Question

Solve the given differential equation by undetermined coefficients.

\(y''' - 2y'' - 4y' + 8y = 6x{e^{2x}}\)

Short answer

The general solution of the given differential equation is

\(y = {c_1}{e^{2x}} + {c_2}x{e^{2x}} + {c_3}{e^{ - 2x}} + \frac{1}{4}{x^3}{e^{2x}} - \frac{3}{{16}}{x^2}{e^{2x}}\)

Step-by-step solution

Note the given data

Given the third order differential equation \(y''' - 2y'' - 4y' + 8y = 6x{e^{2x}}\)

Obtain the general solution of the homogeneous differential equation  

We know a homogeneous differential equation is an equation containing

differentiation and a function with a set of variables.

First, we need to solve the associated homogeneous equation\(y''' - 2y'' - 4y' + 8y = 0\)

Corresponding auxiliary equation is\({m^3} - 2{m^2} - 4m + 8 = 0\)

\(({m^2} - 4)(m - 2) = 0\)

\((m - 2)(m - 2)(m + 2) = 0\)

Solving,

m₁ =2, m₂ =2 and m₃ = -2, which are real and repeated.

So the solution of corresponding homogeneous equation is

\({y_h} = {c_1}{e^{2x}} + {c_2}x{e^{2x}} + {c_3}{e^{ - 2x}}\)

Finding the particular solution

Given non homogeneous differential equation is\(y''' - 2y'' - 4y' + 8y = 6x{e^{2x}}\)

Assume that \({y_p} = A{x^3}{e^{2x}} + B{x^2}{e^{2x}}\) …..(1)is a solution of the non-homogeneous differential equation

Differentiate with respect to x

\({y_p}' = 3A{x^2}{e^{2x}} + 2A{x^3}{e^{2x}} + 2B{x^2}{e^{2x}} + Bx{e^{2x}}\)

Or \({y_p}' = 2A{x^3}{e^{2x}} + (3A + 2B){x^2}{e^{2x}} + 2Bx{e^{2x}}\)……(2)

\({y_p}'' = 4A{x^3}{e^{2x}} + (12A + 4B){x^2}{e^{2x}} + (6A + 8B)x{e^{2x}} + 2B{e^{2x}}\)…..(3)

\({y_p}''' = 8A{x^3}{e^{2x}} + (36A + 8B){x^2}{e^{2x}} + (36A + 24B)x{e^{2x}} + (6A + 12B){e^{2x}}\)……(4)

Substituting (1) ,(2), (3) and (4) in given differential equation

\(\begin{aligned}{l}8A{x^3}{e^{2x}} + (36A + 8B){x^2}{e^{2x}} + (36A + 24B)x{e^{2x}} + (6A + 12B){e^{2x}} - 2[4A{x^3}{e^{2x}} + (12A + 4B){x^2}{e^{2x}} + \\(6A + 8B)x{e^{2x}} + 2B{e^{2x}}] - 4[2A{x^3}{e^{2x}} + (3A + 2B){x^2}{e^{2x}} + 2Bx{e^{2x}}] + 8[A{x^3}{e^{2x}} + B{x^2}{e^{2x}}] = 6x{e^{2x}}\end{aligned}\)

Or, \(24Ax{e^{2x}} + (6A + 8B){e^{2x}} = 6x{e^{2x}}\)

comparing the coefficients,

\(24A = 6\)

\(6A + 8B = 0\)

Solving, \(A = \frac{1}{4},B = \frac{{ - 3}}{{16}}\)

Hence the particular solution is

\({y_p} = \frac{1}{4}{x^3}{e^{2x}} - \frac{3}{{16}}{x^2}{e^{2x}}\)

Finding the solution

Hence the solution of given differential equation is

y=y_h+y_p

\(y = {c_1}{e^{2x}} + {c_2}x{e^{2x}} + {c_3}{e^{ - 2x}} + \frac{1}{4}{x^3}{e^{2x}} - \frac{3}{{16}}{x^2}{e^{2x}}\)