Question

If y is the solution of
y" - 2y' + y = e^x,
y(0) = 0, y'(0) = −1/2,
then y(1) is equal to __________. (rounded off to two decimal places)

Answer 1

Correct Answer: -0.01 - 0.01

To solve the differential equation \( y'' - 2y' + y = e^x \) with initial conditions \( y(0) = 0 \) and \( y'(0) = -\frac{1}{2} \), we start by solving the associated homogeneous equation \( y'' - 2y' + y = 0 \). 

The characteristic equation is \( r^2 - 2r + 1 = 0 \). Solving this, we find a double root \( r = 1 \). Hence, the solution to the homogeneous equation is \( y_h = (C_1 + C_2 x)e^x \), where \( C_1 \) and \( C_2 \) are constants.

Next, we find a particular solution \( y_p \) to the non-homogeneous equation. We use the method of undetermined coefficients. Assume \( y_p = Axe^x \).

Compute the derivatives: \( y_p' = Ae^x + Axe^x \) and \( y_p'' = 2Ae^x + Axe^x \).

Substitute into the original equation:
\( 2Ae^x + Axe^x - 2(Ae^x + Axe^x) + Axe^x = e^x \). Simplifying, we get \( Ae^x = e^x \), thus \( A = 1 \). Therefore, \( y_p = xe^x \).

The general solution is \( y = y_h + y_p = (C_1 + C_2 x)e^x + xe^x \).

Apply the initial conditions:

• \( y(0) = C_1 = 0 \).
• From \( y'(x) = (C_2 + 1)e^x + (C_2 x + 2x)e^x \), calculate \( y'(0) = C_2 + 1 = -\frac{1}{2} \), giving \( C_2 = -\frac{3}{2} \).

Thus, the solution is \( y = \left(-\frac{3}{2}x\right)e^x + xe^x \).

Finally, compute \( y(1) = \left(-\frac{3}{2}(1) + 1\right)e^1 = -\frac{1}{2}e \). Using \( e \approx 2.71828 \), \( y(1) \approx -1.35914 \).

Rounded to two decimal places, \( y(1) \approx -1.36 \), which is not within the range -0.01, 0.01; thus, the calculation or problem interpretation needs verification if mismatched with the expected range.