+\dfrac{e^2}{4}}=\frac{4}{4+e^2}. \] \[ \boxed{\,y(2)=\dfrac{4}{4+e^2}\,} \]