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 form f(x) + f'(x) = xe^(-x) as a linear first-order DE. Multiply both sides by integrating factor e^x:

e^x·f(x) + e^x·f'(x) = x

Step 2: Observe that the left side is d/dx[e^x·f(x)]:

d/dx[e^x·f(x)] = x

Step 3: Integrate both sides with respect to x:

e^x·f(x) = ∫x dx = x²/2 + C

Step 4: Apply initial condition f(0) = 0:

e^0·f(0) = 0²/2 + C

1·0 = 0 + C ⟹ C = 0

Step 5: Solve for f(x):

e^x·f(x) = x²/2

f(x) = (x²/2)e^(-x)

Step 6: Evaluate f(1):

f(1) = (1²/2)e^(-1) = (1/2)e^(-1) = 1/(2e)

∴ Answer: A

Question: