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Question

The solution of dy/dx + y = 0 with initial condition y(0) = 1 is

RankGuru · Accepted Answer

1. **Solve General Equation**: $\frac{dy}{dx} = -y$.
   - $\frac{dy}{y} = -dx \implies \ln y = -x + C$
   - $y = e^{-x+C} = K e^{-x}$.
2. **Apply Initial Condition**: Given $y=1$ when $x=0$.
   - $1 = K e^{0} = K \cdot 1$
   - $K = 1$.
3. **Particular Solution**: Substituting $K=1$ back, we get **$y = e^{-x}$**.

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