Exercise · ZW-FIRST-ORDER-HOMOGENEOUS-EQUATION-009 · ★★★ Advanced
Why does the substitution \(v=y/x\) work in \(y'=F(y/x)\)?
Choose one option.
Show answer
- C. Because it leaves a separable equation in \(x\) and \(v\).
Explanation
We get \(xv'=F(v)-v\), where \(x\) and \(v\) separate: \(\frac{dv}{F(v)-v}=\frac{dx}{x}\).
Review the concept: First-order homogeneous equation
A first-order homogeneous equation is a differential equation in which \(y'\) is a function \(F\) of the quotient \(y/x\).
\[\frac{dy}{dx}=F\!\left(\frac{y}{x}\right),\quad v=\frac{y}{x}\;\Rightarrow\; x\,v'=F(v)-v\]
More practice problems: first-order homogeneous equation
- ZW-FIRST-ORDER-HOMOGENEOUS-EQUATION-001 Which of these equations is first-order homogeneous?
- ZW-FIRST-ORDER-HOMOGENEOUS-EQUATION-002 With y=vx, where v depends on x, what is y'?
- ZW-FIRST-ORDER-HOMOGENEOUS-EQUATION-003 Applying y=vx to y'=(x+y)/(x), what does xv' reduce to?
- ZW-FIRST-ORDER-HOMOGENEOUS-EQUATION-004 Solve y'=(y)/(x).
- ZW-FIRST-ORDER-HOMOGENEOUS-EQUATION-005 Find the general solution of y'=(x+y)/(x) for x>0.
- ZW-FIRST-ORDER-HOMOGENEOUS-EQUATION-006 Solve y'=(y^2+xy)/(x^2) for x>0.
- ZW-FIRST-ORDER-HOMOGENEOUS-EQUATION-007 If y'=(x+y)/(x) and y(1)=2, what is y(e)?
- ZW-FIRST-ORDER-HOMOGENEOUS-EQUATION-008 For y'=(x^2+y^2)/(xy), what is F(v) with v=y/x?
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