Separable ODE: y' + 6xy = 0 with y(0) = 1
Solve the initial value problem $$\frac{dy}{dx} + 6xy = 0, \qquad y(0) = 1.$$
Hints
- Move the $6xy$ term to the right: $\frac{dy}{dx} = -6xy$. The right-hand side factors as (function of $y$) times (function of $x$), so the equation is separable.
- Divide by $y$ and multiply by $dx$: $\frac{dy}{y} = -6x\,dx$. Integrate both sides.
- $\ln|y| = -3x^2 + C$, so $y = A e^{-3x^2}$. Use $y(0) = 1$ to fix $A$.
Worked Solution
How to Think About It: The equation is linear and homogeneous with a non-constant coefficient, but more simply it is separable: $y'/y$ is a pure function of $x$. Integrate, exponentiate, and use the initial condition.
Quick Estimate: At $x = 0$ the slope is $y'(0) = -6 \cdot 0 \cdot 1 = 0$, so the solution starts flat at height $1$ and then decays increasingly fast as $x$ grows in either direction, since $y' = -6xy$ has the opposite sign of $x$. A bell shape centered at $0$ is expected.
Formal Solution:
*Step 1 -- Separate variables.* Rewrite as $\dfrac{dy}{dx} = -6xy$. For $y \ne 0$, $$\frac{dy}{y} = -6x\,dx.$$
*Step 2 -- Integrate.* $$\ln|y| = -3x^{2} + C \quad \Longrightarrow \quad y = A\,e^{-3x^{2}}, \qquad A = \pm e^{C}.$$ (The constant solution $y \equiv 0$ corresponds to $A = 0$, so $y = Ae^{-3x^2}$ is the complete family.)
*Step 3 -- Apply the initial condition.* $y(0) = A e^{0} = A = 1$.
*Step 4 -- Verify.* With $y = e^{-3x^2}$, $y' = -6x e^{-3x^2} = -6xy$, so $y' + 6xy = 0$ and $y(0) = 1$. $\checkmark$
Answer: $y(x) = e^{-3x^{2}}$.
Intuition
A first-order equation of the form $y' = g(x)h(y)$ is solved by moving all $y$-dependence to one side and integrating; here it produces a Gaussian-shaped solution $e^{-3x^2}$. Equations like $y' = -k(x)\,y$ describe anything that decays at a state-dependent proportional rate: discount factors $P' = -r(t)P$, survival probabilities with a time-varying hazard, and the density-proportional decay $\phi'(x) = -x\phi(x)$ that characterizes the normal distribution.