KCET · Maths · Differential Equations
\(\int xf(x)\,dx + \dfrac{f(x)}{2} = 0\), then \(f(x)\) is equal to
- A \(e^{-2x}\)
- B \(e^{2x}\)
- C \(e^{-x^2}\)
- D \(e^{x^2}\)
Answer & Solution
Correct Answer
(C) \(e^{-x^2}\)
Step-by-step Solution
Detailed explanation
Differentiating the given equation with respect to \(x\):
\(x f(x) + \dfrac{1}{2} f'(x) = 0\)
\(\dfrac{f'(x)}{f(x)} = -2x\)
Integrating both sides with respect to \(x\):
\(\int \dfrac{f'(x)}{f(x)} \, dx = \int -2x \, dx\)
\(\ln |f(x)| = -x^2 + C\)
\(f(x) = A e^{-x^2}\)
For \(A = 1\), \(f(x) = e^{-x^2}\).
Answer: \(e^{-x^2}\)
\(x f(x) + \dfrac{1}{2} f'(x) = 0\)
\(\dfrac{f'(x)}{f(x)} = -2x\)
Integrating both sides with respect to \(x\):
\(\int \dfrac{f'(x)}{f(x)} \, dx = \int -2x \, dx\)
\(\ln |f(x)| = -x^2 + C\)
\(f(x) = A e^{-x^2}\)
For \(A = 1\), \(f(x) = e^{-x^2}\).
Answer: \(e^{-x^2}\)
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