Exercise · ZW-SEPARABLE-EQUATION-006 · ★★☆ Intermediate

Solve \(y'=-2xy\), \(y(0)=3\).

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  • C. \(y=3e^{-x^2}\)

Explanation

\(\int dy/y=\int -2x\,dx\) gives \(\ln|y|=-x^2+c\), so \(y=Ce^{-x^2}\) and \(y(0)=C=3\).

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Review the concept: Separable equation

A separable equation is a first-order differential equation whose derivative is the product of a function \(g\) of \(x\) alone and a function \(h\) of \(y\) alone.

\[\frac{dy}{dx}=g(x)\,h(y)\;\Longrightarrow\;\int\frac{dy}{h(y)}=\int g(x)\,dx\]

More practice problems: separable equation

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