Final Thoughts — Question 8

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Question 8

An implicit description of a function is proposed for x≥0x\ge 0: y3+xy=1,y>0.y^3+xy=1,\qquad y>0. The associated initial value problem is y′=−yx+3y2,y(0)=1.y'=-\frac{y}{x+3y^2},\qquad y(0)=1. You may use implicit differentiation where the coefficient of y′y' is nonzero.

Tasks

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

  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≥0x\ge 0 and tends to zero as x→∞x\to\infty.

  4. Prove 2/3<y(1)<3/42/3<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.

Original worksheet page 1: question and worked solution for 1-3-008
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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≥0x\ge 0, let H(y)=y3+xyH(y)=y^3+xy. It is continuous, starts at H(0)=0H(0)=0, and tends to infinity as y→∞y\to\infty. For y>0y>0, H′(y)=3y2+x>0H'(y)=3y^2+x>0, so the Intermediate Value Theorem and strict increase give exactly one positive root of H(y)=1H(y)=1.

Step 2: Verification. Differentiation yields 3y2y′+y+xy′=0,y′=−yx+3y2.3y^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)(0,1). At zero the relation gives y3=1y^3=1, so y(0)=1y(0)=1 and y′(0)=−1/3\boxed{y'(0)=-1/3}.

See the diagram in the original worksheet below.

Step 3: Decrease and limit. Positivity of yy and the denominator implies y′<0y'<0. Also y3+xy=1y^3+xy=1 gives 0<y<1/x0<y<1/x for x>0x>0, so the squeeze principle gives y(x)→0 as x→∞\boxed{y(x)\to 0\text{ as }x\to\infty}.

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

Original worksheet page 2: question and worked solution for 1-3-008

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