Exercise · ZW-SEPARABLE-EQUATION-007 · ★★☆ Intermediate
Solve \(y'=y^2\), \(y(0)=1\).
Choose one option.
Show answer
- A. \(y=\frac{1}{1-x}\)
Explanation
From \(\int y^{-2}dy=\int dx\) we get \(-1/y=x+c\); with \(y(0)=1\), \(c=-1\) and \(y=1/(1-x)\). This solution stops existing at \(x=1\).
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
- ZW-SEPARABLE-EQUATION-001 Which of these equations is separable?
- ZW-SEPARABLE-EQUATION-002 When separating variables in y'=2xy, which equality of integrals results?
- ZW-SEPARABLE-EQUATION-003 Find the general solution of y'=3y.
- ZW-SEPARABLE-EQUATION-004 Separating variables in y'=x/y, which implicit relation do the solutions satisfy?
- ZW-SEPARABLE-EQUATION-005 Solve the problem y'=y, y(0)=2.
- ZW-SEPARABLE-EQUATION-006 Solve y'=-2xy, y(0)=3.
- ZW-SEPARABLE-EQUATION-008 When solving y'=y(y-2) we divide by y(y-2). Which constant solutions might be lost?
- ZW-SEPARABLE-EQUATION-009 Solve y'=-x/y, y(0)=2.