Bernoulli Differential Equations — Question 1

Question 1

A Bernoulli equation has the form $y'+p(x)y=q(x)y^n$. Consider the positive initial-value problem $$y'+\frac 2x y=x^3y^2,\qquad y(1)=1,\qquad x>0.$$

Tasks

1. For positive $y$ and $n\ne 1$, derive the linear equation satisfied by $v=y^{1-n}$ using the chain rule.

2. Apply that reduction to the given IVP and solve the resulting linear equation with an integrating factor.

3. Find the maximal open interval containing $1$, and determine the minimum of the selected solution on that interval.

4. Verify the answer in the original equation and identify a solution omitted by the substitution.

Question 1 – Solution

Strategy. Derive the reduction first, then retain only the interval where its inverse gives the selected positive solution.

Step 1: Reduce to a linear equation. Multiplying by $y^{-n}$ and using $v'=(1-n)y^{-n}y'$ gives $$\boxed{v'+(1-n)pv=(1-n)q}.$$ Here $n=2$, so $v=1/y$ obeys $v'-2v/x=-x^3$. The integrating factor is $x^{-2}$, hence $$(x^{-2}v)'=-x,\qquad v=Cx^2-\frac{x^4}{2}.$$ Since $v(1)=1$, $C=3/2$.

Step 2: Invert and analyze. Thus $$\boxed{y=\frac 2{x^2(3-x^2)},\qquad I=(0,\sqrt 3)}.$$ The denominator $3x^2-x^4$ is positive on $I$, vanishes at its endpoints, and has its unique maximum $9/4$ at $x=\sqrt{3/2}$. The solution therefore has minimum $\boxed{8/9}$ there and tends to $+\infty$ at both ends of $I$.

Step 3: Verify and restore the lost solution. With $D=3x^2-x^4$, $y'=-2(6x-4x^3)/D^2$, so $$y'+2y/x=4x^3/D^2=x^3y^2,\qquad y(1)=1.$$ Also $\boxed{y\equiv 0}$ satisfies the original equation on $x>0$ but is excluded by $v=1/y$; it does not satisfy this IVP.