Question

Let \( y = y(x) \) be the solution of the differential equation: 
\[ \frac{dy}{dx} + \left( \frac{6x^2 + (3x^2 + 2x^3 + 4)e^{-2x}}{(x^3 + 2)(2 + e^{-2x})} \right)y = 2 + e^{-2x}, \quad x \in (-1, 2) \] 
satisfying \( y(0) = \frac{3}{2} \). 
If \( y(1) = \alpha \left(2 + e^{-2}\right) \), then the value of \( \alpha \) is ________.

• A. \(\frac{13}{8}\)
• B. \(\frac{6}{13}\)
• C. \(\frac{12}{13}\)
• D. \(\frac{13}{12}\)

Answer 1

The Correct Option is C

Step 1: Identify the differential equation form
The given equation is a first-order linear differential equation of the form \[ \frac{dy}{dx}+P(x)y=Q(x) \] where \[ P(x)= \frac{6x^2+(3x^2+2x^3+4)e^{-2x}} {(x^3+2)(2+e^{-2x})} \] and \[ Q(x)=2+e^{-2x} \] Step 2: Find the integrating factor (I.F.)
Observe that \[ \frac{d}{dx}\ln(x^3+2)=\frac{3x^2}{x^3+2} \] and \[ \frac{d}{dx}\ln(2+e^{-2x}) = \frac{-2e^{-2x}}{2+e^{-2x}} \] Therefore, \[ \frac{d}{dx} \left[ \ln(x^3+2)-\ln(2+e^{-2x}) \right] \] \[ = \frac{3x^2}{x^3+2} + \frac{2e^{-2x}}{2+e^{-2x}} \] Taking LCM: \[ = \frac{3x^2(2+e^{-2x})+2e^{-2x}(x^3+2)} {(x^3+2)(2+e^{-2x})} \] \[ = \frac{ 6x^2+3x^2e^{-2x}+2x^3e^{-2x}+4e^{-2x} } {(x^3+2)(2+e^{-2x})} \] \[ = \frac{ 6x^2+(3x^2+2x^3+4)e^{-2x} } {(x^3+2)(2+e^{-2x})} \] This is exactly $P(x)$. Hence, \[ \int P(x)\,dx = \ln(x^3+2)-\ln(2+e^{-2x}) \] \[ = \ln\left(\frac{x^3+2}{2+e^{-2x}}\right) \] So the integrating factor is \[ \text{I.F.} = e^{\int P(x)\,dx} = \frac{x^3+2}{2+e^{-2x}} \] Step 3: Multiply the equation by I.F.
Multiplying the differential equation by the integrating factor: \[ \frac{d}{dx} \left[ y\cdot \frac{x^3+2}{2+e^{-2x}} \right] = (2+e^{-2x}) \cdot \frac{x^3+2}{2+e^{-2x}} \] \[ = x^3+2 \] Now integrate both sides: \[ \int \frac{d}{dx} \left[ y\cdot \frac{x^3+2}{2+e^{-2x}} \right]dx = \int (x^3+2)\,dx \] \[ y\cdot \frac{x^3+2}{2+e^{-2x}} = \frac{x^4}{4}+2x+C \] Step 4: Use the initial condition
Given \[ y(0)=\frac{3}{2} \] Substitute $x=0$: \[ \frac{3}{2}\cdot \frac{0^3+2}{2+e^0} = \frac{0^4}{4}+2(0)+C \] \[ \frac{3}{2}\cdot\frac{2}{3}=C \] \[ C=1 \] So the solution becomes \[ y\cdot \frac{x^3+2}{2+e^{-2x}} = \frac{x^4}{4}+2x+1 \] Step 5: Find $y(1)$
Put $x=1$: \[ y(1)\cdot \frac{1^3+2}{2+e^{-2}} = \frac{1^4}{4}+2(1)+1 \] \[ y(1)\cdot \frac{3}{2+e^{-2}} = \frac{1}{4}+2+1 \] \[ = \frac{13}{4} \] Therefore, \[ y(1) = \frac{13}{4}\cdot \frac{2+e^{-2}}{3} \] \[ = \frac{13}{12}(2+e^{-2}) \] Given \[ y(1)=\alpha(2+e^{-2}) \] Comparing, \[ \alpha=\frac{13}{12} \] Final Answer: \[ \boxed{\alpha=\frac{13}{12}} \] \[ \boxed{\text{Option (D)}} \]