Exercise · ZW-BERNOULLI-EQUATION-002 · ★☆☆ Basic
For \(y'+y=xy^2\), which substitution makes it linear?
Choose one option.
Show answer
- C. \(u=1/y\)
Explanation
With \(n=2\) we use \(u=y^{1-n}=y^{-1}=1/y\).
Review the concept: Bernoulli equation
A Bernoulli equation is a first-order linear equation whose independent term \(q\) is multiplied by a power \(y^n\) of the unknown, with \(n\ne0,1\).
\[y'+p(x)\,y=q(x)\,y^n,\quad u=y^{1-n}\;\Rightarrow\; u'+(1-n)p\,u=(1-n)q\]
More practice problems: bernoulli equation
- ZW-BERNOULLI-EQUATION-001 Which of these equations is a (non-linear) Bernoulli equation?
- ZW-BERNOULLI-EQUATION-003 With u=1/y, the equation y'+y=xy^2 becomes:
- ZW-BERNOULLI-EQUATION-004 In y'+p(x) y=q(x) y^n, for which n is the equation already linear?
- ZW-BERNOULLI-EQUATION-005 Solve y'+y=y^2 (with yne0).
- ZW-BERNOULLI-EQUATION-006 Solve y'+(y)/(x)=xy^2 for x>0.
- ZW-BERNOULLI-EQUATION-007 If y'+y=y^2 and y(0)=tfrac12, what is y(ln 3)?
- ZW-BERNOULLI-EQUATION-008 With u=y^-2, the equation y'+y=y^3 becomes:
- ZW-BERNOULLI-EQUATION-009 For n>0, when dividing y'+py=qy^n by y^n, which solution can be lost?