Bernoulli Equations: Power Substitution and Linearization

All flashcards

What defines a Bernoulli differential equation?

It is a first-order equation of the form $y'+P(x)y=Q(x)y^n$, where $P$ and $Q$ are known functions and $n$ is a real number.

What happens in the special cases $n=0$ and $n=1$?

For $n=0$, the equation is linear. For $n=1$, it can be rearranged into a separable equation (and is also linear).

What is the Bernoulli substitution, and what does it achieve?

For $n\neq0,1$, set $u=y^{1-n}$. This converts the nonlinear equation into a first-order linear differential equation in $u$.

What is the transformed linear equation?

The substitution produces $u'+(1-n)P(x)u=(1-n)Q(x)$, equivalently $u'-(n-1)P(x)u=-(n-1)Q(x)$.

Why must possible zero solutions be checked separately?

The substitution may require dividing by $y$ or a power of $y$, which is not valid when $y=0$. A solution such as $y\equiv0$ can therefore be lost and should be tested in the original equation.

Quiz questions

1. What is the standard form of a Bernoulli differential equation?Answer: $y'+P(x)y=Q(x)y^n$
2. What substitution transforms a Bernoulli equation with $n\neq0,1$ into a linear equation?Answer: $u=y^{1-n}$
3. What type of equation results after applying the Bernoulli substitution?Answer: A first-order linear differential equation in $u$
4. For $y'-\frac{2}{x}y=-x^2y^2$, what transformed equation follows from $u=1/y$?Answer: $u'+\frac{2}{x}u=x^2$
5. Why should $y\equiv0$ be checked separately when solving $y'-\frac{2}{x}y=-x^2y^2$ using $u=1/y$?Answer: The substitution requires division by $y$, so it excludes the zero solution.