Bernoulli Differential Equations — Question 7

Question 7 – Solution

Strategy. Affine combinations preserve the common forcing in the transformed linear equation; they do not generally preserve the original cubic equation.

Step 1: Combine the transformed solutions. Set $v_i=y_i^{-2}$. Both satisfy $v_i'-2pv_i=-2q$. Multiplying by $\alpha$ and $1-\alpha$ and adding yields $v'-2pv=-2q$. Positivity of $v_1,v_2$ implies $v>0$, so $y=v^{-1/2}$ is well defined. Differentiating this inverse recovers $y'+py=qy^3$.

Step 2: Prove the pointwise bound. At each point, $v$ lies between $v_1$ and $v_2$. The function $u\mapsto u^{-1/2}$ is decreasing on $u>0$, so reversing their order still puts $y$ between $y_1$ and $y_2$. Equality occurs when the supplied values agree; otherwise the bounds are strict.

Step 3: Apply the construction. For the given equation, the transformed equation is $v'-2xv=-2x$. The supplied transforms are $v_1=1$ and $v_2=1+e^{x^2}$; direct differentiation verifies both. Their weighted combination gives $$v=1+\frac 34e^{x^2},\qquad \boxed{y=\frac 1{\sqrt{1+\frac 34e^{x^2}}}}\quad(x\in\mathbb R).$$ It is positive and lies strictly between the two supplied solutions.

Step 4: Reject ordinary superposition. Because the original functions solve the equation, $$m'+xm=\frac{x}{2}(y_1^3+y_2^3).$$ At $x=1$, put $a=y_1(1)$ and $b=y_2(1)$. They are distinct and positive. The residual is $$m'+m-m^3=\frac{a^3+b^3}{2}-\frac{(a+b)^3}{8} =\frac 38(a+b)(a-b)^2>0.$$ Thus the arithmetic mean fails the nonlinear equation. The valid averaging occurs after the Bernoulli transformation.