Question

If function \(y(x)\) satisfies the differential equation \[ \frac{dy}{dx}+\left[\frac{6x^2+e^{-2x}(3x^2+2x^3+4)}{(x^3+2)(2+e^{-2x})}\right]y = e^{-2x}+2 \] such that \(y(0)=\frac{3}{2}\) and \[ y(1)=\alpha(e^{-2}+2) \] then \(\alpha\) is equal to

• A. \(\frac{13}{12}\)
• B. \(\frac{12}{13}\)
• C. \(\frac{4}{3}\)
• D. \(\frac{17}{13}\)

Hint:

For linear differential equations, always compute the integrating factor \(e^{\int P(x)dx}\) to convert the equation into an exact derivative.

Answer 1

The Correct Option is A

Concept: A first order linear differential equation \[ \frac{dy}{dx}+P(x)y=Q(x) \] is solved using integrating factor (I.F.) \[ I.F.=e^{\int P(x)dx} \]
Step 1: Identify \(P(x)\) \[ P(x)=\frac{6x^2+e^{-2x}(3x^2+2x^3+4)}{(x^3+2)(2+e^{-2x})} \] Thus integrating factor \[ I.F.=e^{\int \frac{6x^2+e^{-2x}(3x^2+2x^3+4)}{(x^3+2)(2+e^{-2x})}dx} \]
Step 2: Substitution Let \[ t=(x^3+2)(2+e^{-2x}) \] Then \[ dt=(6x^2+e^{-2x}(3x^2-2x^3-4))dx \] Using this transformation, \[ I.F.=e^{\ln(x^3+2)+\ln(2+e^{-2x})} \] \[ I.F.=\frac{x^3+2}{2+e^{-2x}} \]
Step 3: Multiply the equation by I.F. \[ y\left(\frac{x^3+2}{2+e^{-2x}}\right) = \int \left(\frac{x^3+2}{2+e^{-2x}}\right)(e^{-2x}+2)dx +C \] \[ y\left(\frac{x^3+2}{2+e^{-2x}}\right) = \int (x^3+2)\,dx +C \] \[ y\left(\frac{x^3+2}{2+e^{-2x}}\right) = \frac{x^4}{4}+2x+C \]
Step 4: Use initial condition At \(x=0\), \[ y(0)=\frac{3}{2} \] \[ \frac{3}{2}\left(\frac{2}{3}\right)=C \] \[ C=1 \] Thus \[ y\left(\frac{x^3+2}{2+e^{-2x}}\right) = \frac{x^4}{4}+2x+1 \]
Step 5: Evaluate at \(x=1\) \[ y\left(\frac{3}{2+e^{-2}}\right) = \frac{1}{4}+2+1 \] \[ y\left(\frac{3}{2+e^{-2}}\right)=\frac{13}{4} \] \[ y=\frac{13}{12}(2+e^{-2}) \] Comparing with \[ y=\alpha(2+e^{-2}) \] \[ \alpha=\frac{13}{12} \]