-2x(y-1) = dy/dx solution y=1+Ae^{-x^2}

Key takeaway: This is a separable differential equation. Separate the variables, integrate both sides, and solve for $y$ carefully, keeping track of the constant of integration.

Start with

$-2x(y-1)=\frac{dy}{dx}.$

Step 1: Separate variables

Divide by $y-1$ and multiply by $dx$:

$\frac{1}{y-1}\,dy=-2x\,dx.$

Step 2: Integrate both sides

$\int \frac{1}{y-1}\,dy=\int -2x\,dx.$

This gives

$\ln|y-1|=-x^2+C.$

Step 3: Exponentiate

$|y-1|=e^{-x^2+C}=e^C e^{-x^2}.$

Let $A=e^C$, where $A$ can absorb the sign as well, so

$y-1=Ae^{-x^2}.$

Therefore,

$\boxed{y=1+Ae^{-x^2}}.$

Step 4: Check the special case

If $A=0$, then $y=1$, which also satisfies the differential equation.

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Pitfalls the pros know 👇 A common mistake is to forget the absolute value after integrating $\frac{1}{y-1}$. Another issue is treating the constant as only positive; after exponentiating, the solution constant should be allowed to be any real number in the final form.

What if the problem changes? If the equation were $\frac{dy}{dx}=2x(y-1)$ instead, the exponential would become $e^{x^2}$ instead of $e^{-x^2}$, so the sign in the exponent matters.

Tags: separable differential equation, integration, initial value