Final Thoughts — Question 8

Question 8

An implicit description of a function is proposed for $x\ge 0$: $$y^3+xy=1,\qquad y>0.$$ The associated initial value problem is $$y'=-\frac{y}{x+3y^2},\qquad y(0)=1.$$ You may use implicit differentiation where the coefficient of $y'$ is nonzero.

Tasks

1. Show that for every fixed $x\ge 0$ the implicit relation has exactly one positive root $y$.

2. Differentiate the relation, verify the ODE and initial value, and calculate the initial slope. Explain why the denominator stays nonzero on this branch.

3. Prove that the branch decreases for $x\ge 0$ and tends to zero as $x\to\infty$.

4. Prove $2/3<y(1)<3/4$ and sketch the positive branch in your solution. Explain why an explicit formula obtained by solving the cubic is unnecessary for these conclusions.

Question 8 – Solution

Strategy. Establish a unique branch before differentiating. Monotonicity and inequalities can describe an implicit solution without an explicit cubic formula.

Step 1: A well-defined positive branch. For fixed $x\ge 0$, let $H(y)=y^3+xy$. It is continuous, starts at $H(0)=0$, and tends to infinity as $y\to\infty$. For $y>0$, $H'(y)=3y^2+x>0$, so the Intermediate Value Theorem and strict increase give exactly one positive root of $H(y)=1$.

Step 2: Verification. Differentiation yields $$3y^2y'+y+xy'=0,\qquad \boxed{y'=-\frac{y}{x+3y^2}.}$$ The denominator is positive on the branch, allowing local differentiable continuation, including near $(0,1)$. At zero the relation gives $y^3=1$, so $y(0)=1$ and $\boxed{y'(0)=-1/3}$.

Step 3: Decrease and limit. Positivity of $y$ and the denominator implies $y'<0$. Also $y^3+xy=1$ gives $0<y<1/x$ for $x>0$, so the squeeze principle gives $\boxed{y(x)\to 0\text{ as }x\to\infty}$.

Step 4: A rigorous bracket. At $x=1$, $$(2/3)^3+2/3=26/27<1,\qquad (3/4)^3+3/4=75/64>1.$$ Strict increase in $y$ proves $\boxed{2/3<y(1)<3/4}$. Root existence, differentiation and inequalities establish these results directly; an explicit cubic formula adds no needed information.