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Find the general solution of y′′ – 5y′ + 4y = x2 – 2x + 1

by RoRi

Find the general solution of the second-order differential equation

    \[ y'' - 5y' + 4y = x^2 - 2x + 1. \]

If the solution is not valid everywhere, describe the interval on which it is valid.

The general solution of the homogeneous equation

    \[ y'' - 5y' + 4y = 0 \]

is given by Theorem 8.7 with a = -5 and b = 4. This gives us d = a^2 - 4b = 9; hence, k = \frac{1}{2} \sqrt{d} = \frac{3}{2}. Thus,

    \begin{align*} y &= e^{-\frac{ax}{2}} (c_1 e^{kx} + c_2 e^{-kx}) \\ &= e^{\frac{5}{2}x} (c_1 e^{\frac{3}{2}x} + c_2 e^{-\frac{3}{2} x} ) \\ &= c_1 e^{4x} + c_2 e^x. \end{align*}

To find a particular solution of y'' - 5y' + 4y = x^2 - 2x + 1 let y_1 = Ax^2 + Bx + C. Then,

    \begin{align*} y_1' &= 2Ax + B \\ y_1'' &= 2A. \end{align*}

Therefore,

    \[ y'' - 4y' + 5y = x^2 - 2x + 1 \quad \implies \quad 2A - 10Ax - 5B + 4Ax^2 + 4Bx + 4C = x^2 - 2x + 1. \]

Setting the coefficients of like powers of x to be equal and solving for the constants we get

    \[ A = \frac{1}{4}, \quad B = \frac{1}{8}, \quad C = \frac{9}{32}. \]

By Theorem 8.8 (page 330 of Apostol) the general solution of the given differential equation is then

    \[ y = c_1 e^{4x} + c_2 e^x + \frac{1}{4}x^2 + \frac{1}{8}x + \frac{9}{32}. \]

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