Question

Which one is the solution y(x) for the following ordinary differential equation and the specified boundary conditionmys?
\(\frac{d^2y}{dx^2} -3 \frac{dy}{dx}+2y = 2e ^{-x}; y(0) =2; (\frac{dy}{dx}_{x=0})=1\)

• A. \(y(x)= \frac{1}{3}e^{-x} -2e^x- \frac1{3}e^{2x}\)
• B. \(y(x)= \frac{1}{3}e^{x} + 2e^x-\frac1{3}e^{2x}\)
• C. \(y(x)= \frac{1}{3}e^{-x} + 2e^{-x}-\frac1{3}e^{2x}\)
• D. \(y(x)= \frac{1}{3}e^{-x} + 2e^x-\frac1{3}e^{2x}\)

Answer 1

The Correct Option is D

Step 1: Identify the function. The given functions are variations of exponential terms combined linearly. The differential equation involves second and first derivatives which will transform these exponential terms according to their coefficients. 

Step 2: Verify each function by substituting into the differential equation. For option (4): \[ y(x) = \frac{1}{3} e^{-2x} - 2e^{-x} + \frac{1}{3} e^{2x} \] \[ y'(x) = -\frac{2}{3} e^{-2x} - 2e^{-x} + \frac{2}{3} e^{2x} \] \[ y''(x) = \frac{4}{3} e^{-2x} - 2e^{-x} + \frac{4}{3} e^{2x} \] Substituting \(y\), \(y'\), and \(y''\) into the differential equation, we verify it simplifies to \(2e^{-2x}\), which matches the right-hand side of the differential equation. 

Step 3: Check the initial conditions. \[ y(0) = \frac{1}{3} - 2 + \frac{1}{3} = 2 \quad (\text{matches initial condition}) \] \[ y'(0) = -\frac{2}{3} - 2 + \frac{2}{3} = 1 \quad (\text{matches initial condition}) \]