Find the general solution of y′′ - 2y′ + 3y = x3

Find the general solution of the second-order differential equation

$y'' - 2y' + 3y = x^3.$

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

The general solution of the homogeneous equation

$y'' - 2y' + 3y = 0$

is given by Theorem 8.7 with $a = -2$ and $b = 3$ . This gives us $d = a^2 - 4b = -8$ ; hence, $k = \frac{1}{2} \sqrt{-d} = \sqrt{2}$ . Thus,

$\begin{align*} y &= e^{-\frac{ax}{2}} (c_1 \cos (kx) + c_2 \sin (kx)) \\ &= e^x (c_1 \cos (\sqrt{2} x) + c_2 \sin (\sqrt{k} x) ). \end{align*}$

To find a particular solution of $y'' - 2y' + 3y = x^3$ let $y_1 = Ax^3 + Bx^2 + Cx + D$ . Then,

$\begin{align*} y_1' &= 3Ax^2 + 2Bx + C \\ y_1'' &= 6Ax + 2B. \end{align*}$

Therefore,

$y'' - 2y' + 3y = x^3 \quad \implies \quad 6Ax + 2B - 6Ax^2 - 4Bx - 2C + 3Ax^3 + 3Bx^2 + 3Cx + 3D = x^3.$

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

$A = \frac{1}{3}, \quad B = \frac{2}{3}, \quad C = \frac{2}{9}, \quad D = -\frac{8}{27}.$

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

$y = e^x (c_1 \cos (\sqrt{2} x ) + c_2 \sin (\sqrt{2} x) ) + \frac{1}{3} x^3 + \frac{2}{3}x^2 + \frac{2}{9} x - \frac{8}{27}.$

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