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Part CCSIR NET December 2023

Existence–uniqueness, Picard, Lipschitz

Consider the initial value problem x2y′′−2x2y′+(4x−2)y=0,y(0)=0x^{2}y'' - 2x^{2}y' + (4x - 2)y = 0, y(0) = 0. Suppose y=φ(x)y = \varphi(x) is a polynomial solution satisfying φ(1)=1\varphi(1) = 1. Which of the following statements are true?

  1. A.φ(4)=16\varphi(4) = 16✓ Correct
  2. B.φ(2)=2\varphi(2) = 2
  3. C.φ(5)=25\varphi(5) = 25✓ Correct
  4. D.φ(3)=3\varphi(3) = 3

The idea, in plain words

You are not asked to solve the equation, only to recognise the solution. Try the simplest square and everything cancels — after which the two requested values are just that square evaluated twice.

Solution

Try y=x2:2x2−4x3+4x3−2x2=0y = x^{2}: 2x^{2} - 4x^{3} + 4x^{3} - 2x^{2} = 0.

So φ(x)=x2\varphi(x) = x^{2}

giving φ(4)=16\varphi(4) = 16 and φ(5)=25\varphi(5) = 25.

Why each option is right or wrong

Checked against 6 symbolic computations
  • A.Correct

    True.

    Try y=x2y = x^{2}: SymPy substitutes it into x2y′′−2x2y′+(4x−2)yx^{2}y'' - 2x^{2}y' + (4x - 2)y and gets exactly 0

    so x2x^{2} is a solution.

    It also satisfies y(0) = 0 and φ(1)=1\varphi(1) = 1

    so φ(x)=x2\varphi(x) = x^{2} and φ(4)=16(\varphi(4) = 16 (verified).

    The trigger: with polynomial coefficients of mixed degree, test low-degree monomials before attempting a series — here the degrees balance only for x2x^{2}.

  • False: φ(2)=22=4\varphi(2) = 2^{2} = 4, not 2.

    This option (and (D)) is what you get by assuming φ(x)=x\varphi(x) = x

    which fits φ(1)=1\varphi(1) = 1 but solves nothing — SymPy substitutes y = x and obtains 2x(x − 1)

    which is not identically zero.

    Matching the single condition φ(1)=1\varphi(1) = 1 is not the same as solving the equation.

  • C.Correct

    True.

    φ(5)=52=25(\varphi(5) = 5^{2} = 25 (verified).

    Note that (A) and (C) are consistent with each other and with φ(x)=x2\varphi(x) = x^{2}

    while (B) and (D) are consistent with the wrong guess φ(x)=x\varphi(x) = x — the four options split cleanly into two families, and identifying φ\varphi decides both at once.

  • False: φ(3)=32=9\varphi(3) = 3^{2} = 9, not 3.

    Same underlying error as (B) — the linear guess φ(x)=x\varphi(x) = x.

2 of these 2 wrong options correspond to a named reasoning error — the same errors recur across subjects.

Related counterexample: Every IVP has a unique solution

More on this topic

The chapter behind this: Existence, uniqueness and how far the solution extends — free to read

From Ordinary Differential Equations › Existence–uniqueness, Picard, Lipschitz

Last revised . Found a mistake? Tell us — corrections are the fastest thing we act on.

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