Differential Equations Question - None

Question

Let $y = f(x)$ be a differentiable function satisfying $f(x) + f'(x) = xe^{-x}$ for all values of real $x$. If $f(0) = 0$, then the value of $f(1)$ equals:

Accepted Answer

Step 1: Recognize the differential equation form: f(x) + f'(x) = xe^{-x}

Step 2: Multiply both sides by the integrating factor e^x (since d/dx[e^x] = e^x):
e^x·f(x) + e^x·f'(x) = x

Step 3: Notice the left side is the derivative of a product:
d/dx[e^x·f(x)] = x

Step 4: Integrate both sides with respect to x:
e^x·f(x) = ∫x dx = x²/2 + C

Step 5: Apply initial condition f(0) = 0:
e^0·f(0) = 0²/2 + C
0 = 0 + C
C = 0

Step 6: Therefore: e^x·f(x) = x²/2
f(x) = (x²/2)·e^{-x}

Step 7: Calculate f(1):
f(1) = (1²/2)·e^{-1} = (1/2)·e^{-1} = 1/(2e)
∴ Answer: A

Question:

Step-by-Step Solution

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