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Find a polynomial satisfying given conditions

  1. Find a polynomial satisfying

        \[ P'(x) - 3P(x) = 4 - 5x + 3x^2. \]

    Prove that there is only one such polynomial.

  2. Given a polynomial Q(x), prove there is exactly one polynomial P(x) such that

        \[ P'(x) - 3P(x) = Q(x). \]


  1. Proof. (Finding the polynomial will prove that it is unique since we will not have any choices to make while deriving the polynomial P(x).) First, we write

        \begin{align*}   P(x) = \sum_{k=0}^n c_k x^k && \implies && P'(x) &= \sum_{k=1}^n c_k (k) x^{k-1} \\  &&&&& = \sum_{k=0}^{n-1} (k+1)c_{k+1} x^k. \end{align*}

    Thus, we have

        \[ P'(x) - 3P(x) = \sum_{k=0}^{n-1} (k+1)c_{k+1} x^k - \sum_{k=0}^n 3c_k x^k = \sum_{k=0}^{n-1} ((k+1)c_{k+1} - 3c_k)x^k - 3c_n x^n. \]

    Setting this equal to 4 - 5x + 3x^2 we have

        \[ \sum_{k=0}^{n-1} ((k+1)c_{k+1} - 3c_k)x^k - 3c_n x^n = 4 - 5x + 3x^2. \]

    But, this implies n = 2 and c_2 = -1 since -3c_n x^n is the only x^n term on the left (so if n > 2, then we couldn’t have x^2 the largest power of x on the right). Therefore, P(x) is a degree polynomial and with c_2 = -1, so we have

        \[ P(x) = c_0 + c_1 x - x^2 \quad \implies \quad P'(x) = c_1 - 2x. \]

    Hence we have

        \begin{align*}  P'(x) - 3P(x) = 4 - 5x + 3x^2 && \implies && c_1 - 2x - 3c_0 - 3c_1 x + 3x^2 &= 4 - 5x + 3x^2 \\  && \implies && (c_1 - 3c_0) - (2+3c_1)x &= 4 - 5x. \end{align*}

    Thus we have the equations

        \[ c_1 - 3c_0 = 4 \quad \text{and} \quad 2+3c_1 = 5.\]

    These uniquely determine c_0 and c_1,

        \[ c_1 = 1,\quad c_0 = -1. \]

    Hence, there is a unique P(x) satisfying this equation,

        \[ P(x) = -1 + x - x^2.  \qquad \blacksquare \]

  2. Proof. Let Q(x) be a given polynomial and suppose there exist two polynomials P_1 (x) and P_2(x) such that

        \[ P'(x) - 3P(x) = Q(x) \quad \text{and} \quad R'(x) - 3R(x) = Q(x). \]

    This implies

        \[ (P'(x) - R'(x)) - 3(P(x) - R(x)) = 0 \quad \implies \quad (P(x) - R(x))' - 3(P(x) - R(x)) = 0. \]

    Now, if P(x) - R(x) \neq 0 then it is of degree n for some n \geq 1. We know its derivative has degree n-1 (Apostol, Page 166). But then, this would imply

        \[ (P(x) - R(x))' - 3(P(x) - R(x)) \]

    has degree n (since the coefficient of x^n in (P(x) - R(x))' is zero since it is degree n-1, and the coefficient of 3(P(x) - R(x)) is nonzero since it has degree n). But we know this difference is 0, which means it cannot have degree n for any n \geq 1. Thus, we must have P(x) - R(x) = 0 or P(x) = R(x). \qquad \blacksquare

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