Question

Consider how the methods of undetermined coefficients and variation of parameters can be combined to solve the given differential equation. Carry out your ideas to solve the differential equation.

y'' − 2y' + y = 4x^2 − 1 + x^−1e^x

y(x) =____

Answer 1

Given differential equation is $y'' -2y' +y = 4x^2 - 1 + \frac{e^x}{x}$

Auxiliary equation is $m^2 - 2m + 1 = 0 \implies m =1,1$

so $y = C_1e^x+C_2xe^x$

Let us solve for ($4x^2-1)$ by undetermined coefficients and for $\frac{e^x}{x}$ by variation of parameters

(a) undetermined coefficients:

let $y = Ax^2 +Bx+C$ be solution of the equation $y'' -2y' +y = 4x^2 - 1$

then differentiating y first and second time with respect to x, and putting in differential equation, then equation will be ,

$2A-4Ax-2B+Ax^2 +Bx+C = 4x^2-1 \implies A = 4, B=16, C=23$

then $y = C_1e^x+C_2xe^x+ 4x^2 +16x+23$

(b) method of variation of parameters

Differential equation is $y'' -2y' +y = \frac{e^x}{x}$

since $y = C_1e^x+C_2xe^x$ for auxiliary equation

Here $u = e^x, v=xe^x, R =\frac{e^x}{x}$

Then W, Wronskian of u and v will be

$W = \begin{vmatrix} e^x & xe^x \\ e^x & e^x(x+1) \end{vmatrix} = e^{2x}$

then $P.I. = -u\int\frac{vR}{W}dx + v\int\frac{uR}{W}dx$

$P.I. = e^x\int\frac{xe^xe^x}{xe^{2x}}dx + xe^x\int\frac{e^{2x}}{xe^{2x}}dx$

$P.I. = -e^x\int dx + xe^x\int \frac{dx}{x} = xe^x(ln(x) - 1)$

Hence, final solution of the equation will be

$y = C_1e^x+C_2xe^x+ 4x^2 +16x+23 + xe^x(ln(x)-1)$