Exercise · ZW-SEPARABLE-EQUATION-009 · ★★★ Advanced
Solve \(y'=-x/y\), \(y(0)=2\).
Choose one option.
Show answer
- B. \(y=\sqrt{4-x^2}\)
Explanation
\(y\,dy=-x\,dx\) gives \(y^2=C-x^2\); with \(y(0)=2\), \(C=4\). Since \(y(0)=2>0\) we take the positive root, and the solution exists only for \(|x|<2\).
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-007 Solve y'=y^2, y(0)=1.
- ZW-SEPARABLE-EQUATION-008 When solving y'=y(y-2) we divide by y(y-2). Which constant solutions might be lost?